ass 


' 


^    • 


* 


WORKS  OF  PROF.  I.  P.  CHURCH 

PUBLISHED    BY 

JOHN   WILEY   &  SONS. 


Mechanics  of  Engineering. 

Comprising  Statics  and  Dynamics  of  Solids,  the 
Mechanics  of  the  Materials  of  Construction  or 
Strength  and  Elasticity  of  Beams,  Columns,  Shafts, 
Arches,  etc.,  and  the  Principles  of  Hydraulics  and 
Pneumatics  with  Applications.  For  the  Use  of 
Technical  Schools.  8vo,  xxvi  +  §34  pages,  647 
figures,  cloth,  $6.00. 

Notes  and  Examples  in  Mechanics. 

With  an  Appendix  on  the  Graphical  Statics  of 
Mechanism.  8vo,  167  pages,  178  figures,  cloth,  $2.00. 

Diagrams  of  Mean  Velocity  of  Water  in  Open 
Channels. 

Based  on  the  Formula  of  Ganguillet  and  Kutter. 
9  X  12  inches,  paper,  $1.50. 

Hydraulic  flotors  ;  with  related  subjects  ;  includ- 
ing Centrifugal  Pumps,  Pipes,  and  Open 
Channels.  8vo,  279  pages,  125  figures,  cloth, 
$2.00. 


MECHANICS  OF  ENGINEERING. 


COMPRISING 

STATICS  AND   DYNAMICS   OF  SOLIDS  ;   THE   MECHANICS  OF 

MATERIALS  OF  CONSTRUCTION,  OR  STRENGTH  AND  ELASTICITY 

OF  BEAMS,  COLUMNS,  SHAFTS,  ARCHES,  ETC.  ;  AND  THE 

PRINCIPLES  OF  HYDRAULICS  AND  PNEUMATICS, 

WITH  APPLICATIONS. 


FOR    USE  IN  TECHNICAL  SCHOOLS. 


BY 

IRVING  P.  CHURCH,  C.E., 

PROFESSOR  OF  APPLIED  MECHANICS  AND  HYDRAULICS,  COLLKGE  OF  CIVIL  ENGINEERING, 
CORNELL  UNIVERSITY. 


TWELFTH  THOUSAND. 


V\B*A 
orrHt 

(  UNfYERSfTY  } 

or 


NEW  YORK: 

JOHN   WILEY   &  SONS 
LONDON  .   CHAPMAN  &  HALL,  LIMITW* 

1905. 


Copyright,  1890, 

BT 

IRTING  P.  CHURCH. 


BOBIRT  DRUMMOND,   ELHXJTHOIYPER  AND  PRINTER.  NKW  YOABL. 


PREFACE. 


For  the  engineering  student,  pursuing  the  study  of  Applied  Mechan- 
ics as  part  of  his  professional  training,  and  not  as  additional  mathe- 
mathical  culture,  not  only  is  a  thoroughly  systematic,  clear,  and 
consistent  treatment  of  the  subject  quite  essential,  but  one  which  pre- 
sents the  quantities  and  conceptions  involved  in  as  practical  and  con- 
crete  a  form  as  possible,  with  all  the  aids  of  the  printer's  and  engraver's 
arts ;  and  especially  one  which,  besides  showing  the  derivation  of 
formulae  from  principles,  illustrates,  inculcates,  and  lays  stress  on 
correct  numerical  substitution  and  the  consistent  and  proper  use  of 
units  of  measurement;  for  without  this  no  reliable  results  can  be 
reached,  and  the  principal  ot ject  of  these  formulae  is  frustrated. 

With  these  requirements  in  view,  and  aided  by  the  experience  of  ten 
years  in  teaching  the  Mechanics  of  Engineering  at  this  institution,  the 
writer  has  been  led  to  prepare  the  present  work,  in  which  attention  is 
called  to  the  following  features  : 

The  diagrams  are  very  numerous  (about  one  to  every  page ;  an  appeal 
to  the  eye  is  often  worth  a  page  of  verbal  description). 

The  symbols  for  distances,  angles,  forces,  etc.,  used  in  the  algebraic 
work  are,  as  far  as  possible,  inserted  directly  in  the  diagrams,  render- 
ing the  latter  full  and  explicit,  and  thus  saving  time  and  mental  effort 
to  the  student.  In  problems  in  Dynamics  three  kinds  of  arrows  are 
used  to  distinguish  forces,  velocities,  and  accelerations,  respectively, 
and  thus  to  prevent  confusion  of  ideas. 

Illustrations  and  examples  of  a  practical  nature,  both  algebraic  and 
numerical,  are  of  frequent  occurrence. 

Formulae  are  divided  into  two  classes  ;  those  (homogeneous)  admit* 
ting  of  the  use  of  any  system  of  units  whatever  for  measurements  of 
force,  space,  mass,  and  time,  in  numerical  substitution;  and  those 
which  are  true  for  specified  units  only.  Attention  is  repeatedly  di- 
rected to  the  matter  of  correct  numerical  substitution,  especially  in 
Dynamics,  where  time  and  mass,  as  well  as  force  and  space,  are  among 
the  quantities  considered.  The  importance,  in  this  connection,  of 
frequent  mention  of  the  quality  of  the  various  kinds  of  quantity  em- 
ployed, is  also  recognized,  and  a  corresponding  phraseology  adopted. 

The  definition  of  force  (§3)  is  made  to  include  and  illustrate  Newton's 
law  of  action  and  reaction,  the  misconception  of  which  leads  to  such 
lengthy  discussions  in  technical  journals  every  few  years. 

In  the  matter  of  "Centrifugal  force,"  the  artificial  method,  so  com- 
monly  adopted,  of  regarding  a  particle  moving  uniformly  in  a  circle 
as  in  equilibrium,  i.  e.,  acted  on  by  a  balanced  system  of  forces,  one  of 
which  is  the  "  Centrifugal  force,"  has  been  avoided,  as  being  at  vari- 
ance with  a  system  of  Mechanics  founded  on  Newton's  laws,  according 
to  the  first  of  which  a  particle  moving  in  any  other  than  a  straight  line 
cannot  be  in.  equilibrium.  In  such  a  system  of  Mechanics  nothing  can 


iv  PREFA  CE. 

be  recognized  as  a  force  which  is  not  a  definite  pull,  push,  pressure, 
rub,  attraction  or  repulsion,  of  one  body  upon,  or  against,  another. 

It  is  true  that  the  artificial  nature  of  the  method  referred  to  is  in 
some  text-books  fully  explained  in  the  context,  (in  Goodeve's  Steam 
Engine,  for  instance,  in  treating  the  governor  ball,)  but  is  too  often  not 
mentioned  at  all,  so  that  the  student  risks  being  led  into  error  in 
attempting  kindred  problems  by  what  would  then  seem  to  him  correct; 
methods. 

The  general  theorem  of  Work  and  Energy  in  machines  is  developed 
gradually  by  definite  and  limited  steps,  in  preference  to  giving  a  single 
demonstration  which,  from  its  generality,  might  be  too  vague  and  ab- 
struse to  be  readily  grasped  by  the  student. 

In  the  use  of  the  Calculus,  (in  the  elements  of  which  the  student  is 
supposed  to  have  had  the  training  usually  given  in  technical  schools  by 
the  end  of  the  second  year)  the  integral  sign  is  always  used  to  indicate 
summation  (except  on  p.  357)  while  the  name  of  anti-derivative  of  a 
given  function  (of  one  variable)  has  been  adopted  for  that  function  whose 
derivative,  or  differential  co-efficient,  is  the  given  function  (see  §253. ) 

The  signs  -\  and  ||  are  used  for  perpendicular  and  parallel,  respect- 
ively. 

In  Torsion  and  Flexure  of  Beams,  the  well  worn  and  simple  theories 
of  Navier  have  been  thought  sufficient  for  establishing  practical  for- 
mulae for  safe  loads  and  deflections  of  beams  and  shafts ;  and  promi 
aence  has  been  given  to  the  methods  of  designing  the  cross-sections 
and  riveting  of  built-beams  and  plate-girders,  forming  the  basis  of  the 
tables  and  rules  usually  given  in  the  pocket-books  of  our  iron  and  steel 
inanufacturers. 

The  analytical  treatment  of  the  continuous  girder  is  not  presented  in 
the  general  case,  preference  being  given  to  the  graphic  method  by 
Mohr,  as  greatly  superior  in  simplicity,  directness,  and  interest.  For 
similar  reasons  the  graphics  of  the  arch  of  masonry  is  to  be  preferred  to 
the  analytical  chapter  on  Linear  Arches,  whose  insertion  is  chiefly  a 
concession  to  the  mathematical  student,  as  are  also  §§  119, 1985  234,  235, 
264,  265,  266,  284,  287,  291,  and  297. 

The  graphics  of  curved  beams  or  arch  ribs  is  made  to  precede  that  of 
the  straight  girder,  since  the  treatment  of  the  latter  as  a  particular  case 
of  the  former  is  then  a  comparatively  simple  matter.  Hence  Prof. 
Eddy's  methods  *  (inserted  by  his  kind  permission)  for  the  arch  rib  of 
hinged  ends,  and  also  that  of  fixed  ends,  are  presented  as  special  geo- 
metrical devices,  instead  of  being  based  on  Prof.  Eddy's  general  theorem 
(involving  a  straight  girder  of  the  same  section  and  mode  of  support). 

Acknowledgment  is  also  due  Prof.  Burr  and  Prof.  Robinson,  for 
their  cordial  consent  to  the  use  of  certain  items  and  passages  from 
their  works;  (see  §§206,  212,  220,  and  297.) 

*  See  \>p.  14  and  25  of  "  Researches  in  Graphical  Statics."  by  Prof.  H.  T.  Eddy,. 
C.E.,  Ph.  D.;  published  by  D.  Van  Nostrand,  New  York,  1878  :  reprinted  from  Van  Nos- 
trand's  Magazine  for  1877. 


PREFACE.  T 

Advantage  has  been  taken  of  the  results  of  the  most  recent  experimental 
investigations  in  Hydraulics  in  assigning  values  of  the  numerous  coeffi- 
cients necessary  in  this  science.  The  researches  of  Messrs.  Fteley  and 
Stearns  in  1880  and  of  M.  Bazin  in  1887  on  the  flow  of  water  over  weirs, 
and  of  Mr.  Clemens  Herschel  in  testing  his  invention  the  "  Venturi  Water- 
meter,"  are  instances  in  point;  as  also  some  late  experiments  on  the 
transmission  of  compressed  air  and  of  natural  gas,  and  Mr.  Freeman's 
extensive  investigations  in  the  Hydraulics  of  Fire-streams  and  resistance 
of  Fire-hose,  p.  832.  (See  Transac.  Am.  Soc.  Civ.  Eng.  for  Nov.  1889.) 

In  dealing  with  fluid  tension  care  has  been  taken  to  use  the  absolute 
pressure  and  not  simply  the  excess  over  atmospheric,  thus  avoiding  the 
occurrence  oftthe  term  "  negative  pressure;"  this  precaution  being  specially 
necessary  in  the  treatment  of  gaseous  fluids. 

Though  space  has  forbidden  dealing  at  any  great  length  with  the  action  of 
fluid  motors,  sufficient  matter  is  given  in  treating  of  the  mode  of  working 
of  steam,  gas,  and  hot-air  engines,  air-compressors,  and  pumping-engines, 
together  with  numerical  examples,  to  be  of  considerable  advantage,  it  is 
thought,  to  students  not  making  a  specialty  of  mechanical  engineering. 

Special  acknowledgment  is  due  to  Col.  J.  T.  Fanning,  the  well-known 
author  of  "  Hydraulic  and  Water-supply  Engineering,"  for  his  consent  to 
the  use  of  an  abridgment  of  the  table  of  coefficients  for  friction  of  water 
in  pipes,  given  in  that  work;  and  to  Prof.  C.  L.  Crandall,  of  this  univer- 
sity, for  permission  to  incorporate  the  chapter  on  Retaining- Walls. 

References  to  original  research  In  the  Hydraulic  Laboratory  of  the 
Civil  Engineering  Department  at  this  institution  will  be  found  on  pp.  694 
and  729. 

CORNELL  UNIVERSITY,  ITHACA,  N.  Y., 
January,  1890. 


NOTE. — Additional  matter  involving  many  examples  and  forming  an 
appendix  to  the  present  work,  but  too  bulky  to  be  incorporated  with  it, 
was  issued  in  a  separate  volume  in  1892  and  entitled  "  Notes  and  Examples 
in  Mechanics."  A  second  edition,  revised  and  enlarged,  was  published  in 
1897. 

ITHACA,  N  Y.,  May,  1900. 


TABLE  OF  CONTENTS, 

[MECHANICS  OF  SOLIDS.] 


PRELIMINARY  CHAPTER. 

§§  1-15*.         Definitions.     Kinds  of  Quantity.    Homogeneous  Equa- 
tions.    Parallelogram  of  Forces J 

PAKT  I.    STATICS. 

CHAPTER  I.    STATICS  OF  A  MATERIAL  POINT. 
§§  16-19.          Composition  and  Equilibrium  of  Concurrent  Forces. ...      8 

CHAPTER  II.    PARALLEL  FORCES  AND  THE  CENTRE  OF  GRAVITY. 

§§  20-22.          Parallel  Forces 13 

§§  23-276.        Centre  of  Gravity.     Problems.     Centrobaric  Method. . .     18 

CHAPTER  III.    STATICS  OF  A  RIGID  BODY. 

§§28-34.          Couples '.. 27 

§§  35-39.          Composition  and  Equilibrium  of  Non-concurrent  Forces.    31 

CHAPTER  IV.     STATICS  OF  FLEXIBLE  CORDS. 

§§  40-48.          Postulates.    Suspended  Weights.    Parabolic  Cord.   Cat- 

enary 42 

PAET  II.    DYNAMICS. 
CHAPTER  I.      RECTILINEAR  MOTION  OF  A  MATERIAL  POINT. 

§§  49-55.  Uniform  Motion.  Falling  Bodies.  Newton's  Laws. 

Mass ,,....  49 

§§  56-60c  Uniformly  Accelerated  Motion.  Graphic  Representa- 
tions. Variably  Accelerated  Motions.  Impact. . .  54 

CHAPTER  II.    VIRTUAL  VELOCITIES. 

§§  61-6&,  Definitions  and  Propositions.  Connecting-rod.  Prob- 
lems   67 

vii 


CONTENTS. 


CHAPTER  III.    CURVILINEAR  MOTION  OP  A  MATERIAL  POINT. 

PAGB 

§§  70-74.  Composition  of  Motions,  of  Velocities,  etc.  General 

Equations  ........  .  ............................  73 

§§  75-84  Normal  Acceleration.  Centripetal  and  Centrifugal 
Forces.  Simple  Pendulums.  Projectiles.  Rela- 
tive Motion  .................................  ....  77 

CHAPTER  IV.    MOMENT  OF  INERTIA. 

§§85-94.          Plane  Figures.     Rigid  Bodies.     Reduction  Formulae. 
The  Rectangle,  Triangle,  etc.      Compound  Plane 
Figures.     Polar  Moment  of  Inertia  ...............     91 

§S    95-104       Rod.     Thin  Plates.     Geometric  Solids  ................     98 

§§  105-107.      Numerical  Substitution.     Ellipsoid  of  Inertia  .........  1C- 

CHAPTER  V.    DYNAMICS  OP  A  RIGID  BODY. 

§§  108-115  Translation.  Rotation  about  a  Fixed  Axis.  Centre  of 

Percussion  ...........................  »  ..........  105 

§§  116-121.  Torsion  Balance.  Compound  Pendulum.  The  Fly- 

wheel ..........................................  116 

§§122-123.       Uniform  Rotation.     "  Centrifugal  Action."    Free  Axes.  125 

§§  124-126,   '  Rolling  Motions.    Parallel  Rod  of  Locomotive  .........  130 

CHAPTER  VI.    WORK,  ENERGY,  AND  POWER. 

§§  127-134.       Work.     Power.     Horse-power.     Kinetic  Energy.  .....  133 

§§135-138.       Steam-hammer.     Pile-driving.     Inelastic  Impact  .....     138 

§§  139-141.  Rotary  Motion.  Equivalent  Systems  of  Forces.  Any 

Motion  of  a  Rigid  Body  ....................  .  ____  143 

§§  142-146.  Work  and  Kinetic  Energy  in  a  Moving  Machine  of 

Rigid  Parts  .................................  147 

§§  147-155.  Power  of  Motors.  Potential  Energy.  Heat,  etc.  Dy- 

namometers.    Atwood's  Machine.      Boat-rowing. 

Examples  .......................................  153 

CHAPTER  VII.     FRICTION. 

§§  156-164      Sliding  Friction.     Its  Laws.     Bent  Lever  .............  164 

§§  165-171.  Axle  -  friction.  Friction  Wheels.  Pivots.  Belting. 

Transmission  of  Power  by  Belting  ...........  ______  175 

§§  172-177.  Rolling  Friction.  Brakes.  Engine  friction.  Anoma- 

lie»  in  Friction.     Rigidity  of  Cordage  .     Examples.  186 


CONTENTS. 


IX 


PAET  III.     STRENGTH  OF  MATERIALS. 

(OR  MECHANICS  OF  MATEKIALS.) 

CHAPTER  I.    ELEMENTARY  STRESSES  AND  STRAINS. 

§§  178-182.  Stress  and  Strain  ;  of  Two  Kinds.  Oblique  Section 

of  Rod  in  Tension " 195 

§§  182a-190.  Hooke's  Law.  Elasticity.  Safe  Limit.  Elastic 
Limit.  Rupture.  Modulus  of  Elasticity. 
Isotropes.  Resilience.  Internal  Stress.  Tem- 
perature Stresses 201 

TENSION. 

§§  191-195.  Hooke's  Law  by  Experiment.  Strain  Diagrams. 

Lateral  Contraction.  Modulus  of  Tenacity 207 

§§  196-199.  Resilience  of  Stretched  Prism.  Load  Applied  Sud- 
denly. Prism  Under  Its  Own  Weight.  Solid 
of  Uniform  Strength.  Temperature  Stresses ..  213 

COMPRESSION  OF  SHORT  BLOCKS. 

§§  200-202.        Short  and  Long  Columns.     Remarks  on  Crushing. . .  218 

EXAMPLES  IN  TENSION  AND  COMPRESSION. 

§§  203-206.        Tables.     Examples.      Factor  of  Safety.      Practical 

Notes. .     220 

SHEARING. 

§§  207-213.        Rivets.      Shearing  Distortion.     Table.      Punching. 

Examples 225 

CHAPTER  II.     TORSION. 

§§  214-220.        Angle  and  Moment  of  Torsion.    Torsional  Strength, 

Stiffness,  and  Resilience.     Non-Circular  Shafts 233 

§§  221-223.  Transmission  of  Power,  Autographic  Testing  Ma- 
chine. Examples 238 


CHAPTER  III.     FLEXURE  OF  HOMOGENEOUS  PRISMS  UNDER 
PERPENDICULAR  FORCES  IN  ONE  PLANE. 


224-232a. 


The  Common  Theory.  Elastic  Forces.  Neutral 
Axis.  The  "Shear"  and  " Moment."  Flex- 
ural  Strength  and  Stiffness.  Radius  of  Curva- 
ture. Resilience 244 

ELASTIC  CURVES. 

Single  Central  Load  ;  at  Rest,  and  Applied  Suddenly. 

Eccentric  Load.    Uniform  Load.    Cantilever. .  252 


X  CONTENTS. 

SAFE  LOADS. 

§§  239-346.  Maximum  Moment.  Shear  =  x-Derivative  of  the 
Moment.  Simple  Beams  With  Various  Loads. 
Comparative  Strength  and  Stiffness  of  Rectan- 
gular Beams 262 

§§  247-252.  Moments  of  Inertia.  "  Shape  Iron."  I-Beams,  Etc. 

Cantilevers.  Tables.  Numerical  Examples..  273 

SHEARING  STRESSES  IN   FLEXURE. 

§§  253-257.  Shearing  Stress  Parallel  to  Neutral  Surface  ;  and  in 
Cross  Section.  Web  of  I-Beam.  Riveting  of 
Built  Beams 284 

SPECIAL  PROBLEMS  IN  FLEXURE. 

§§  258-265.  Designing  Sections  of  Built  Beams.  Moving  Loads. 
Special  Cases  of  Quiescent  Loads.  Hydrostatic 
Load 295 

§§  266-270.  Derivatives  of  Ordinate  of  Elastic  Curve.  Weight 

Falling  on  Beam.  Crank  Shaft.  Other  Shafts.  310 


271-276. 


CHAPTER  IV.     FLEXURE;    CONTINUED. 

CONTINUOUS  GIRDERS. 

Analytical  Treatment  of  Symmetrical  Cases  of  Beams 

on  Three  Supports  ;  also,  Built  in  ;  (see  §  391.). .  320 


THE  DANGEROUS  SECTION  IN   NON-PRISMATIC  BEAMS. 

§§  277-279.        Double  Truncated  Wedge,  Pyramid,  and  Cone 332 

NON-PRISMATIC  BEAMS   OF  UNIFORM  STRENGTH. 

§§  280-289.        Parabolic    and    Wedge-Shaped     Beams.       I-Beam. 

Elliptical  Beam  and  Cantilevers 335 

DEFLECTION  OF  BEAMS  OF  UNIFORM  STRENGTH. 

§§  290-293.  Parabolic  and  Wedge-Shaped  Beams.  Special  Prob- 
lems  342 

CHAPTER    V.     FLEXURE    OF  PRISMATIC    BEAMS   UNDER 
OBLIQUE  FORCES. 

§§  294-301.  Gravity  and  Neutral  Axes.  Shear,  Thrust,  and 
Stress-Couple.  Strength  and  Stiffness  of  Ob- 
lique Cantilever.  Hooks.  Crane 347 


CONTENTS.  xj 

CHAPTER  VI.      FLEXURE  OF  LONG  COLUMNS. 

§§  302-308.  End-Conditions.  Euler's  Formula.  "  Incipient 
Flexure."  Hodgkinson's  Formulae.  Rankine's 
Formula . .  363 

§§  309-314.  Radii  of  Gyration.  Built  Columns.  Trussed  Gird- 
ers. Buckling  of  Web-Plates.  Examples 376 

CHAPTER  VII.     LINEAR     ARCHES  (OF  BLOCKWORK.) 

§§  315-323.     Inverted  Catenary.     Parabolic  Arch.     Circular  Arch. 

Transformed  Catenary  as  Arch 386 

CHAPTER  VIII.     ELEMENTS  OF  GRAPHICAL  STATICS 

§§  321-326.  Force  Polygons.  Concurrent  and  Non-Concurrent 
Forces  in  a  Plane.  Force  Diagrams.  Equili- 
brium Polygons 397 

§§  327-332.  Constructions  for  Resultant,  Pier-Reactions,  and 
Stresses  in  Roof  Truss.  Bow's  Notation.  The 
Special  Equilibrium  Polygon 402 

CHAPTER  IX.     GRAPHICAL  STATICS  OF  VERTICAL  FORCES. 

§§  333-336.        Jointed  Rods.     Centre  of  Gravity 412 

§§  337-343.        Useful  Relations  Between  Force  Diagrams  and  Their 

Equilibrium  Polygons 415 

CHAPTER  X.     RIGHT  ARCHES  OF  MASONRY. 

§§  344-352.  Definitions.  Mortar  and  Friction.  Pressure  in  Joints. 
Conditions  of  Safe  Equilibrium.  True  Linear 

Arch 421 

§§  353-357.        Arrangement  of  Data  for  Graphic  Treatment 428 

§§  358-362.        Graphical  Treatment   of  Arch.     Symmetrical  and 

Unsymmetrical  Cases 431 

CHAPTER  XI.     ARCH-RIBS. 

§§  364-374.  Mode  of  Support.  Special  Equilibrium  Polygon  and 
its  Force  Diagram.  Change  in  Angle  Between 
Rib  Tangents.  Displacement  of  Any  Point  on 
Rib 438 

§§  374o-378a.  Graphical  Arithmetic.  Summation  of  Products. 
Moment  of  Inertia  by  Graphics.  Classification 
of  Arch-Ribs 450 

§§  379-388.  Prof.  Eddy's  Graphical  Method  for  Arch-Ribs  of 
Hinged  Ends ;  and  of  Fixed  Ends.  Stress 
Diagrams.  Temperature  Stresses.  Braced 
Arches . .  461 


£ii  CONTENTS. 

HORIZONTAL    STRAIGHT    GIRDERS. 

§§  389-390.        Prismatic  Girder  Supported  at  the  Extremities,  also 

with  Ends  Fixed  Horizontally 479 

CHAPTER  XII.      GRAPHICS  OF  CONTINUOUS  GIRDERS. 

§§  391-397.  The  Elastic  Curve  an  Equilibrium  Polygon.  Mohr's 
Theorem.  Example  and  Numerical  Case.  End- 
Tangents.  Re-arrangement  of  Moment- Area. .  484 

§§  398-405.  Positive  Moment- Areas  ;  Amount  and  Gravity  Ver- 
tical. Construction  of  the  "False  Polygons" 
in  Any  Given  Case.  Moment  Curves,  Shears, 
and  Reactions.  Numerical  Example  in  Detail. 
Remarks..  49T 


[CONTENTS  OF  "MECHANICS  OF  FLUIDS] 

PAKT  IV.— HYDKAULICS. 

CHAPTER  I.— DEFINITIONS.    FLUID  PRESSURE.    HYDRO- 
STATICS  BEGUN. 

PAOK 

§§  406-417.  Perfect  fluids.  Liquids  and  Gases.  Principle  of  ' '  Equal 

Transmission  of  Pressure."  Non-planar  Pistons. . .  616 

§§  418-427.  Hydraulic  Press.  Free  Surface  of  Liquid.  Barometers 
and  Manometers.  Safety-valves.  Strength  of  Thin 
Hollow  Cylinders  against  Bursting  and  Collapse. . .  526 

CHAPTER  II.— HYDROSTATICS    CONTINUED.     PRESSURE    OF 
LIQUIDS  IN  TANKS  AND  RESERVOIRS. 

§§428-434.  Liquid  in  Motion,  but  in  "  Relative  Equilibrium." 
Pressure  on  Bottom  and  Sides  of  Vessels.  Centre 
of  Pressure  of  Rectangles,  Triangles,  etc 540 

§§  435-444.  Stability  of  Rectangular  and  Trapezoidal  Walls  against 
Water  Pressure.  High  Masonry  Dams.  Proposed 
Quaker  Bridge  Dam.  Earthwork  Dam.  Water 
Pressure  on  both  Sides  of  a  Gate 554 

CHAPTER  III.— EARTH  PRESSURE  AND  RETAINING  WALLS. 

§§  445-455.  Angle  of  Repose.  Wedge  of  Maximum  Thrust.  Geo- 
metrical Constructions.  Resistance  of  Retaining 
Walls.  Results  of  Experience 678 

CHAPTER  IV.— HYDROSTATICS  CONTINUED.    IMMERSION 
AND  FLOTATION. 

§§466-460.  Buoyant  Effort.  Examples  of  Immersion.  Specific 

Gravity.  Equilibrium  of  Flotation.  Hydrometer..  58*. 

§§  461-466.  Depth  of  Flotation.  Draught  and  Angular  Stability  of 

Ships.  The  Metacentre 59t 


XIV 


CONTENTS. 


CHAPTER  V.— HYDROSTATICS  CONTINUED. 
FLUIDS. 


GASEOUS 


§§466-478.  Thermometers.  Absolute  Temperature.  Gases  and  Va- 
pors. Critical  Temperature.  Law  of  Charles. 
Closed  Air-manometer.  Mariotte's  Law.  Mixture  of 
Gases.  Barometric  Levelling.  Adiabatic  Change..  604 

§§479-489.  Work  Done  in  Steam-engine  Cylinders.  Expanding 
Steam.  Graphic  Representation  of  Change  of  Statp 
of  Gas.  Compressed-air  Engine.  Air-compressor. 
Hot-air  Engines.  Gas-engines.  Heat- efficiency. 
Duty  of  Pumping-engines.  Buoyant  Effort  of  the 
Atmosphere, 634 


CHAPTER  VI.— HYDRODYNAMICS  BEGUN.     STEADY  FLOW 
OF  LIQUIDS  THROUGH  PIPES  AND  ORIFICES. 


Phenomena  of  a  "  Steady  Flow."  Bernoulli's  Theorem 
for  Steady  Flow  without  Friction,  and  Applications. 
Orifice  in  "  Thin  Plate"  ..........................  646 

§§  496-500.  Rounded  Orifice.  Various  Problems  involving  Flow 
through  Orifices.  Jet  from  Force-pump.  Velocity 
and  Density;  Relation.  Efflux  under  Water.  Efflux 
from  Vessel  in  Motion.  Barker's  Mill 

§§  501-508.  Efflux  from  Rectangular  and  Triangular  Orifices.  Pon- 
celet's  Experiments.  Perfect  and  Complete  Con- 
traction, etc.  Overfall  Weirs.  Experiments  of 
Francis,  Fteley  and  Stearns,  and  Bazin.  Short 
Pipes  or  Tubes 

§§  509-513.  Conical  Tubes.  Venturi's  Tube.  Fluid  Friction. 
Froude's  Experiments.  Bernoulli's  Theorem  with 
Friction.  Hydraulic  Radius.  Loss  of  Head.  Prob- 
lems involving  Friction  Heads  in  Pipes.  Accumu- 
lator ...........................  '  ................ 

§§  513^518.  Loss  of  Head  in  Orifices  and  Short  Pipes.  Coefficient 
of  Friction  of  Water  in  Pipes.  Fanning's  Table. 
Petroleum  Pumping.  Flow  through  Long  Pipes  .. 

§§  519-526.  Chezy's  Formula.  Fire-engine  Hose.  Pressure-energy. 
Losses  of  Head  due  to  Sudden  Enlargement  of  Sec- 
tion; Borda's  Formula.  Diaphragm  in  Pipe.  Ven- 
turi  Water-meter  ............................  .  ____  714 

§§  527-536.  Sudden  Diminution  of  Section.  Losses  of  Head  due  to 
Elbows,  Bends,  Valve-gates,  and  Throttle-valves. 
Examples,  Prof.  Bellinger's  Experiments  on  Elbows. 
Siphons.  Branching  Pipes.  Time  of  Emptying 
Vessels  of  Various  Forms;  Prisms,  Wedges.  Pyra- 
mids, Cones,  Paraboloids,  Spheres,  Obelisks,  and 
Volumes  of  Irregular  Form  using  Simpson's  Rule.  .  727 


663 


673 


692 


703 


CONTENTS.  XV 

PAGB 

CHAPTER  VII.— HYDRODYNAMICS,  CONTINUED;  STEADY 
FLOW  OF  WATER  IN  OPEN  CHANNELS. 

§§  538-S4&&.  Nomenclature.  Velocity  Measurements  and  Instru- 
ments for  the  same.  Ritchie-Haskell  Direction 
Current-meter.  Change  of  Velocity  with  Depth. 
Pilot's  Tube.  Hydrometric  Pendulum.  Wolt- 
mann's  Mill.  Gauging  Streams.  Chezy's  Formula 
for  Uniform  Motion  in  Open  Channel.  Experi- 
ments   749 

§§  5436-547. ,  Gutter's  Formula.  Sections  of  Least  Resistance.  Trape- 
zoidal Section  of  Given  Side  Slope  and  Minimum 
Friction.  Variable  Motion  in  Open  Channel. 
Bends.  Formula  introducing  Depths  at  End  Sec- 
tions. Backwater 759 

CHAPTER  VIII.— DYNAMICS  OF  GASEOUS  FLUIDS. 

§§  548-556.  Theorem  for  Steady  Flow  of  Gases  without  Friction. 
Flow  through  Orifices  by  Water-formula;  with 
isothermal  Expansion;  with  Adiabatic  Expansion. 
Maximum  Flow  of  Weight.  Experimental  Co- 
efficients for  Orifices  and  Short  Pipes.  Flow  con- 
sidering Velocity  of  Approach 773 

§§  557-5610.  Transmission  of  Compressed  Air  through  Long  Pipes. 
Experiments  in' St.  Gothard  Tunnel.  Pipes  of  Vari- 
able Diameter.  The  Piping  of  Natural  Gas 786 

CHAPTER  IX.— IMPULSE  AND  RESISTANCE  OF  FLUIDS. 

§§  563-569.  Reaction  of  a  Jet  of  Liquid.  Impulse  of  Jet  on  Curved 
Vanes,  Fixed  and  in  Motion.  Pitot's  Tube.  The 
California  "Hurdy-gurdy."  Impulse  on  Plates. 
Plates  Moving  in  a  Fluid.  Plates  in  Currents  of 
Fluid " 798 

.§§  570-675.  Wind-pressure.  Smithsonian  Scale.  Mechanics  of  the 
Sail-boat.  Resistance  of  Still  Water  to  Immersed 
Solids  in  Motion.  Spinning  Ball,  Deviation  from 
Vertical  Plane.  Robinson's  Cup  anemometer.  Re- 
sistance of  Ships.  Transporting  Power  of  a  Cur- 
rent. Fire-streams,  Hose- friction,  etc 818 


(  UNIVERSITY 

>»»C* 

MECHANICS  OF 


PEELIMINAEY  CHAPTEE. 

1.  Mechanics  treats  of  the  mutual  actions  and  relative  mo- 
tions of   material   bodies,  solid,  liquid,  and  gaseous;   and  by 
Mechanics  of  Engineering  is  meant  a  presentment  of  those 

principles  of  pure  mechanics,  and  their  applications,  which  are 
of  special  service  in  engineering  problems. 

2.  Kinds  of  Quantity. — Mechanics   involves   the  following 
fundamental  kinds  of  quantity  :    Space,  of  one,  two,  or  three 
dimensions,  i.e.,  length,  surface,  or  volume,  respectively ;  time, 
which  needs  no  definition  here;  force  and  mass,  as  defined  be- 
low;  and  abstract  numbers,  whose  values  are  independent  of 
arbitrary  units,  as.  for  example,  a  ratio. 

3.  Force. — A  force  is  one  of  a  pair  of  equal,  opposite,  and 
simultaneous  actions  between  two  bodies,  by  which  the  state* 
of  their  motions  is  altered  or  a  change  of  form  in  the  bodies 
themselves   is   effected.     Pressure,   attraction,   repulsion,  and 
traction  are  instances  in  point.     Muscular  sensation    conveys 
the  idea  of  force,  while  a  spring-balance  gives  an  absolute 
measure   of   it.  a  beam-balance  only  a  relative   measure.     In 
accordance  with  Newton's  third  law  of  motion,  that  action  and 
reaction  are  equal,  opposite,  and  simultaneous,  forces  always 
occur  in  pairs;  thus,  if  a  pressure  of  4:0  Ibs.  exists  between 
bodies  A  and  B,  if  A  is  considered 'by  itself  (i.e.,  "free"), 
apart  from  all  other  bodies  whose  actions  upon  it  are  called 
forces,  among  these  forces  will  be  one  of  40  Ibs.  directed  from 
B  toward  A.     Similarly,  if  B  is  under  consideration,  a  force 

*  The  state  of  motion  of  a  small  body  under  the  action  of  no  force,  or  of 
balanced  forces,  is  either  absolute  rest,  or  uniform  motion  in  a  right  line. 
If  the  motion  is  different  from  this,  the  fact  is  due  to  the  action  of  an  un- 
balanced force  (§  54). 


2  MECHANICS   OF   ENGINEERING. 

of  4:0  Ibs.  directed  from  A  toward  B  takes  its  place  among  the 
forces  acting  on  B.  This  is  the  interpretation  of  Newton's 
third  law. 

In  conceiving  of  a  force  as  applied  at  a  certain  point  of  a 
body  it  is  useful  to  imagine  one  end  of  an  imponderable  spiral 
spring  in  a  state  of  compression  (or  tension)  as  attached  at  the 
given  point,  the  axis  of  the  spring  having  the  given  direction 
of  the  force. 

4.  Mass  is  the  quantity  of  matter  in  a  body.     The  masses  of 
several  bodies  being  proportional  to  their  weights  at  the  same 
locality  on  the  earth's  surface,  in  physics  the  weight  is  taken 
as  the  mass,  but  in  practical  engineering  another  mode  is  used 
for  measuring  it  (as  explained  in  a  subsequent  chapter),  viz.: 
the  mass  of  a  body  is  equal  to  its  weight  divided  by  the  ac- 

.celeration  of  gravity  in  the  locality  where  the1  weight  is  taken, 
or,  symbolically,  M  =  G  -f-  g.  This  quotient  is  a  constant 
quantity,  as  it  should  be,  since  the  mass  of  a  body  is  invariable 
wherever  the  body  be  carried. 

5.  Derived  Quantities. — All   kinds  of  quantity  besides   the 
fundamental  ones  just  mentioned  are  compounds  of  the  latter, 
formed  by  multiplication  or  division,  such  as  velocity,  accele- 
ration, momentum,  work,  energy,  moment,  power,  and  force- 
distribution.     Some    of    these   are   merely   names   given    for 
convenience  to   certain  combinations  of  factors  which  come 
together  not  in  dealing  with  first  principles,  but  as  a  result  of 
common  algebraic  transformations. 

6.  Homogeneous  Equations  are  those  of  such  a  form  that  they 
are  true  for  any  arbitrary  system  of  units,  and  in  which  all 
terms  combined  by  algebraic  addition  are  of  the  same  kind. 

it 

Thus,  the  equation  s  =  ~  (in  which  g  =  the  acceleration  of 

gravity  and  t  the  time  of  vertical  fall  of  a  body  in  vacuo, 
from  rest)  will  give  the  distance  fallen  through,  $,  whatever 
units  be  adopted  for  measuring  time  and  distance.  But  if  for 


PRELIMINARY    CHAPTER.  3 

g  we  write  the  numerical  value  32.2,  which  it  assumes  when 
time  is  measured  in  seconds  and  distance  in  feet,  the  equation 
s  —  16.1£a  is  true  for  those  units  alone,  and  the  equation  is  not 
ot  homogeneous  form.  Algebraic  combination  of  homogeneous 
equations  should  always  produce  homogeneous  equations  ;  if 
not,  some  error  has  been  made  in  the  algebraic  work.  If  any 
equation  derived  or  proposed  for  practical  use  is  not  homogene- 
ous, an  explicit  statement  should  be  made  in  the  context  as  to 
the  proper  units  to  be  employed. 

7.  Heaviness. — By  heaviness  of  a  substance  is  meant  the 
weight  of  a  cubic  unit  of  the  substance.  E.g.  the  heaviness  of 
fresh  water  is  62.5,  in  case  the  unit  of  force  is  the  pound, 
and  the  foot  the  unit  of  space;  i.e.,  a  cubic  foot  of  fresh 
water  weighs  62.5  Ibs.  In  case  the  substance  is  not  uniform 
in  composition,  the  heaviness  varies  from  point  to  point.  If 
the  weight  of  a  homogeneous  body  be  denoted  by  #,  its  volume 
by  F",  and  the  heaviness  of  its  substance  by  y,  then  G  =  Vy. 

WEIGHT  IN  POUNDS  OF  A  CUBIC  .FOOT  (i.e.,  THE  HEAVINESS)  OP  VARIOUS 

MATERIALS. 


Anthracite,  solid 100 

broken 57 

Brick,  common  hard 125 

"  soft 100 

Brick-work,  common 112 

Concrete 125 

Earth,  loose  72 

"  as  mud 102 

Granite 164  to  172 

Ice 58 

Iron,  cast 450 

"     wrought 480 


Masonry,  dry  rubble. , 138 

dressed  granite  or 

limestone 165 

Mortar 100 

Petroleum 55 

Snow 7 

"  wet 15  to  50 

Steel ." .  490 

Timber 25  to  60 

Water,  fresh 62 . 5 

sea..  ..64.0 


8.  Specific  Gravity  is  the  ratio  of  the  heaviness  of  a  material 
to  that  of  water,  and  is  therefore  an  abstract  number. 

9.  A  Material  Point  is  a  solid  body,  or  small  particle,  whose 
dimensions  are  practically  nothing,  compared  with  its  range  of 
motion. 


4  MECHANICS    OF   ENGINEERING. 

10.  A  Rigid  Body  is  a  solid,  whose  distortion  or  change  of 
form  under  any  system  of  forces  to  be  brought  upon  it  in 
practice  is,  for  certain  purposes,  insensible. 

11.  Equilibrium. — When  a  system  of  forces  applied  to  a 
body  produces  the  same  -effect  as  if  no  force  acted,  so  far  as 
the  state  of  motion  of  the  body  is  concerned,  they  are  said  to 
be  balanced,  or  to  be  in  equilibrium.     [If  no  force  acts  on  a 
material  point  it  remains   at  rest  if  already  at  rest ;  but  if 
already  in    motion   it   continues   in  motion,  and  uniformly 
(equal  spaces   in   equal  times),  in  a   right   line  in  direction 
of  its  original  motion.     See  §  54.] 

12.  Division  of  the  Subject. — Statics  will  treat  of  bodies  at 
rest,  i.e.,  of  balanced   forces   or   equilibrium ;    dynamics,  of 
bodies  in  motion  ;  strength  of  materials  will  treat  of  the  effect 
of  forces  in  distorting  bodies;  hydraulics,  of  the  mechanics 
of  liquids  and  gases  (thus  including  pneumatics). 

13.  Parallelogram  of  Forces. — Duchayla's  Proof.     To  fully 
determine  a  force  we  must  have  given  its  amount,  its  direc- 
tion, and  its  point  of  application  in  the  body.     It  is  generally 
denoted  in  diagrams  by  an  arrow.     It  is  a  matter  of  experience 
that  besides  the  point  of  application  already  spoken  of  any 
other  may  be  chosen  in  the  line  of  action  of  the  force.     This 
is  called  the  transmissibility  of  force;  i.e.,  so  far  as  the  state  of 
motion  of  the  body  is  concerned,  a  force  may  be  applied  any- 
where in  its  line  of  action. 

The  Resultant  of  two  forces  (called  its  components)  applied 
at  a  point  of  a  body  is  a  single  force  applied  at  the  same  point, 
'which  will  replace  them.  To  prove  that  this  resultant  is  given 
in  amount  and  position  by  the  diagonal  of  the  parallelogram 
formed  on  the  two  given  forces  (conceived  as  laid  off  to  some 
scale,  so  m,any  pounds  to  the  inch,  say),  Duchayla's  method 
.  requires  four  postulates,  viz. :  (1)  the  resultant  of  two  forces 
must  lie  in  the  same  plane  with  them ;  (2)  the  resultant  of  two 
equal  forces  must  bisect  the  angle  between  them ;  (3)  if  one  of 
the  two  forces  be  increased,  the  angle  between  the  other  force 
and  the  resultant  will  be  greater  than  before;  and  (4)  the  trans- 
missibility of  force,  already  mentioned.  Granting  these,  we 
proceed  as  follows  (Fig.  1)  :  Given  the  two  forces  P  and  Q  =- 


PRELIMINARY    CHAPTER. 


p,  _|_  pn  (p,  and  pn 

applied  at  <9.  Transmit  P"  to  J..  Draw  the  parallelograms 
OB  and  AD\  OD  will  also  be  a  parallelogram.  By  postulate 
(2),  since  OB  is  a  rhombus,  P  and  P'  at  <9  may  be  replaced  by 
a  single  force  R'  acting  through  B.  Transmit  R'  to  B  and 
replace  it  by  P  and  P'.  Transmit  P  from  B  to  J.,  P'  from 
B  to  I>.  Similarly  P  and  P",  at  ^L,  may  be  replaced  by  a 
single  force  R"  passing  through  _Z> ;  transmit  it  there  and  re- 
solve  it  into  P  and  P" .  Pf  is  already  at  D.  Hence  P  and 
P'  +  P",  acting  at  _Z>,  are  equivalent  to  P  and  P'  -\-  P"  act- 
ing at  0,  in  their  respective  directions.  Therefore  the  result- 
ant of  P  and  P'  +  P"  must  lie  in  the  line  OD,  the  diagonal 
of  the  parallelogram  formed  on  P  and  Q  =  2P  at  0.  Similarly 


SLfiLfl 


FIG.  i.  FIG.  2. 

this  may  be  proved  (that  the  diagonal  gives  the  direction  of 
the  resultant)  for  any  two  forces  P  and  mP ;  and  for  any  two 
forces  nP  and  mP,  m  and  ?i  being  any  two  whole  numbers, 
i.e.,  for  any  two  commensurable  forces.  When  the  forces  are 
incommensurable  (Fig.  2),  P  arid  Q  being  the  given  forces, 
we  may  use  a  reductio  ad  dbsurdum,  thus :  Form  the  parallelo- 
gram OD  on  P  and  Q  applied  at  0.  Suppose  for  an  instant 
that  R  the  resultant  of  P  and  Q  does  not  follow  the  diagonal 
OD,  but  some  other  direction,  as  OD' .  Note  the  intersection 
H,  and  draw  HC  parallel  to  DB.  Divide  P  into  equal  parts, 
each  less  than  HD ;  then  in  laying  off  parts  equal  to  these  from 
0  along  OB,  a  point  of  division  will  come  at  some  point  F 
between  C  and  B.  Complete  the  parallelogram  OFEG.  The 
force  Q"  =  OF  is  commensurable  with  P,  and  hence  their 


6  MECHANICS   OF  ENGINEERING. 

resultant  acts  along  OE.  Now  Q  is  greater  than  Q"  ',  while  R 
makes  a  less  angle  with  P  than  OE,  which  is  contrary  to  pos- 
tulate (3);  therefore  R  cannot  lie  outside  of  the  line  OD* 
Q.  E.  D. 

It  still  remains  to  prove  that  the  resultant  is  represented  in 
amount,  as  well  as  position,  by  the  diagonal.     OD  (Fig.  3)  is 
X/R'  the  direction  of  R  the  resultant  of  P  and 

/F  \N          _      .     Q  ;  required  its  amount.     If  R'  be  a  force 
-  ^t  equal  and  opposite  to  It  it  will  balance  P 
\P/1    al)d   Q  ?  i-e?  tne  resultant   of  R'  and   P 
p  """Jj*     must  lie  in  the  line  QO  prolonged  (besides 

being  equal  to  Q).  "We  can  therefore  de- 
termine R  by  drawing  BA  parallel  to  DO  to  intersect  QO 
prolonged  in  A  ;  and  then  complete  the  parallelogram  on 
BA  and  BO.*  Since  OFABis  a  parallelogram  R  must  =BA, 
and  since  OABD  is  a  parallelogram  BA=OD\  therefore 
and  also  72=  <9lX  Q.  E.  D. 


Corollary.  —  The  resultant  of  three  forces  applied  at  the  same 
point  is  the  diagonal  of  the  parallelepiped  formed  on  the  three 
forces. 

14.  Concurrent  forces  are  those  whose  lines  of  action  intersect 
in  a  common  point,  while  non-concurrent  forces  are  those  which 
do  not  so  intersect  ;  results  obtained  for  a  system  of  concurrent 
forces  are  really  derivable,  as  particular  cases,  from  those  per- 
taining to  a  system  of  non-concurrent  forces. 

15.  Resultant.  —  A  single  force,  the  action  of  which,  as  re- 
gards the  state  of  motion  of  the  body  acted  on,  is  equivalent  to 
that  of  a  number  of  forces  forming  a  system,  is  said  to  be  the 
Resultant  of  that  system,  and  may  replace  the  system  ;  and  con- 
versely a  force  which  is  equal  and  opposite  to  the  resultant  of 
a  system  will  balance  that  system,  or,  in  other  words,  when  it 
is  combined  with  that  system  there  will  result  a  new  system  in 
equilibrium  ;  this  (ideal)  force  is  called  the  Anti-resultant. 

In  general,  as  will  be  sesn,  a  given  system  of  forces  can  al- 

*  R  must  =  ~OF;  for  if  R'  >  or  <  ~OF,  the  diagonal  formed  on  R'  and  P 
cannot  take  the  direction  of  QO  prolonged. 


PRELIMINARY    CHAPTER.  7 

ways  De  replaced  by  two  single  forces,  but  these  two  can  be 
combined  into  a  single  resultant  only  in  'particular  cases. 

15a.  Equivalent  Systems  are  those  which  may  be  replaced  by 
the  same  set  of  two  single  forces  —  or,  in  other  Mords,  those 
which  have  the  same  effect,  as  to  state  of  motion,  upon  the 
given  body. 

15b.  Formulae.  —  If  in  Fig.  3  the  forces  P  and  Q  and  the  angle  a  = 
P  0  Q  are  given,  we  have,  for  the  resultant, 


Pz  +  Q'  +  2PQcosa. 

(If  a  is  >  90°  its  cosine  is  negative.)  In  general,  given  any  three  parts 
of  either  plane  triangle  0  D  Q,  or  0  1)  B,  the  other  three  may  be  obtained 
by  ordinary  trigonometry.  Evidently  if  or  =  0,  R  =  P  +  Q  ;  if  a  = 
180°,  #  =  P-  £  ;  and  if  a:  =  90°,  R  =  V  F*  +  Q"- 

15c.  Varieties  of  Forces,  —  Great  care  should  be  used  in  deciding 
what  may  properly  be  called  forces.  The  latter  may  be  divided  into  ac- 
tions by  contact,  and  actions  at  a  distance.  If  pressure  exists  between  two 
bodies  and  they  are  perfectly  smooth  at  the  surface  of  contact,  the  pressure 
(or  thrust,  or  compressive  action),  of  one  against  the  other  constitutes  a  force, 
whose  direction  is  normal  to  the  tangent  plane  at  any  point  of  contact  (a 
matter  of  experience)  ;  while  if  those  surfaces  are  not  smooth  there  may  also 
exist  mutual  tangential  actions  or  friction.  (If  the  bpdies  really  form  a 
continuous  substance  at  the  surface  considered,  these  tangential  actions  are 
called  shearing  forces.}  Again,  when  a  rod  or  wire  is  subjected  to  tension, 
any  portion  of  it  is  said  to  exert  a  pull  or  tensile  force  upon  the  remainder  ; 
the  ability  to  do  this  depends  on  the  property  of  cohesion.  The  foregoing 
are  examples  of  actions  by  contact. 

Actions  at  a  distance  are  exemplified  in  the  mysterious  attractions,  or  re- 
pulsions, observable  in  the  phenomena  of  gravitation,  electricity,  and  mag- 
netism, where  the  bodies  concerned  are  not  necessarily  in  contact.  By  the 
term  weight  we  shall  always  mean  the  force  of  the  earth's  attraction  on  the 
body  in  question,  and  not  the  amount  of  matter  in  it. 

[NOTE.—  In  some  common  phrases,  such  as  "  The  tremendous  force"  of  a  heavy  body  in 
rapid  motion,  the  word  force  is  not  used  in  a  technical  sense,  but  signifies  energy  (as  ex- 
plained in  Chap.  VI.).  The  mere  fact  that  a  hody  is  in  motion,  whatever  its  mass  and 
velocity,  does  not  imply  that  it  is  under  the  action  of  any  force,  necessarily.  For  instance, 
at  any  point  in  the  path  of  a  cannon  ball  through  the  air,  the  only  forces  acting  on  it  are 
the  resistance  of  the  air  and  the  attraction  of  the  earth,  the  latter  having  a  vertica  And 
downward  direction.] 


PART  I.-STATICS. 


CHAPTEK  L 

STATICS  OF  A  MATERIAL  POINT. 

16.  Composition  of  Concurrent  Forces. — A  system  of  forces 
acting  on  a  material  point  is  necessarily  composed  of  concurrent 
forces. 

CASE  I. — All  the  forces  in  One  Plane.  Let  0  be  the 
material  point,  the  common  point  of  application  of  all  the 
forces ;  P^  P»  etc.,  the  given  forces,  making 
angles  or,,  <*2,  etc.,  with  the  axis  X.  By  the 
parallelogram  of  forces  P.  may  be  resolved 
into  and  replaced  by  its  components,  P1  cos  a, 

-> •—-  acting  along  X>   and   Pl   sin   al   along    Y, 

FIG.  4.  Similarly  all  the  remaining  forces  may  be  re- 

placed by  their  X  and  Y  components.  We  have  now  a  new 
system,  the  equivalent  of  that  first  given,  consisting  of  a  set  of 
X  forces,  having  the  same  line  of  application  (axis  X),  and  a 
set  of  Y  forces,  all  acting  in  the  line  Y.  The  resultant  of  the 
X  forces  being  their  algebraic  sum  (denoted  by  2X)  (since 
they  have  the  same  line  of  application)  we  have 

2X  =  Pl  cos  a,  +  P,  cos  <+  etc.  =  2(P  cos  a), 
and  similarly 

2  Y  =  Pl  sin  a ,  +  P9  sin  «2  +  etc.  =  2(P  sin  a). 

These  two  forces,  2X  and  2  Y,  may  be  combined  by  the 
parallelogram  of  forces,  giving  R  —  |/(^"X)a  +  (2Y)* as  tne 
single  resultant  of  the  whole  system,  and  its  direction  is  deter- 

2Y 
mined  by  the  angle  or;  thus,  tan  a  =  -^p-;  see  Fig.  5.     For 

equilibrium  to  exist,  R  must  =  0.  which  requires,  separately^ 


STATICS    OF    A    MATERIAL    POINT. 


=0,  and  2Y=0  (for  the  two  squares  (2X)*  and 
can  neither  of  them  be  negative  quantities). 

CASE  II. — The  forces  having  any  directions  in  space, 
but  all  applied  at  ,<9,  the  material  point.  Let  P19  P^ 
etc.,  be  the  given  forces,  P1  making  the  angles  <*,,  /?1?  and  ya 
respectively,  with  three  arbitrary  axes,  X,  Y,  and  Z  (Fig.  6), 
at  right  angles  to  each  other  and  intersecting  at  0,  the  origin. 
Similarly  let  <*„  /?3,  y^  be  the  angles  made  by  Pz  with  these 
axes,  and.  so  on  for  all  the  forces.  By  the  parallelepiped  of 
forces,  Pl  may  be  replaced  by  its  components. 

Xl  =  P1  cos  flfj,  Yl  =  Pl  cos  ftv  and  Zl  —  Pl  cos  yl ;  and 


<"'2Y 


FIQ.  7. 


similarly  for  all  the  forces,  so  that  the  entire  system  is  now 
replaced  by  the  three  forces, 

=  P,  cos  or,  +  P,  cos  at  +  etc  ; 
=  P,  cos  A  +  P2  cos  #,  +  etc; 
=  P,  cos  ^  +  P,  cos  /2  +  etc  ; 

and  finally  by  the  single  resultant 


Therefore,  for  equilibrium  we  must  have  separately, 

2Z=  0,  2Y=  0,  and  ^Z^  0. 
position  may  be  determined  by  its  direction  cosines,  viz., 


cos  a  =  -Q-  ;  cos  /?  = 


cos      = 


17.  Conditions  of  Equilibrium. — Evidently,  in  dealing  with 
a  system  of  concurrent  forces,  it  would  be  a  simple  matter  to 


10         •'   •<    •       MECHANICS   OF   ENGINEERING. 

replace  any  two  of  the  forces  by  their  resultant  (diagonal 
formed  on  them),  then  to  combine  this  resultant  with  a  third 
force,  and  so  on  until  all  the  forces  had  been  combined,  the 
last  resultant  being  the  resultant  of  the  whole  system.  The 
foregoing  treatment,  however,  is  useful  in  showing  that  for 
equilibrium  of  concurrent  forces  in  a  plane  only  two  conditions 
are  necessary,  viz.,  2X '=  0  and  2  Y  =  0;  while  in  space 
there  are  three,  2X=  0,  2  Y  =  0,  and  2Z  —  0.  In  Case  I., 
then,  we  have  conditions  enough  for  determining  two  unknown 
quantities ;  in  Case  II.,  three. 

18.  Problems  involving  equilibrium  of  concurrent  forces. 
(A  rigid  body  in  equilibrium  under  no  more  than  three  forces 
may  be  treated  as  a  material  point,  since  the  (two  or)  three 
forces  are  necessarily  concurrent.)* 

PROBLEM  1. — A  body  weighing  G  Ibs.  rests  on  a  horizontal 

table :  required  the  pressure  between  it  and  the  table.     Fig.  8. 

Consider  the  body  free,  i.e.,  conceive  all  other  bodies  removed 

,  (the  table  in  this  instance),  being  replaced  by  the 

i  forces  which  they  exert  on  the  first  body.     Taking 

G  •  the  axis  ^vertical  and  positive  upward,  and  not 
+X  assuming  in  advance  either  the  amount  or  direc- 
tion  of  JV,  the  pressure  of  the  table  against  the 
body,  but  knowing  that  6r,  the  action  of  the  earth 
on  the  body,  is  vertical  and  downward,  we  have 
here  a  system  of  concurrent  forces  in  equilibrium,  in  which 
the  X  and  Y  components  of  G  are  known  (being  0  and  - 
G  respectively),  while  those,  jVx  and  N^  of  N  are  unknown. 
Putting  2X  =  0,  we  have  N^  +  0  =  0 ;  i.e.,  N  has  no  hori- 
zontal component,  /.  N  is  vertical.  Putting  2  Y  =  0,  we 
have  N^  —  G  =  0,  .*.  N?  =  -\-  G ;  or  the  vertical  component 
of  N)  i.e.,  N  itself,  is  positive  (upward  in  this  case),  and  is 
numerically  equal  to  G. 

PKOBLEM  2. — Fig.  9.  A  body  of  weight  G  (Ibs.)  is  moving 
in  a  straight  line  over  a  rough  horizontal  table  with  a  uniform 
velocity  v  (feet  per  second)  to  the  right.  The  tension  in  an 
oblique  cord  by  which  it  is  pulled  is  given,  and  =  P  (Ibs.), 

*  Three  parallel  forces  form  an  exception  ;  see  §§20,  21,  etc. 


STATICS    OF   A    MATERIAL    POINT.  11 

which  remains  constant,  the  cord  making  a  given  angle  of 
elevation,  <*,  with  the  path  of  the  body.  Required  the  vertical 
pressure  ffl  (Ibs.)  of  the  table,  and  also  its 
horizontal  action  F  (friction)  (Ibs.)  against 
the  body 

Referring  by  anticipation  to  Newton's  first 
law  of  motion,  viz.,  a  material  point  acted 
on  by  no  force  or  by  balanced  forces  is  either  FIG.  9. 

at  rest  or  moving  uniformly  in  a  straight  line,  we  see  that  this 
problem  is  a  case  of  balanced  forces,  i.e.,  of  equilibrium.  Since 
there  are  only  two  unknown  quantities,  N  and  F,  we  may 
determine  them  by  the  two  equations  of  Case  I.,  taking  the 
axes  X  and  Y  as  before.  Here  let  us  leave  the  direction  of 
N  as  well  as  its  amount  to  be  determined  by  the  analysis.  As 
j^must  evidently  point  toward  the  left,  treat  it  as  negative  in 
summing  the  X  components  ;  the  analysis,  therefore,  can  be 
expected  to  give  only  its  numerical  value. 
2X  =  0  gives  P  cos  a  —  F—  0.  .-.  F  =  P  cos  a. 

2  Y  =  0  gives  AT+  P  sin  a—j&  =  0.  .'.  N  =  G  -  Pein  a. 
,'.  N  is  upward  or  downward  according  as  G  is  >  or  <  P 
sin  a.  For  N  to  be  a  downward  pressure  upon  the  body  would 
require  the  surface  of  the  table  to  be  above  it.  The  ratio  of  the 
friction  F  to  the  pressure  N  which  produces  it  can  now  be 
obtained,  and  is  called  the  coefficient  of  friction.  It  may  vary 
slightly  with  the  velocity. 

This  problem  may  be  looked  upon  as  arising  from  an  experi- 
ment made  to  determine  the  coefficient  of  friction  between  the 
given  surfaces  at  the  given  uniform  velocity. 

19.  The  Free-Body  Method.  —  The  foregoing  rather  labored  so- 
lutions of  very  simple  problems  have  been  made  such  to  illus- 
trate what  may  be  called  the  free-body  method  of  treating  any 
problem  involving  a  body  acted  on  by  a  system  of  forces.  It 
consists  in  conceiving  the  body  isolated  from  all  others  which 
act  on  it  in  any  way,  those  actions  being  introduced  as  so  many 
forces,  known  or  unknown,  in  amount  and  position.  The  sys- 
tem of  forces  thus  formed  may  be  made  to  yield  certain  equa- 


Of  THE 

UNIVERSITY 


12  MECHANICS   OF   ENGINEERING. 

tions,  whose  character  and  number  depend  on  circumstances, 
such  as  the  behavior  of  the  body,  whether  the  forces  are  con- 
fined to  a  plane  or  not,  etc.,  and  which  are  therefore  theoreti- 
cally available  for  determining  an  equal  number  of  unknown 
quantities,  whether  these  be  forces,  masses,  spaces,  times,  or 
abstract  numbers.  Of  course  in  some  instances  the  unknown 
quantities  may  enter  these  equations  with  such  high  powers 
that  the  elimination  may  be  impossible  ;  but  this  is  a  matter 
of  algebra,  not  of  mechanics. 

(Addendum  to  §  49a  of  page  49.)  Numerical  Example.— A  set  of  light 
screens  is  set  up  at  intervals  of  100  feet  apart  in  the  horizontal  path  of  a 
cannon-ball,  with  the  object  of  determining  its  velocity,  and  also  the  rate  of 
change  (or  negative  acceleration)  of  that  velocity,  as  due  to  the  resistance 
of  the  air. 

By  electrical  connection  the  time  of  passing  each  screen  is  noted,  and 
the  intervals  of  time  are  given  in  this  diagram  for  four  of  the  screens, 
A,  B,  C,  and  D. 


100' 

..0.0621  sec.. 


....100' 

.0.0632  sec.. 


100' 

..0.0643  sec.. 


A  1  B  2  C  3  D 

.  From  these  data  it  is  required  to  compute,  as  nearly  as  the  circumstances 
allow,  the  velocity  and  acceleration  (negative)  of  the  ball  at  various  points 
(the  ball  moves  from  left  to  right). 

Solution. — In  passing  from  A  to  B  the  ball  has  an  average  velocity  of 
1610  ft.  per  second,  obtained  by  dividing  the  distance  of  100  feet  by  the 
time  of  passage,  0.0621  second.  Similarly  we  find  the  average  velocity 
between  B  and  G  to  be  1582  ft.  per  second,  and  that  between  G  and  D  to 
be  1554  ft.  per  second. 

As  the  velocity  is  not  changing  very  rapidly,  we  may  claim  that  the  ball 
actually  possesses  the  velocity  Vi  =  1610  ft.  per  second  at  the  point  1,  mid- 
way between  A  and  B,  or  very  near  that  point ;  and  similarly  the  velocity 
$2  =  1582  ft.  per  second  at  point  2,  midway  between  B  and  G ;  and 
V3  =  1554  ft.  per  second  at  point  3,  midway  between  C  and  D. 

Hence  the  total  gain  of  velocity  from  1  to  2  is  1582-1610  =  —  28  ft.  per 
second;  and  the  time  in  which  this  gain  is  made  is  one  half  of  the  0  )621 
second  plus  one  half  of  the  0.0632  second,  i.e.,  0.0626  second.  Therefore 
an  approximate  value  for  the  average  acceleration  between  points  1  and  2 
is  found  by  dividing  the  —28  ft.  per  second  gain  in  velocity  by  the 
time  0.0626  second  occupied  in  acquiring  the  gain.  This  gives  —  447  ft. 
per  second  per  second  average  acceleration  for  portion  1. .  .2  of  path,  and 
since  screen  B  lies  at  the  middle  of  this  portion,  the  actual  acceleration  of 
the  ball's  motion  as  it  passes  the  screen  B  is  very  nearly  equal  to  this,  viz.: 
—  447  ft.  per  second  per  second  (or  "  ft.  per  square  second  "). 

By  a  similar  process  the  student  may  compute  the  acceleration  at  screen 
C.  Of  course  the  reason  why  these  results  are  merely  approximate  is  that 
the  spaces  and  times  concerned,  though  relatively  small ,  are  not  infinitesimal. 

[A  recent  English  writer  calls  a  unit  of  velocity  a  "speed;"  and  a  unit 
of  acceleration,  a  "hurry."] 


PARALLEL  FORCES   ANJJ   THE   CENTRE   OF   GRAVITY.   13 


CHAPTER  II. 
PARALLEL  FORCES  AND  THE  CENTRE  OF  GRAVITY. 

20.  Preliminary  Remarks. — Although  by  its  title  this  section 
should  be  restricted  to  a  treatment  of  the  equilibrium  of  forces, 
certain  propositions  involving  the  composition  and  resolution 
of  forces,  without  reference  to  the.  behavior  of  the  body  under 
their  action,  will  be  found  necessary  as  preliminary  to  the  prin- 
cipal object  in  view. 

As  a  rigid  body  possesses  extension  in  three  dimensions,  to 
deal  with  a  system  of  forces  acting  on  it  we  require  three  co- 
ordinate axes :  in  other  words,  the  system  consists  of  "  forces 
in  space,"  and  in  general  the  forces  are  non-concurrent.  In 
most  problems  in  statics,  however,  the  forces  acting  are  in  one 
plane:  we  accordingly  begin  by  considering  non-concurrent 
forces  in  a  plane,  of  which  the  simplest  case  is  that  of  two 
parallel  forces.  For  the  present  the  body  on  which  the  forces 
act  will  not  be  shown  in  the  figure,  but  must  be  understood  to 
be  there  (since  we  have  no  conception  of  forces  independently 
of  material  bodies).  The  device  will  frequently  be  adopted  of 
introducing  into  the  given  system  two  opposite  and  equal  forces 
acting  in  the  same  line :  evidently  this  will  not  alter  the  effect 
of  the  given  system,  as  regards  the  rest  or  motion  of  the  body. 

21.  Resultant    of    two     Parallel 
Forces. 

CASE  I. — The  two  forces  have 
the  same  direction.  Fig.  10. 
Let  P  and  Q  be  the  given  forces, 
and  AB  a  line  perpendicular  to 
them  (P  and  Q  are  supposed  to  have 
been  transferred  to  the  intersections 
A  and  B}.  Put  in  at  A  and  B  two  equal  and  opposite 
forces  8  and  $,  combining  them  with  P  and  Q  to  form  P' 


14  MECHANICS    OF    EN  GIN  EE  KING. 

and  Q'.  Transfer  P'  and  Q'  to  their  intersection  at  C,  and  there 
resolve  them  again  into  8  and  P,  /Sand  Q.  S  and  S  annul  each 
other  at  C\  therefore  P  and  Q,  acting  along  a  common  line  CD, 
replace  the  P  and  Q  first  given  ;  i.e.,  the  resultant  of  the  origi- 
nal two  forces  is  a  force  R  —P-\-  Q,  acting  parallel  to  them 
through  the  point  D,  whose  position  must  now  be  determined. 
The  triangle  CAD  is  similar  to  the  triangle  shaded  by  lines, 
/.  P  :  8  ll  CD  :  a?;  and  CDB  being  similar  to  the  triangle 
shaded  by  dots,  /.  8  :  Q  ::  a  —  x  :  CD.  Combining  these,  we 

:^^-^    and    .-.  x  =          -  =  -f     Now  write  this 


—  Qcd,  and  add  Re,  i.e.,  Pc-\-  Qc,to  each  member,  c  being 
the  distance  of  0  (Fig.  10),  any  point  in  AB  produced,  from 
A.  This  will  give  R(x  +  c)  =  PC  -\-Q(a  +  c),  in  which  c, 
a  -f-  c,  and  x  -f-  c  are  respectively  the  lengths  of  perpendiculars 
let  fall  from  0  upon  P,  Q,  and,  their  resultant  R.  Any  one  of 
these  products,  such  as  PC,  is  for  convenience  (since  products  of 
this  form  occur  so  frequently  in  Mechanics  as  a  result  of  alge- 
braic transformation)  called  the  Moment  of  the  force  about  the 
arbitrary  point  0.  Hence  the  resultant  of  two  parallel  forces  of 
the  same  direction  is  equal  to  their  sum,  acts  in  their  plane,  in 
a  line  parallel  to  them,  and  at  such  a  distance  from  any  arbi- 
trary point  0  in  their  plane  as  may  be  determined  by  writing 
its  moment  about  0  equal  to  the  sum  of  the  moments  of  the 
two  forces  about  0.  0  is  called  a  centre  of  moments,  and  each 
of  the  perpendiculars  a  lever-arm. 

CASE  II.  —  Two  parallel  forces  P  and   Q  of  opposite  direc- 
tions.     Fig.  11.     By  a  process  similar  to  the  foregoing,  we 

obtain  R  =  P  -  Q  and  (P  -  Q)x 
=  Qa,  i.e.,  Rx  =  Qa.  Subtract 
each  member  of  the  last  equation 
from  Re  (i.e..  Pc—Qc],  in  which  c 
is  the  distance,  from  A,  of  any  arbi- 
trary point  O  in  AB  produced.  This 
gives  R(c  —  x)  =  PC  —  Q(a  +  c). 
But  (c  —  x),  c,  and  (a  -\-  c)  are  re- 
11.  spectively  the  perpendiculars,  from 

*  That  is,  the  resultant  of  two  parallel  forces  pointing  in  the  same  direc- 
tion divides  the  distance  between  them  in  the  inverse  ratio  of  those  forces. 


PARALLEL  FORCES  AND  THE  CENTRE  OF  GRAVITY.  15 


0,  upon  P,  P,  and  Q.  That  is,  E(c  —  x)  is  the  moment  of  It 
about  <9;  PC,  that  of  P  about  (9;  and  $(#+c)>  that  of  Q 
about  (9.  But  the  moment  of  Q  is  subtracted  from  that  of  P, 
which  corresponds  with  the  fact  that  Q  in  this  figure  would 
produce  a  rotation  about  0  opposite  in  direction  to  that  of  P. 
HavLig  in  view,  then,  this  imaginary  rotation,  we  may  define 
the  moment  of  a  force  as  positive  when  the  indicated  direction 
about  the  given  point  is  against  the  hands  of  a  watch;  as  nega- 
tive when  with  the  hands  of  a  watch.* 

Hence,  in  general,  the  resultant  of  any  two  parallel  forces  is, 
in  amount,  equal  to  their  algebraic  sum,  acts  in  a  parallel  direc- 
tion in  the  same  plane,  while  its  moment,  about  any  arbitrary 
point  in  the  plane,  is  equal  to  the  algebraic  sum  of  the  mo- 
ments of  the  two  forces  about  the  same  point. 

Corollary. — If  each  term  in  the  preceding  moment  equations 
be  multiplied  by  the  secant  of  an  angle  (or,  Fig.  12)  thus: 


n 

.         u,         ^ 

«  6l"2'     "" 

FIG.  12. 


FIG.  13. 


(using  the  notation  of  Fig.  12),  we  have  Pa  sec  a  =  P^ 
sec  a  4-  P2aa  sec  <*,  i.e.,  Pb  =  Plb,  -f-  Pa£a,  in  which  J,  5,, 
and  5a  are  the  oblique  distances  of  the  three  lines  of  action 
from  any  point  0  in  their  plane,  and  lie  on  the  same  straight 
line ;  P  is  the  resultant  of  the  parallel  forces  P,  and  P,. 

22.  Resultant  of  any  System  of  Parallel  Forces  in  Space.— 

Let  Pj,  P,,  fv  etc.,  be  the  forces  of  the  system,  and  x^  y^ 
%v  x»  y»  z»  etc-5  tne  co-ordinates  of  their  points  of  application 
as  referred  to  an  arbitrary  set  of  three  co-ordinate  axes  X,  Y^ 
and  Z,  perpendicular  to  each  other.  Each  force  is  here  re- 

*  These  two  directions  of  rotation  are  often  called  counter  clockwise,  and 
clockwise,  respectively. 


16*  MECHANICS    OF    ENGINEERING. 

stricted  to  a  definite  point  of  application  in  its  line  of  action 
(with  reference  to  establishing  more  directly  the  fundamental 
equations  for  the  co-ordinates  of  the  centre  of  gravity  of  a 
body).  The  resultant  P'  of  any  two  of  the  forces,  as 
PI  and  PV  is  =  Pl  -\-  P^  and  may  be  applied  at  (7,  the  in- 
tersection of  its  own  line  of  action  with  a  line  BD  joining 
the  points  of  application  of  Pl  and  P»  its  components. 
Produce  the  latter  line  to  A,  where  it  pierces  the  plane  XYy 
and  let  #„  5',  and  &2,  respectively,  be  the  distances  of  B,  Cy 
D,  from  A  }  then  from  the  corollary  of  the  last  article  we  have 


but  from  similar  triangles 

V  :  5,  :  5,  ::  «'  :  *,  :  zt,        .:  P'z'  = 

Now  combine  P\  applied  at  (7,  with  P^  applied  at  E,  calling 
their  resultanjt  P"  and  its  vertical  co-ordinate  zh  ',  and  we  obtain 


P"z"  =  P'z'  +  PA,  i.e.,  P"z"  ==  PA 
also 


Proceeding  thus  until  all  the  forces  have  been  considered,  we 
shall  have  finally,  for  the  resultant  of  the  whole  system, 


and  for  the  vertical  co-ordinate  of   its  point  of  application, 
which  we  may  write  z, 

Rz  =  P,e>  +  PA  +  PA  +  etc  ; 


•      -_  ..         .,---_ 

P,  +  P,+  P,+  ....    '       2P 

nd  similarly  for  the  other  co-ordinates. 
- 

*  = 


In  these  equations,  in  the  general  case,  such  products  as  P&, 
etc.,  cannot  strictly  be  called  moments.     The  point  whose  co« 


or"   +x" 

r^H* 


PARALLEL  FORCES  AND  THE  CENTRE  OF  GRAVITY.  1? 

orciinates  are  the  x,  y,  and  3,  just  obtained,  is  called  the  Centre 
of  Parallel  Forces,  and  its  position  is  independent  of  the  (com- 
mon) direction  of  the  forces  concerned. 

Example. — If  the  parallel  forces  are  contained  in  one  plane, 
and  the  axis  l^be  assumed  parallel  to  the  direction  of  the 
forces,  then  each  product  like  Plxl  will  be  a  moment,  as  de- 
fined in  §  21 ;  and  it  will  be  noticed  in  the  accompanying  nu- 
merical example,  Fig.  14,  that  a  detailed  substitution  in  the 
equation  P3  P  |Y  f^  p* 

having  regard  to  the  proper  sign  of  each 
force  and  .of  each  abscissa,  gives  the  same  ^FIG.  14. 

result  as  if  each  product  Px  were  first  obtained  numerically, 
and  a  sign  affixed  to  the  product  considered  as  a  moment 
about  the  point  0.  Let  Pl  =  —  1  lb.;  P3  =  +  2  Ibs.;  P3  = 
+  3  Ibs.;  P.  =  -  6  Ibs.;  x,  =  +  1  ft.;  x,  =  +  3  ft.;  x,  =  -  2 
ft.;  and  #4  =  —  1  ft.  Required  the  amount  and  position  of  the 
resultant  R.  In  amount  R  =  2P  =-1  +  2  +  3  —  6^  —  2 
Ibs.;  i.e.,  it  is  a  downward  force  of  2  Ibs.  As  to  its  position, 
Rx=  2(Px)  gives  ( -  2>  =  ( -  1)  X  (+  lj  +  2  X  3  + 
3  x  (-  2)  +  (-6)  X  (-  1)  =  -  1  +  6  -  6  +  6.  Now  from 
the  figure,  by  inspection,  it  is  evident  that  the  moment  of  Pl 
about  0  is  negative  (with  the  hands  of  a  watch),  and  is  numer- 
ically =  1,  i.e.,  its  moment  =  —  1  ;  similarly,  by  inspection, 
that  of  P9  is  seen  to  be  positive,  that  of  P3  negative,  that  of 
P4  positive;  which  agree  with  the  results  just  found,  that 
(—  2)  a?  =  —  1  +  6  —  6  +  6  =  +  5  ft.  Ibs.  (Since  a  moment 
is  a  product  of  a  force  (Ibs.)  by  a  length  (ft.),  it  may  be  called 
so  many  foot-pounds.)  Next,  solving  for  x,  we  obtain 
x  =  (+  5)  -r-  (—  2)  =  —  2.5  ft.;  i.e.,  the  resultant  of  the  given 
forces  is  a  downward  force  of  2  Ibs.  acting  in  a  vertical  line 
"2.5  ft.  to  the  left  of  the  origin.  Hence,  if  the  body  in  question 
be  a  horizontal  rod  whose  weight  has  been  already  included  in 
the  statement  of  forces,  a  support  placed  2.5  ft.  to  the  left  of 
0  and  capable  of  resisting  at  least  2  Ibs.  downward  pressure 
will  preserve  equilibrium ;  and  the  pressure  which  it  exerts 


18  MECHANICS    OF   ENGINEERING. 

against  the  rod  must  be  an  upward  force,  P^  of  2  Ibs.,  i  e.  the 
equal  and  opposite  of  the  resultant  of  P1?  P9,  P8,  P4. 

Fig.  15  shows  the  rod  as  a  free  body  in  equilibrium  under 

the  live  forces.     P5  =  +  2  Ibs.  =  the  reaction  of  the  support. 

I  i  Of  course  P5  is  one  of  a  pair  of  equal 

f  3     4  !      i1       2  and   opposite   forces  ;  the   other  one 

C—  —  j  is  the  pressure  of  the  rod  against  the 

F|t  —  2.S-  .....  io  support,  and  would  take  its  place  among 

FIG.  is.  the  forces  acting  on  the  support. 

23.  Centre  of  Gravity.  —  Among  the  forces  acting  on  any 
rigid  body  at  the  surface  of  the  earth  is  the  so-called  attraction 
of  the  latter  (i.e.,  gravitation),  as  shown  by  a  spring-balance, 
which  indicates  the  weight  of  the  body  hung  upon  it.  The 
weights  of  the  different  particles  of  any  rigid  body  constitute  a 
system,  of  parallel  forces  (practically  so,  though  actually  slightly 
convergent).  The  point  of  application  of  the  resultant  of  these 
forces  is  called  the  centre  of  gravity  of  the  body,  and  may  also 
be  considered  the  centre  of  mass,  the  body  being  of  very  small 
dimensions  compared  with  the  earth's  radius. 

If  a?,  y,  and  z  denote  the  co-ordinates  of  the  centre  of  gravity 
of  a  body  referred  to  three  co-ordinate  axes,  the  equations 
derived  for  them  in  §  22  are  directly  applicable,  with  slight 
changes  in  notation. 

Denote  the  weight  of  any  particle  *  of  the  body  by  dG>  its 
volume  by  dV,\)j  yita  heaviness  (rate  of  weight,  see  §  7)  and 
its  co-ordinates  by  aj,  y,  and  0;  then,  using  the  integral  sign  as 
indicating  a  summation  of  like  terms  for  all  the  particles  of  the 
body,  we  have,  for  heterogeneous  bodies, 

-_fyxdV.  -  _fyydV_.  -  _fr*dV.          m 

'  fydV'  y  "  fydV'        ~  fydV 

while,  if  the  body  is  homogeneous,  y  is  the  same  for  all  its  ele- 
ments, and  being  therefore  placed  outside  the  sign  of  summa- 
tion, is  cancelled  out,  leaving  for  homogeneous  bodies  (  V  de- 
noting the  total  volume) 

=J—~-  .    .    (2) 


*  Any  subdivision  of  the  body  may  be  adopted  for  use  of  equations  (1) 
and  (2),  etc.;  but  it  must  be  remembered  that  the  x  (or  y,  or  2)  in  each  term 
of  the  summations,  or  integrals,  is  the  co-ordinate  of  the  center  of  gravity  of 
the  subdivision  employed. 


PARALLEL    FORCES    AND    THE    CENTRE    OF    GRAVITY.    19 

Corollary. — It  is  also  evident  that  if  a  homogeneous  body  is 
for  convenience  considered  as  made  up  of  several  finite  parts, 
whose  volumes  are  Fn  Fa,  etc.,  and  whose  gravity  co-ordinates 
are  x»  y»  ^  ;  a?a,  ya,  z^ ;  etc.,  we  may  write 


+  v,  + 


(3) 


If  the  body  is  heterogeneous,  put  6r,  (weights),  etc.,  instead 
of  "Fj,  etc.,  in  equation  (3). 

If  the  body  is  an  infinitely  thin  homogeneous  shell  of  uni- 
form thickness  =  A,  then  d  V=  hdF(dF  denoting  an  element, 
and  F  the  whole  area  of  one  surface)  and  equations  (2)  become, 
after  cancellation, 

-    fxdF    -    fydF   -     fzdF 

/y,  -  •/.  _  .        n,   -  «/   «?  •        9    -  «/  _  /  A\ 

F  '  y-     p  >  •       p  •  •  •  •   w 

Similarly,  for  a  homogeneous  wire  of  constant  small  cross- 
section  (i.e..  a  geometrical  line,  having  weight),  its  length 
being  s,  and  an  element  of  length  ds,  we  obtain 


It  is  often  convenient  to  find  the  centre  of  gravity  of  a  thin 
plate  by  experiment,  balancing  it  on  a  needle-point;  other 
shapes  are  not  so  easily  dealt  with. 

24.  Symmetry.  —  Considerations  of  symmetry  of  form  often 
determine  the  centre  of  gravity  of  homogeneous  solids  without 
analysis,  or  limit  it  to  a  certain  line  or  plane.  Thus  the  centre 
of  gravity  of  a  sphere,  or  any  regular  polyedron,  is  at  its  centre 
of  figure  ;  of  a  right  cylinder,  in  the  middle  of  its  axis  ;  of  a 
thin  plate  of  the  form  of  a  circle  or  regular  polygon,  in  the 
centre  of  figure  ;  of  a  straight  wire  of  uniform  cross-section,  in 
the  middle  of  its  length. 

Again,  if  a  homogeneous  body  is  symmetrical  about  a  plane, 
the  centre  of  gravity  must  lie  in  that  plane,  called  a  plane  of 


20  MECHANICS    OF   ENGINEERING. 

gravity ;  if  about  a  line,  in  that  line  called  a  line  of  gravity  ; 
if  about  a  point,  in  that  point. 

25.  By  considering  certain  modes  of  subdivision  of  a  homo- 
geneous body,  lines  or  planes  of  gravity  are  often  made  appar- 
ent.    E.g.,  a  line  joining  the  middle  of  the  bases  of  a  trape- 
zoidal plate  is  a  line  of  gravity,  since  it  bisects  all  the  strips 
of   uniform  width   determined  by  drawing   parallels   to   the 
bases;  similarly,  a  line  joining  the  apex  of  a  triangular  plate  to 
the  middle  of  the  opposite  side  is  a  line  of  gravity.     Other 
cases  can  easily  be  suggested  by  the  student. 

26.  Problems. — (1)  Kequired  the  position  of  the  centre  of 

gravity  of  a  fine  homogeneous  wire  of  the 
l~dyformofa'  circular  arc,  AB,  Fig.  16.  Take 
the  origin  0  at  the  centre  of  the  circle,  and 
the  axis  X  bisecting  the  wire.  Let  the 
length  of  the  wire,  s,  —  2«t ;  ds  =  ele- 
ment of  arc.  We  need  determine  only  the 
x,  since  evidently  y  =  0.  Equations  (5), 

//vayy  o 

§  23,  are  applicable  here,  i.e.,  x  = . 

s 

From  similar  triangles  we  have 

rdy 
ds  :  dy  ::  r  :  a?;  /.  as  =  -  f--9 


X 
+  a 


—        r         -          ra 
.-.  x  =  —  /  dy  —  TT— ,  i.e.,  =   chord  X  radius  ~  length  of 

^Slt/y  -  _  a        ZSl 

wire.     For  a  semicircular  wire,  this  reduces  to  x  =  %r  -4-  n. 

,  PROBLEM  2.  Centre  of  gravity  of  trapezoidal  (and  trian- 
gular] thin  plates,  homogeneous,  etc. — Prolong  the  non-parallel 
sides  of  the  trapezoid  to  intersect  at  0,  which  take  as  an  origin, 
making  the  axis  X  perpendicular  to  the  bases  b  and  £>,.  We 
may  here  use  equations  (4),  §  23,  and  may  take  a  vertical  strip 
for  our  element  of  area,  dF,  in  determining  a?;  for  each  point 
of  such  a  strip  has  the  same  x.  Now  dF  —  (y  -f-  y')dx,  and 


PARALLEL   FORCES   AND   THE   CENTRE   OF   GRAVITY.   21 

from  similar  triangles  y  +  y'  =  j  x.  Now  F,  —  -(bh  —  bji^ 

can  be  written  —  ,   (A2  —  Aj8),  and  x  =  -—= —  becomes 
—  fii  _f 

I — 7»          /»/  — II?  C\       7  3  I 

=  LAt/h       ;. .  j,- :  2.  A  .^  "~  ^  =  3  A2  -  A/ 

for  the  trapezoid. 

For  a  triangle  A,  =  0,  and  we  have  x=   gA;  that  is,  the 

centre  of  gravity  of  a  triangle  is  one  third  the  altitude  from  the 
base.     The  centre  of  gravity  is  finally  determined  by  knowing 


FIG.  17. 


FIG.  18. 


that  a  line  joining  the  middles  of  b  and  5,  is  a  line  of  gravity; 
or  joining  O  and  the  middle  of  b  in  the  case  of  a  triangle. 

PROBLEM  3.  Sector  of  a  circle.  Thin  plate,  etc.  —  Let  the 
notation,  axes,  etc.,  be  as  in  Fig.  18.  Angle  of  sector  =  2ar; 
x  =  ?  Using  polar  co-ordinates,  the  element  of  area  dF  (a 
small  rectangle)  =  pdcp  .  dp,  and  its  x  —  p  cos  (p  ;  hence  the 
total  area  = 


i.e.,  F=  r*a.     From  equations  (4),  §  23,  we  have 


x  = 


= i/yv* 


cos  9 


22  MECHANICS   OF   ENGINEERING. 

(Note  on  double  integration. — The  quantity 
cos  cp  J  p*  dp  \dcp, 

is  that  portion  of  the  summation    /   /  cos  cpp*dpdcp  which 

belongs  to  a  single  elementary  sector  (triangle),  since  all  its 
elements  (rectangles),  from  centre  to  circumference,  have  the 
same  cp  and  dcp.) 
That  is, 

-  1    r3  /*+*  r*    F+a  2    r  sin  a 

*  =  ^-3  7_a  cos  cpdcp  =  ^-a  [sin  p  =  3--^— ; 

—       4  r  sin  £  ft 
or,  putting  p  =  2a  =  total  angle  of  sector,  x  =  -^ -g — • 

-       4r 
For  a  semicircular  plate  this  reduces  to  x  =  •$—. 

O7t 

[Note. — In  numerical  substitution  the  arcs  a  and  ft  used 
above  (unless  sin  or  cos  is  prefixed)  are  understood  to  be  ex- 
pressed in  circular  measure  (ar-measure) ;  e.g.,  for  a  quad- 
rant, ft  =  |  =  1.5707*+ ;  for  30°,  ft  =  ^ ;  or,  in  general,  if  ft 

180°  n  ~\ 

in  degrees  = ,  then  p  in  ^r-measure  =  -. 

\  PROBLEM  4.  Sector  of  a  flat  ring ;  thin 

\          plate,  etc. — Treatment  similar  to  that  of 
Problem  3,  the  difference  being  that  the 

C, 

\.     Result, 


ha. 

instead  of 


Fie.  19. 

-       4     r*  —  r,3     sin  £/? 
x  =  -. 


*  "Radians." 


PARALLEL  FORCES  AND  THE  CENTRE  OF  GRAVITY.  23 

PROBLEM  5.  —  Segment  of  a  circle;  thin  plate,  etc.  —  Fig.  20, 
Since  each  rectangular  element  of  any  ver- 
tical strip  has  the  same  #,  we  may  take  the 
strip  as  dF  in  finding  a?,  and   use  y  as  the 
half-height  of  the  strip.     dF  —  %ydx,  and  0 
from  similar  triangles  x  :  y  :  :  (—  dy)  :  dx, 
i.e.,  xdx  =  —  ydy.      Hence  from   eq.  (4), 
§23, 


_  —  Zjay^dy  _  _2_  R         3,  __  2    a\ 


^ 


but  a  =  the  half-chord,  hence,  finally,  x  = 

PROBLEM  6.  —  Trapezoid  ;  thin  plate,  etc., 
by  the  method  in  the  corollary  of  §  23  ;  equa- 
tions (3).     Eequired  the  distance  x  from  the 
base  AB.    Join  DB,  thus  dividing  the  trape- 
zoid  ABCD  into  two  triangles  ADB  =  F, 
and  DEC  '  =  FV  whose  gravity  a?'s  are,  re- 
spectively,  x,  =  -JA  and  #2  =  |A.     Also,  F, 
and  T^7  (area   of   trape- 
.     Eq.  (3)  of  §23  gives 
hence,  substituting,^,  + 


zoid)  = 
Fx=  Flxl 


Fi«.  21. 


=  Ift^  + 


~  _  ^    (fti  +    2  J,] 


The  line  joining  the  middles  of  5j  and  J3  is  a  line  of  gravity,  and 
is  divided  in  such  a  ratio  by  the  centre  of  gravity  that  the  fol- 
lowing construction  for  finding  the  latter  holds  good  :  Prolong 
each  base,  in  opposite  directions,  an  amount  equal  to  the  other 
base;  join  the  two  points  thus  found:  the  intersection  with 
the  other  line  of  gravity  is  the  centre  of  gravity  of  the  trape- 
zoid.  Thus,  Fig/ 21,  with  BE  ^  bt  and*  DF=  £„  join  FE, 
etc. 


24  MECHANICS   OF   ENGINEERING. 

PROBLEM  Y.  Homogeneous  oblique  cone  or  pyramid. — • 
Take  the  origin  at  the  vertex,  and  the  axis  X  perpendicular  to 
the  base  (or  bases,  if  a  frustum).  In  finding  x  we  may  put 
d  V  =  volume  of  any  lamina  parallel  to  YZ<  F  being  the  base 
of  such  a  lamina,  each  point  of  the  lamina  having  the  same  x. 
Hence,  (equations  (2),  §  23), 


V=fdV=/Fdx; 

r 

but 

_^; 

and 


q       14    74 

For  a  frustum,  a?  =  — .  -^ A ;  while  for  a  pyramid,  A,,  be- 

4    Aa     —  Aj 

Q 

ing  =  0,  x  =  jA.     Hence  the  centre  of  gravity  of  a  pyramid 

is  one  fourth  the  altitude  from  the  base.  It  also  lies  in  the  line 
joining  the  vertex  to  the  centre  of  gravity 
of  the  base. 

PROBLEM  8. — If  the  heaviness  of  the  ma- 
terial of  the  atjove  cone  or  pyramid  varied 
directly  as  a?,  y^  being  its  heaviness  at  the 
Flo.  22.  base  F»  we  would  use  equations  (1),  §  23, 

*¥ 

putting  y  =  j*  x ;  and  finally,  for  the  frustum, 
fit 

_      4    A,6 -A,6 

x  =  =-. 


5  •  A,4 -A,4' 
and  for  a  complete  cone  a?  =  -?  At. 

27.  The  Centrobarie  Method.— If  an  elementary  area  dF  be 
revolved  about  an  axis  in  its  plane,  through  an  angle  a  <  2,T, 


PARALLEL   FORCES   AND   THE   CENTRE   OF   GRAVITY.    25 


Fio.  23. 


the  distance  from  the  axis  being  =  x,  the  volume  generated  is 
dV  =  axdF,  and  the  total  volume  generated  by  all  the 
of  a  finite  plane  figure  whose  plane  con- 
tains the  axis  and  which  lies  entirely  on  one 
side  of  the  axis,  will  be  V  =  fd  V  = 
afxdF.  Bat  from  §23,  otfxdF=  aFw\ 
ax  being  the  length  of  path  described  by 
the  centre  of  gravity  of  the  plane  figure, 
we  may  write  :  The  volume  of  a  solid  of  revolution  generated 
~by  a  plane  figure,  lying  on  one  side  of  the  axis,  equals  the 
area  of  the  figure  multiplied  ~by  the  length  of  curve  described 
by  the  centre  of  gravity  of  the  figure. 

A  corresponding  statement  may  be  made  for  the  surface 
generated  by  the  revolution  of  a  line.  The  arc  a.  must  be  ex- 
pressed in  n  measure  in  numerical  work. 

27a.  Centre  of  Gravity  of  any  duadrilateral.— Fig.  230. 
Construction'  ABGD  being  any  quad- 
rilateral. Draw  the  diagonals.  On  the 
long  segment  DK  of  DB  lay  off  DE  = 
BK,  the  shorter,  to  determine  E\  simi- 
larly, determine  N  on  the  other  diagonal, 
by  making  GN  =  AK.  Bisect  EK  in  H 
and  KN  \\\  M.  The  intersection  of  EM 
and  Nil  is  the  centre  of  gravity,  C. 
Proof.— H  being  the  middle  of  DB,  and  AH  and  EG 
having  been  joined,  /  the  centre  of  gravity  of  the  triangle 
ABD  is  found  on  AH,  by  making  HI  =  \AH\  similarly,  by 
making  HL  =  %HG,  L  is  the  centre  of  gravity  of  triangle 
BDG.  . ' .  IL  is  parallel  to  AG  and  is  a  gravity-line  of  the 
whole  figure ;  and  the  centre  of  gravity  C  may  be  found  on  it 
if  we  can  make  CL  :  CI  ::  area  ABD  :  area  BDG  (§  21). 
But  since  these  triangles  have  a  common  base  DB,  their  areas 
are  proportional  to  the  slant  heights  (equally  inclined  to  DB) 
AK  and  KG,  i.e.,  to  GN  and  NA.  Hence  HN,  which  di- 
vides IL  in  the  required  ratio,  contains  C,  and  is  .'.  a  gravity- 
line.  By  similar  reasoning,  using  the  other  diagonal,  AG,  and 


FIG.  23«. 


26  MECHANICS   VF   ENGINEERING. 

the  two  triangles  into  which  it  divides  the  whole  figure,  we 
may  prove  EM.  to  be  a  gravity-line  also,  llence  the  construc- 
tion is  proved. 

27b.  EXAMPLES.-  —  1.  Required  the  volume  of  a  sphere  by 
the  centrobaric  method. 

A  sphere  may  be  generated  by  a  semicircle  revolving  about 
its  diameter  through  an  arc  a  =  %TT.  The  length  of  the  path 

4/* 

described  by  its  centre  of  gravity  is  =  2?r  ^—  (see  Prob.  3,  § 
26),  while  the  area  of  the  semicircle  is  %?rr\  Hence  by  §  27, 

4r  4: 

Volume  generated  —  %n  .  77-  .  %7tr*  =  ^  nr*. 

O7t  O 

2.  Required  the  position  of  the  centre  of  gravity  of  the  sector 
of  a  flat  ring  in  which  rl  =  21  feet,  r^  =  20  feet,  and  ft  —  80° 
(see  Fig.  19,  and  §  26,  Prob.  4). 

ft 
sin  —  =  sin  40°  =  0.64279,  and  /3  in  circular  measure  = 

a 

rrrp:  7t  =  —  n  =  1.3962.*  By  using  rl  and  r^  in  feet,  x  will  be 
lou  y 


.    ft 

fin2         4    12610.64279 


obtained  in  feet. 


*"  Radians.' 


STATICS    OF   A    U1G1D    BODY.  27 


CHAPTER  III. 

STATICS  OF  A  RIGID  BODY. 

28.  Couples. — On  account  of   the  peculiar  properties  and 
utility  of  a  system  of  two  equal  forces  acting  in  parallel  lines 
and  in  opposite  directions,  it  is  specially 

considered,  and  called  a  Couple.  The 
arm  of  a  couple  is  the  perpendicular 
distance  between  the  forces ;  its  moment, 
the  product  of  this  arm,  by  one  of  the 
forces.  The  axis  of  a  couple  is  an 
imaginary  line  drawn  perpendicular  to 
its  plane  on  that  side  from  which  the  rotation  appears  positive 
(against  the  hands  of  a  watch).  (An  ideal  rotation  is  meant, 
suggested  by  the  position  of  the  arrows ;  any  actual  rotation 
of  the  rigid  body  is  a  subject  for  future  consideration.)  In 
dealing  with  two  or  more  couples  the  lengths  of  their  axes  are 
made  proportional  to  their  moments;  in  fact,  by  selecting  a 
proper  scale,  numerically  equal  to  these  moments.  E.g.,  in  Fig. 
24,  the  moments  of  the  two  couples  there  shown  are  Pa  and 
Qb',  their  axes  p  and  q  so  laid  oil  that  Pa  :  Qb  '.'. p  \  q,  and 
that  the  ideal  rotation  may  appear  positive,  viewed  from  the 
outer  end  of  the  axis. 

29.  No  single  foree  can  balance  a  couple. — For  suppose  the 
couple  P,  P,  could  be  balanced  by  a  force  JR',  then  this,  acting 

Pf  at  some  point  C,  ought  to  hold  the  couple 

OJ.~ &*<-£• £--•  in  equilibrium.  Draw  CO  through  0,  the 

JT  7p  f}j  centre  of  symmetry  of  the  couple,  and 

FI«.  as.  make  OD  =  OC.  At  D  put  in  two  op- 

posite and  equal  forces,  S  and  T,  equal  and  parallel  to  Rf. 
The  supposed  equilibrium  is  undisturbed.  But  if  R',  Pr  and 


28  MECHANICS   OF   ENGINEERING. 

P  are  in  equilibrium,  so  ought  (by  symmetry  about  0)  S,  P, 
and  P  to  be  in  equilibrium,  and  they  may  be  removed  without 
disturbing  equilibrium.  But  we  have  left  T&ud  R' ,  which  are 
evidently  not  in  equilibrium;  /.  the  proposition  is  proved  by 
this  reductio  ad  absurdum.  Conversely  a  couple  has  no  single 
resultant. 


30.  A  couple  may  he  transferred  anywhere  in  its  own  plane. 
— First,  it  may  be  turned   through  any  angle  a,  about  any 

pi  point  of  its  arm,  or  of  its  arm   produced. 

G^r  •• J£—  Let  (P,  P')be  a  couple,  G  any  point  of  its 

fp'    arm  (produced),  and  a  any  angle.      Make 
_  .       v  ^  GC  =  GA,  CD  =  AB,  and  put  in  at   <7, 

^    \  2         \(  |       P1  and  P3  equal  to  P  (or  P'\  opposite  to 

0\    PI --:-j\''7  each  other  and  perpendicular  to  GC\  and 

:'  R   Ps  and  PI  similarly  at  D.     Now  apply  and 

FIG.  26.  combine  P  and  Pl  at  6>,  P'  and  P4  at  0'; 

then  evidently^  and  Rr  neutralize  each  other,  leaving  P3  and 
P,  equivalent  to  the  original  couple  (P,  P'}.  The  arm' 
CD  =  AB.  Secondly,  if  G  be  at  infinity,  and  a  =  0,  the 
same  proof  applies,  i.e.,  a  couple  may  be  moved  parallel  to 
itself  in  its  own  plane.  Therefore,  by  a  combination  of  the 
two  transferrals,  the  proposition  is  established  for  any  trans- 
ferral  in  the  plane. 

31.  A  couple  may  ~be  replaced  ~by  another  of  equal  moment 
in  a  parallel  plane. — Let  (P,  P')  be  a  couple.     Let  CD,  in  a 
parallel  plane,  be  parallel  to  AB.     At  D  put  in  a  pair  of  equal 

and  opposite  forces,  Ss  and  #4,  parallel  to  P  and  each  =  =P. 

ED 


Similarly  at  (7,  Sl  and  x9,  parallel  to  P  and   each  =  ==P. 
But,  from  similar  triangles, 

AE  _  BE 
ED  ~~  ' 


STATICS    OF    A    RIGID    BODY.  29 

[NOTE. — The  above  values  are  so  chosen  that  the  intersection  point  E 
may  be  the  point  of  application  of  (P  -j-  £„),  the  resultant  of  P  and  &,; 
and  also  of  (P-j-/S3),  the  resultant  of  Pand  S9,  as  follows  from  §  21;  thus 
(Fig.  28),  R,  the  resultant  of  the  two  parallel  forces  Pand  #,,  is  =  P-+-S3, 
and  its  moment  about  any  centre  of  moments,  as  E,  its  own  point  of  ap- 
plication, should  equal  the  (algebraic)  sum  of  the  moments  of  its  com- 

AW 

ponents  about  E\  i.e.,  R  X  zero  =  P .  AE  —  S3 .  DE\  .-.£,  =  =  .  P.] 

DE 


R4 

•• ! i 


FIG.  27.  FIG.  28. 


Replacing  P'  and  £,  by  (Pf  +  £„),  and  P  and  S9  by 
(P  -)-  $,),  the  latter  resultants  cancel  each  other  at  .Z^,  leaving 
the  couple  (&„  S4)  with  an  arm  CD,  equivalent  to  the  original 


couple  P,  Px  with  an  arm  AB.     But,  since  8^  =  =  .  P  = 


j    D  ^ 

^=^  .  P,  we  have  Sl  X  S5  =  PyjAB\  that  is,  their  moments 


are  equal. 

32.  Transferral  and  Transformation  of  Couples.  —  In  view  of 
the  foregoing,  we  may  state,  in  general,  that  a  couple  acting  on 
a  rigid  body  may  be  transferred  to  any  position  in  any  parallel 
plane,  and  may  have  the  values  of  its  forces  and  arm  changed 
in  any  way  so  long  as  its  moment  is  kept  unchanged,  and  still 
have  the  same  effect  on  the  rigid  body  (as  to  rest  or  motion, 
not  in  distorting  it). 

Corollaries.  —  A  couple  may  be  replaced  by  another  in  any 
position  so  long  as  their  axes  are  equal  and  parallel  and  simi- 
larly situated  with  respect  to  their  planes. 

A  couple  can  be  balanced  only  by  another  couple  whose  axis 
is  equal  and  parallel  to  that  of  the  first,  and  dissimilarly  situ- 
ated. For  example,  Fig.  29,  Pa  being  =  Qb,  the  rigid  body 
AB  (here  supposed  without  weight)  is  in  equilibrium  in  each 


30 


MECHANICS   OF   ENGINEERING. 


case  shown.  By  "reduction  of  a  couple  to  a  certain  arm  a" 
is  meant  that  for  the  original  couple  whose  arm  is  a' ',  with 
forces  each  =  P',  a  new  couple  is  substituted  whose  arm  shall 
be  =  a,  and  the  value  of  whose  forces  P  and  P  must  be  com- 
puted from  the  condition 

Pa  =  PV,      i.e.,      P  =  P'a'  -5-  a. 


FIG.  29. 


FIG.  30. 


33.  Composition  of  Couples.— Let  (P,  P')  and  (Q,  Qf]  be  two 
couples  in  different  planes  reduced  to  the  same  arm  AB  =  a, 
which  is  a  portion  of  the  line  of  intersection  of  their  planes. 
That  is,  whatever  the  original  values  of  the  individual  forces 
and  arms  of  the  two  couples  were,  they  have  been  transferred 
and  replaced  in  accordance  with  §  32,  so  that  P.  AB,  the 
moment  of  the  first  couple,  and  the  direction  of  its  axis, p, 
have  remained  unchanged ;  similarly  for  the  other  couple. 
Combining  P  with  Q  and  P'  with  Q',  we  have  a  resultant 
couple  (R,  Rf)  whose  arm  is  also  AB.  The  axes  p  and  q  of 
the  component  couples  are  proportional  to  P .  AB  and  Q .  AB, 
i.e.,  to  P  and  Q,  and.  contain  the  same  angle  as  P  and  Q. 
Therefore  the  parallelogram  p .  .  .  q  is  similar  to  the  parallelo- 
gram P  . . .  Q ;  whence  p  \  q  \  r : :  P  :  Q  \  R,  or  p  :  q  :  r : : 
Pa  :  Qa  :  Ra.  Also  r  is  evidently  perpendicular  to  the  plane 
of  the  resultant  couple  (R,  R'\  whose  moment  is  Ra.  Hence 
r,  the  diagonal  of  the  parallelogram  on  p  and  q,  is  the  axis  of 
the  resultant  couple.  To  combine  two  couples,  therefore,  we 
have  only  to  combine  their  axes,  as  if  they  were  forces,  by  a 
parallelogram,  the  diagonal  being  the  axis  of  the  resultant 
couple;  the  plane  of  this  couple  will  be  perpendicular  to  the 


STATICS    OF   A    KIGID    BODY.  31 

axis  just  found,  and  its  moment  bears  the  same  relation  to  the 
moments  of  the  component  couples  as  the  diagonal  axis  to  the 
two  component  axes.  Thus,  if  two  couples,  of  moments  POL 
and  Qb,  lie  in  planes  perpendicular  to  each  other,  their  result- 
ant couple  has  a  moment  EG  =  \/(paf  -\-  (Qb)''- 

If  three  couples  in  different  planes  are  to  be  combined,  the 
axis  of  their  resultant  couple  is  the  diagonal  of  the  parallele- 
piped formed  on  the  axes,  laid  off  to  the  same  scale  &\\&  point- 
ing in  the  proper  directions,  the  proper  direction  of  an  axis 
being  away  from  the  plane  of  its  couple,  on  the  side  from 
which  the  couple  appears  of  positive  rotation. 

34.  If  several  couples  lie  in  the  same  plane  their  axes  are 
parallel  and  the  axis  of  the  resultant  couple  is  their  algebraic 
sum  ;  and  a  similar  relation  holds  for  the  moments  :  thus,  in 
Fig.  24,  the  resultant  of  the  two  couples  has  a  moment  =  Qb 
—  Pa,  which  shows  us  that  a  convenient  way  of  combining 
couples,  when  all  in  one  plane,  is  to  call  the  moments  positive 
or  negative,  according  as  the  ideal  rotations  are  against,  or  with, 
the  hands  of  a  watch,  as  seen  from  the  same  side  of  the  plane  ; 
the  sign  of  the  algebraic  sum  will  then  show  the  ideal  rotation 
of  the  resultant  couple. 

35.  Composition  of  Non-concurrent  Forces  in  a  Plane.  —  Let 
PV  P<»  etc.,  be  the  forces  of  the  system  ;  &„  ylt  a?,,  ya,  etc.,  the 


x, 


FIG.  31. 

co-ordinates  of  their  points  of  application ;  and  <*„  <*„  .  .  .  etc., 
their  angles  with  the  axis  X.  Replace  Pl  by  its  components 
X^  and  T"n  parallel  to  the  arbitrary  axes  of  reference.  At  the 
origin  put  in  two  forces,  opposite  to  each  other  and  equal  and 
parallel  to  X^ ;  similarly  for  Yr  (Of  course  Xl  =  Pl  cos  a  and 
Yl  =  Pl  sin  a.)  We  now  have  P1  replaced  by  two  forces  X^ 


32  MECHANICS   OF   ENGINEERING. 

and  Yl  at  the  origin,  and  two  couples,  in  the  same  plane,  whose 
moments  are  respectively  —  Xlyl  and  +  Ytx»  and  are  there- 
fore (§34)  equivalent  to  a  single  couple,  in  the  same  plane  with 
a  moment  =  (Y.x,—  X,y:). 

Treating  all  the  remaining  forces  in  the  same  way,  the  whole 
system  of  forces  is  replaced  by 

the  force  2(Z)  —X^  +  X,  +  ...  at  the  origin,  along  the  axis  X; 
the  force  2(  Y)  =  Y,+  Y*  +  •  •  •  at  the  origin,  along  the  axis  Y\ 

and  the  couple  whose  mom.  G  =  2  (  Yx  —  Xy\  which  may  be 
called  the  couple  G  (see  Fig.  32),  and  may  be  placed  anywhere 
in  the  plane.  Now  2(X)  and  2(  Y)  may  be  combined  into  a 
force  JK  ;  i.e., 


23  =  4/(^X)2  +  2  YJ1  and  its  direction-cosine  is  cos  a  =  —  T 

Since,  then,  the  whole  system  reduces   to  C  and  It,  we  must 

have  for  equilibrium  R  =  0,  and  G  —  0  ;   i.e.,  for  equilibrium 

2X  =  0,  2Y=  0,  and  2(Yx-Xy)  =  0.    .  eq.  (1) 

If  R  alone  =  0,  the  system  reduces  to  a  couple  whose  mo- 
ment is  G  =  2(  Yx  —  Xy)  ;  and  if  G  alone  =  0  the  system  re- 
duces to  a  single  force  R,  applied  at  the  origin.  If,  in  general, 
neither  It  nor  G  =  0,  the  system  is  still  equivalent  to  a  single 
force,  but  not  applied  at  the  origin  (as  could  hardly  be  ex- 
pected, since  the  origin  is  arbitrary)  ;  as  follows  (see  Fig.  33)  : 

Replace  the  couple  O  by  one  of  equal  moment,  G,  with  each 

force  =  R.     Its  arm  will  therefore  be  -p.    Move  this  couple 

in  the  plane  so  that  one  of  its  forces  R  may  cancel  the  R  al- 
ready at  the  origin,  thus  leaving  a  single  resultant  R  for  the 
whole  system,  applied  in  a  line  at  a  perpendicular  distance, 

=  -r>  ,  from  the  origin,  and  making  an  angle  a  whose  cosine  = 


o  = 

axis  X. 


D 

36.  More  convenient  form  for  the  equations  of  equilibrium 
of  non-concurrent  forces  in  a  plane.  —  In  (I.),  Fig.  34,  0  being 


STATICS    OF    A    RIGID    BODY.  33 

any  point  and  a  its  perpendicular  distance  from  a  force  P; 
put  in  at  0  two  equal  and  opposite  forces  P  and  P'  =  and  || 
to  P,  and  we  have  P  replaced  by  an  equal  single  force  P'  at 
O,  and  a  couple  whose  moment  is  -f-  Pa.  (II.)  shows  a  simi- 
lar construction,  dealing  with  the  JTand  T" components  of  P, 
to  that  in  (II.)  P  is  replaced  by  single  forces  X'  and  Y'  at  0 


.-  /I:..,B 


FIG.  33.  FIG. 


(and  they  are  equivalent  to  a  resultant  P',  at  <9,  as  in  (I.),  and 
two  couples  whose  moments  are  +  Yx  and  —  Xy. 

Hence,  0  being  the  same  point  in  both  cases,  the  couple  Pa 
is  equivalent  to  the  two  last  mentioned,  and,  their  axes  being 
parallel,  we  must  have  Pa  =  Yx  —  Xy  Equations  (1), 
§  35,  for  equilibrium,  may  now  be  written 

2X  -  0,  2  Y  =  0,  and  2(Pa)  =  0.     .     .     (2) 

In  problems  involving  the  equilibrium  of  non-concurrent 
forces  in  a  plane,  we  have  three  independent  conditions,  or 
equations,  and  can  determine  at  most  three  unknown  quantities. 
For  practical  solution,  then,  the  rigid  body  having  been  made 
free  (by  conceiving  the  actions  of  all  other  bodies  as  repre- 
sented by  forces),  and  being  in  equilibrium  (which  it  must  be 
if  at  rest),  we  apply  equations  (2)  literally  ;  i.e.,  assuming  an 
origin  and  two  axes,  equate  the  sum  of  the  X  components  of 
all  the  forces  to  zero;  similarly  for  the  ^components;  and 
then  for  the  "moment-equation,"  having  dropped  a  perpen- 
dicular from  the  origin  upon  each  force,  write  the  algebraic 
sum  of  the  products  (moments)  obtained  by  multiplying  each 
force  by  its  perpendicular,  or  "  lever-arm"  equal  to  zero,  call- 
ing each  product  -f-  or  —  according  as  the  ideal  rotation  ap- 
pears against,  or  with,  the  hands  of  a  watch,  as  seen  from  the 
same  side  of  the  plane.  (The  converse  convention  would  do  as 
well.) 


34  MECHANICS    OF   ENGINEERING. 

Sometimes  it  is  convenient  to  use  three  moment  equations, 
taking  a  new  origin  each  time,  and  then  the  2£X '=  0  and  2  Y 
=  0  are  superfluous,  as  they  would  not  be  independent  equa- 
tions. 

37.  Problems  involving  Non-concurrent  Forces  in  a  Plane.— 

Remarks.  The  weight  of  a  rigid  body  is  a  vertical  force 
through  its  centre  of  gravity,  downwards. 

If  the  surface  of  contact  of  two  bodies  is  smooth  the  action 
(pressure,  or  force)  of  one  on  the  other  is  perpendicular  to  the 
surface  at  the  point  of  contact.  If  a  cord  must  be  imagined 
cut,  to  make  a  body  free,  its  tension  must  be  inserted  in  the 
line  of  the  cord,  and  in  such  a  direction  as  to  keep  taut  the 
small  portion  still  fastened  to  the  body.  In  case  the  pin  of 
a  hinge  must  be  removed,  to  make  the  body  free,  its  pressure 
against  the  ring  being  unknown  in  direction  and  amount,  it  is 
most  convenient  to  represent  it  by  its  unknown  components  X 
and  Y9  in  known  directions.  In  the  following  problems  there 
is  supposed  to  be  no  friction.  If  the  line  of  action  of  an  un- 
known force  is  known,  but  not  its  direction  (forward  or  back 
ward),  assume  a  direction  for  it  and  adhere  to  it  in  all  the  three 
equations,  and  if  the  assumption  is  correct  the  value  of  the 
force,  after  elimination,  will  be  positive  ;  if  incorrect,  negative. 
Problem  1. — Fig.  35.  Given  an  oblique  rigid  rod,  with  two 
loads  6^  (its  own  weight)  and  #, ;  required  the  reaction  of  the 
smooth  vertical  wall  at  A,  and  the  direction  and  amount  of  the 
np  Am<7<2- pressure  at  0.  The  reaction  at  A 
'—I  must  be  horizontal  ;  call  it  X'.  The  pres- 
sure  at  0,  being  unknown  in  direction,  will 
have  both  its  X  and  Y  components  un- 
known. The  three  unknowns,  then,  are 
X0,  X',  and  Y0,  while  6^,  £a,  «,,  «„  and 
A  are  known.  The  figure  shows  the  rod 
FI&.  85.  ag  a  j?ree  ijojiy^  a]i  the  forces  acting  on  it 

have  been  put  in,  and,  since  the  rod  is  at  rest,  constitute  a  sys- 
tem of  non-concurrent  forces  in  a  plane,  ready  for  the  condi- 
tions of  equilibrium.  Taking  origin  and  axes  as  in  the  figure. 


STATICS   OF   A    RIGID   BODY. 


35 


=  0  gives  +X0  -  X'  =  0  ;  2Y=  0  gives  +  Y0  -  Gk 
-  £a  =  0  ;  while  2(Pa)  =  0,  about  0,  gives  +  X'h  — 
Glal  —  <?X  =  0.  (The  moments  of  X0  and  Y0  about  0 
are,  each,  =  zero.)  By  elimination  we  obtain  Y9  =  6rl  -{- 
#3;  X0  =  X'  ==C  [<?,«,  +  6*X]  -T-  A;  while  the  pressure  at 
0  = 


T02  -f-  J7*,  and  makes  with  the  horizontal  an  angle 
whose  tan  =  Y0  -~  X^ 

[N.B.  A  special  solution  for  this  problem  consists  in  this,  that  the  result- 
ant of  the  two  known  forces  GI  and  O-t  intersects  the  line  of  X'  in  a  point 
which  is  easily  found  by  §  21.  The  hinge-pressure  must  pass  through  this 
point,  since  three  forces  in  equilibrium  must  be  concurrent.] 

We  might  vary  this  problem  by  limiting  X1  to  a  safe  value, 
depending  on  the  stability  of  the  wall,  and  making  h  an  un- 
known. The  three  unknowns  would  then  be  JT0,  Y^  and  A. 

Problem  2.  —  Given  two  rods  with  loads,  three  hinges  (or 
"pin-joints"),  and  all  dimensions:  required  the  three  hinge- 


Fio.  36. 


FIG.  87. 


pressures;  i.e.,  there  are  six  unknowns,  viz.,  three  .5Tand  three 
Y  components.  We  obtain  three  equations  from  each  of  the 
two  free  bodies  in  Fig.  37.  The  student  may  fill  out  the  de- 
tails. Notice  the  application  of  the  principle  of  action  arid 
reaction  at  B  (see  §  3). 

Problem  3. — A  Warren  bridge-truss  rests  on  the  horizontal 
smooth  abutment-surfaces  in  Fig.  38.  It  is  composed  of  equal- 
isosceles  triangles  ;  no  piece  is 
continuous  beyond  a  joint,  each 
of  which  is  &pin  connection.  All 
loads  are  considered  as  acting  at 
the  joints,  so  that  each  piece  will 
be  subjected  to  a  simple  tension 
or  compression. 


Fio.  38- 


36  MECHANICS   OF   ENGINEERING. 

First,  required  the  reactions  of   the  supports  Vl  and  Fa 
these  and  the  loads  are  called  the  external  forces. 
about  0  =  0  gives  (the  whole  truss  is  the  free  body) 


Pt  .  4«  —  P,  .  f*  —  P,  .  f*  =  0; 
while  2(Pa)  about  K  =  0  gives 

-  V,  .  Za  +  P,  .  4*  +  P$a  +  P,fa  =  0; 
.'.  V,  =  i[5Pt  +  3Pa  +  PJ  ; 
and  Fa  -  4[P,  +  3Pa  +  5PJ. 


Secondly,  required  the  stress  (thrust  or  pull,  compression  or 
tension)  in  each  of  the  pieces  A,  B,  and  (7cut  by  the  imaginary 
line  DE.     The  stresses  in  the  pieces  are  called  'W^m^/  forces. 
These  appear  in  a  system  of  forces  acting  on  a  free  body  only 
when  a  portion  of  the  truss  or  frame  is  conceived  separated 
from  t1         -Tiainder  in  such  a  way  as  to  expose  an  internal 
of  one    r  more  pieces.     Consider  as  a  free  body  the  por- 
Jn  the  lei  c  of  DE  (that  on  the  right  would  serve  as  well, 
ip  p  but  the  pulls  or  thrusts  in  A,  B,  and 

A  C  would  be  found  to  act  in  directions 
opposite  to  those  they  have  on  the 
other  portion  ;  see  §3).  Fig.  39.  The 
arrows  (forces)  A,  B,  and  C  are  not 
pointed  yet.  They,  with  F15  Plt  and 
Pa,  form  a  system  in  equilibrium. 
2(P<i)  about  O  =  0  gives 

(Ah)  - 


Therefore  the  moment  (Ah)  =  i»[4:Ft  —  3Pt  —  PJ,  which 
is  positive,  since  (from  above)  4  Ft  is  >  3P,  +  P3.  Hence 
^L  must  point  to  the  left,  i.e.,  is  a  thirst  or  compression,  and  is 


-PJ. 

Similarly,  taki       moments  about  O^  ti  e  intersection  of  ^1 
and  ^,  we  have  nation  in  which  the  only  unknown  is  (7, 

viz,  (Gh)~  F,f*  +  >,a  =  0.     .-.  (CK)  =  |*p  F,  -  2P,], 


STATICS   OF   A    RIGID   BODY. 

a  positive  moment,  since  3  Vl  is  >2P,  ;  .-.  C 


37 

point  te  the 

right,  i.e.,  is  a  tension,  and  =  ^[3  Vl  —  2PJ. 

Finally,  to  obtain  B,  put  2(vert.  comps.)  =  0;  i.e.  (JJcos  cp) 
+  F,  -  P,  -  P,  =  0.  .:£  cos  9  =  P,  +  P3  -  Vi  ;  but 
(see  foregoing  value  of  "Pi)  we  may  write 

F,  -  (P,  +  PJ  -  (iPt  +  iPJ  +  *P,. 

/.  J?  cos  cp  will  be  +  (upward)  or  —  (downward),  and  B  will 
be  compression  or  tension,  as  \PZ  is  <  or  > 

B  -  [P,+  P,  -  FJ  -  cos  9>  = 

Problem  4.  —  Given  the  weight  #,  of  rod,  the  weight  O-M 

and  all  the  geometrical  elements  (the  student  will  assume  a 

W\|P, 


FIG.  40.  Fio.  41. 

convenient  notation);  required  the  tension  Li  the  cord,  and  the 
amount  and  direction  of  pressure  on  hinge-pin. 

Problem  5. — Roof -truss ;  pin-connection  ;  all  loads  at  joints; 
wind-pressures  W  and  TF,  normal  to  OA ;  required  the  three 
reactions  or  supporting  forces  (of  the  two  horizontal  surfaces 
and  one  vertical  surface),  and  the 
stress  in  each  piece.  All  geomet- 
rical elements  are  given ;  also  P, 
PaPaW  (Fig.  40). 


38.  Composition  of  Non-concur- 
rent Forces  in  Space. — Let  P^  P^ 
etc.,  be  the  given  forces,  and  a?,, 
2,,  xn  y^  z»  etc.,  their  points  of  ap- 
plication referred  to  an  arbitrary 
origin  and  axes ;  al9  /?n  y^  etc.,  FIG.  42. 

the  angles  made  by  their  lines  of  application  with  X^.Y*  and  Z. 


38 


MECHANICS    OF   ENGINEERING. 


Considering  the  first  force  JP,,  replace  it  by  its  three  com- 
ponents parallel  to  the  axes,  Xl  =  Pl  cos  or,;  Yl  —  P^  cos  ft^\ 
and  Z,  =  PI  cos  yl  (Pl  itself  is  not  shown  in  the  figure).  At 
0,  and  also  at  A,  put  a  pair  of  equal  and  opposite  forces, 
each  equal  and  parallel  to  Z,;  Zl  is  now  replaced  by  a  single 
force  Zi  acting  upward  at  the  origin,  and  two  couples,  one 
in  a  plane  parallel  to  YZ  and  having  a  moment  =  —  Zlyl  (as 
we  see  it  looking  toward  O  from  a  remote  point  on  the  axis 
-f-  X),  the  other  in  a  plane  parallel  to  XZ  and  having  a  mo- 
ment =  -f-  ZjOJj  (seen  from  a  remote  point  on  the  axis  -\-  Y). 
Similarly  at  0  and  C  put  in  pairs  of  forces  equal  and  parallel 
to  JTj,  and  we  have  X^  at  B^  replaced  by  the  single  force  X^ 
at  the  origin,  and  the  couples,  one  in  a  plane  parallel  to  XYl 
and  having  a  moment  -|-  X^y^  seen  from  a  remote  point  on 
the  axis  -\-  Z,  the  other  in  a  plane  parallel  to  XZ,  and  of  a 
moment  =  —  X^^  seen  from  a  remote  point  on  the  axis  -\-Y, 
and  finally,  by  a  similar  device,  Yl  at  B  is  replaced  by  a  force 
Yl  at  the  origin  and  two  couples,  parallel  to  the  planes  XY 
and  yZ,  and  having'  moments  —  Y&  and  +  Yfa  respective- 
ly. (In  Fig.  42  the  single  forces  at  the  origin  are  broken 
lines,  while  the  two  forces  constituting  any  one  of  the  six 

couples  may  be  recognized  as  being 
equal  and  parallel,  of  opposite  di- 
rections, and  both  continuous,  or 
both  dotted.)  We  have,  therefore, 
replaced  the  force  JP,  by  three 
forces  X^  yj,  Z»  at  <9,  and  six 
couples  (shown  more  clearly  in 
Fig.  43;  the  couples  have  been 
transferred  to  symmetrical  posi- 
tions). Combining  each  two  couples- 
FlQ  43  whose  axes  are  parallel  to  JT,  y, 

or  Z,  they  can  be  reduced  to  three,  viz., 

one  with  an  X  axis  and  a  moment  =  Ylzl  — 
one  with  a  !Faxis  and  a  moment  =  Zlxl  — 
one  with  a  Z  axis  and  a  moment  —  Xlyl  — 


STATICS    OF    A    RIGID   BODY.  39 

Dealing  with  each  of  the  other  forces  P^  Pn  etc.,  in  the  same 
manner,  the  whole  system  may  finally  be  replaced  by  three 
forces  2X,  ^  Y,  and  2Z,  at  the  origin  and  three  couples 
whose  moments  are,  respectively, 

L  =  2(Yz  —  Ziy)  with  its  axis  parallel. to  JT; 
M  =  2(Zx  —  Xz)  with  its  axis  parallel  to  Y\ 
N  =  ^(Xy  —  Yx)  with  its  axis  parallel  to  Z. 

The  "  axes"  of  these  couples,  being  parallel  to  the  respective 
co-ordinate  axes  JT,  Y,  and  Z,  and  proportional  to  the  mo- 
ments Z,  Mj  and  jV,  respectively,  the  axis  of  their  resultant 
(7,  whose  moment  is  G,  must  be  the  diagonal  of  a  parallelo- 
pipedon  constructed  on  the  three  component  axes  (propor- 
tional to)  Z,  M,  and  &  Therefore,  O  = 
while  the  resultant  of  2X,  2  Y,  and  2Z  is 


E  =  V(2X?  +  (2  Y)* 

acting  at  the  origin.     If  oc,  /?,  and  y  are  the  direction-angles 

^X  2  Y  ~S7 

of  ^,  we  have  cos  a  =  -^-,  cos  ft  =  -^-,  and  cos  y  =  -^  ; 

while  if  A-,  >w,  and  v  are  those  of  the  axis  of  the  couple  (7,  we 

L  M  N 

have  cos  A  =  ^,  cos  /*  =  -^,  and  cos  Y  =  -g. 

For  equilibrium  we  have   both   (7  =  0  and  It  =  0;   i.e., 
separately,  m?  conditions,  viz., 

J2X=  0,  2  T  =  0,  ^Z=0 ;     and    Z=0,  Jf=0,  ^T=0   .   (1) 

Now,  noting  that  ^"^  =  0,  2T  =  0,  and  ^(Xy  —  Tx)  =  0 
are  the  conditions  for  equilibrium  of  the  system  of  non-concur- 
rent forces  which  would  be  formed  by  projecting  each  force  of 
our  actual  system  upon  the  plane  X Y,  and  similar  relations 
for  the  planes  YZ  and  XZ,  we  may  restate  equations  (1)  in 
another  form,  more  serviceable  in  practical  problems,  viz. : 
U ote. — If  a  system  of  non-concurrent  forces  in  space  is  in 
equilibrium,  the  plane  systems  formed  ly  projecting  the  given 
system  upon  each  of  three  arbitrary  co-ordinate  planes  will  each 
ue  in  equilibrium.  But  we  cau  obtain  only  six  independent 


40  MECHANICS    OF   ENGINEERING. 

equations  in  any  case,  available  for  six  unknowns.  If  E  alone 
=  0,  we  have  the  system  equivalent  to  a  couple  (7,  whose 
moment  =  G ;  if  G  alone  =  0,  the  system  has  a  single  re- 
sultant R  applied  at  the  origin.  In  general,  neither  R  nor  G 
being  —  0,  we  cannot  further  combine  R  and  0  (as  was  done 
with  non-concurrent  forces  in  a  plane)  to  produce  a  single  re- 
sultant unless  R  and  C are  in  the  same  plane;  i.e.,  when  the 
angle  between  R  and  the  axis  of  C  is  =  90°.  Call  that  angle 
6.  If,  then,  cos  6  =  cos  a  cos  A  -|-  cos  ft  cos  jn  -\-  cos  y  cos  * 
is  =  0  =  cos  90°,  we  may  combine  R  and  C  to  produce  2 
single  resultant  for  the  whole  system ;  acting  in  a  plane  con- 
taining It  and  parallel  to  the  plane  of  £7  in  a  direction  parallel 

to  j£,  at  a  perpendicular  distance  c  =*  -&  from  the  origin  and 

=  R  in  intensity.  The  condition  that  a  system  of  forces  in 
space  have  a  single  resultant  is,  therefore,  substituting  the 
previously  derived  values  of  the  cosines,  (2^)  .  L  -\-  (2  Y)  .  M 
+  (2Z-) .  N  =  0. 

This  includes  the  cases  when  R  is  zero  and  when  the  system 
reduces  to  a  couple. 

To  return  to  the  general  case,  R  and  C  not  being  in  the 
same  plane,  the  composition  of  forces  in  space  cannot  b( 
further  simplified.  Still  we  can  give  any  value  we  please  to 
P,  one  of  the  forces  of  the  couple  (7,  calculate  the  correspond- 

C1 ' 
ing  arm  a  =  -p,  then  transfer  C  until  one  of  the  .P's  has  the 

same  point  of  application  as  R,  and  combine  them  by  the 
parallelogram  of  forces.  We  thus  have  the  whole  system 
equivalent  to  two  forces,  viz.,  the  second  P,  and  the  resultant 
of  R  and  the  first  P.  These  two  forces  are  not  in  the  same 
plane,  and  therefore  cannot  be  replaced  by  a  single  resultant. 

39.  Problem.  (Non-concurrent  forces  in  space.) — Given  all 
geometrical  elements  (including  «,  /?,  y,  angles  of  P),  also  the 
weight  of  Q,  and  weight  of  apparatus  G ;  A  being  a  hinge  whose 
pin  is  in  the  axis  I7,  0  a  ball-and-socket  joint :  required  the 
amount  of  P  (Ibs.)  to  preserve  equilibrium,  also  the  pressures 


STATICS   OF   A   RIGID   BODY. 


41 


(amount  ant*  direction)  at  A  and  0  ;  no  friction.     Eeplace  P 
by  its  X,  Z,  and  Z  components.     The  pressure  at  A  will  have 


FIG.  44. 


Z  and.X  components  ;  that  at  (9,  X,  Y,  and  Z  components. 

The  body  is  now  free,  and  there  are  six  unknowns. 
^X,  ^Y,  and  -S'Zgive,  respectively, 

P  cos  or  +  X,  +  XQ  =  0  ; 

P  cos  /?  +  JT0  =  0  ;  and  Z,  +  Z0  —  Q  —  G  -  P  cos  y  =  0. 

As  for  moment-equations  (see  note  in  last  paragraph),  project- 
ing the  system  upon  YZ  and  putting  2(Pa)  about  (9  =  0, 
we  have 

-Z,l+Qd+  Ge  +  (P  co& y)b  +  (P cos /3)c  =  0; 
projecting  it  upon  XZ,  and  putting  2 (Pa)  about  (9  =  0,  we 
have  Qr  —  (P  cos  a)c  —  (P  cos  ;/)#  =  0  ; 

projecting  on  XY,  moments  about  0  give 

X,l  +  (P  cos  «)&  -  (P  cos  /?>  =  0. 

From  these  six  equations  we  may  obtain  the  six  unknownSy 
P,  X^  7^  Z0,  JT,,  and  Zr  If  for  any  one  of  these  a  negative 
result  is  obtained,  it  shows  that  its  direction  in  Fig.  4.4.  should 
be  reversed. 


42 


MECHANICS   OF   ENGINEERING. 


CHAPTER  IT. 


STATICS  OF  FLEXIBLE  CORDS. 

40.  Postulate  and  Principles. — The  cords  are  perfectly  flexi- 
ble and  inextensible.  All  problems  will  be  restricted  to  one 
plane.  Solutions  of  problems  are  based  on  three  principles, 
viz.: 

PRIN.  I. — The  strain  on  a  cord  at  any  point  can  act  only 
along  the  cord,  or  along  the  tangent  if  it  be  curved. 

PRIN.  II. — We  may  apply  to  flexible  cords  in  equilibrium  all 
the  conditions  for  the  equilibrium  of  rigid  bodies  ;  since,  if  the 
system  of  cords  became  rigid,  it  would  still,  with  greater  rea- 
son, be  in  equilibrium. 

PRIN.  III. — The  conditions  of  equilibrium  cannot  be  applied, 
of  course,  unless  the  system  can  be  considered  a  free  body^ 
which  is  allowable  only  when  we  conceive  to  be  .put  in,  at  the 
points  of  support  or  fastening,  the  reactions  (upon  the  cord) 
of  those  points  and  the  supports  removed.  These  reactions 
having  been  put  in,  then  consider  the  case  in  Fig.  45  in  one 
plane.  If  we  take  any  point,  p,  on  the  cord  as  a  centre 
of  moments,  knowing  that  the  resultant  R,  of  the  forces  P^ 
P»  and  JP8,  situated  on  one  side  of  p,  must  act  along  the  cord 
R  if>  &  through  p  (by  Prin.  1),  therefore 

n  we  have  Plal  —  JP^at  —  P3a3 
=  E  X  zero  =  0,  and  (equally 

That  is,  in  a  system  of  cords  in 
Fia  45.  equilibrium  in  a  plane,  if  a  centre 

of  moments  be  taken  on  the  cord,  the  algebraic  sum  of  the  ?no- 


STATICS    OF   FLEXIBLE   OOKDS. 


43 


ments  of  those  forces  situated  on  one  side  (either']  of  this  point 
will  equal  zero. 

41.  The  Pulley. — A  cord  in  equilibrium  over  a  pulley  whose 
axle  is  smooth  has  the  same  tension  on  both  sides ;  for,  Fig.  46, 

C; 

\      A    ' 

P 

FIG.  46.  FIG.  47. 

considering  the  pulley  and  its  portion  of  cord  free  2(Pa)  =  0 
about  the  centre  of  axle  gives  P'r  —  Pr,  i.e.,  P'  —  P  =  ten- 
sion in  the  cord.  Hence  the  pressure  R  at  the  axle  bisects 
the  angle  a,  and  therefore  if  a  weighted  pulley  rides  upon  a 
cord  ABC,  Fig.  47,  its  position  of  equilibrium,  B,  may  be 
found  by  cutting  the  vertical  through  A  by  an  arc  of  radius 
CD  —  length  of  cord,  and  centre  at  <7,  and  drawing  a  horizon- 
tal through  the  middle  of  AD  to  cut  CD  in  B.  A  smooth 
ring  would  serve  as  well  as  the  pulley  ;  this  would  be  a  slip- 
knot. 

42.  If  three  cords  meet  at  a  fixed  knot,  and  are  in  equilib- 
rium, the  tension  in  any  one  is  the  equal  and 

opposite  of  the  resultant  of  those  in  the  other 
two. 

43.  Tackle. — If  a  cord  is  continuous  over  a 
number  of  sheaves  in  blocks  forming  a  tackle, 
neglecting  the  weight  of  the  cord  and  blocks  and 
friction  of  any  sort,  we  may  easily  find  the  ratio 
between  the  cord-tension  P  and  the  weight  to  be 
sustained.  E.g.,  Fig.  48,  regarding  all  the  straight 
cords  as  vertical  and  considering  the  block  B 
free,  we  have,  Fig.  49  (from  2  T  =  0),  4P  —  G 

S7 

=  0,  .*.  P  =  — -.     The  stress  on  the  support  C  will  =  5P. 


44  MECHANICS    OF   ENGINEERING. 

44.  Weights  Suspended  by  Fixed  Knots. — Given  all  the  geo- 

metrical elements  in  Fig.  50,  and 

H0|o°  x  OIie  weight,  6^,;"  required  the  re- 

maining weights  and  the  forces 
ZT0,  F0,  Hn  and  Vw  at  the  points 
of  support,  that  equilibrium  may 
obtain.  Jf0  and  F0  are  the  hori- 
zontal and  vertical  components  of 
FIG.  50.  the  tension  in  the  cord  at  0 '; 

similarly  Hn  and  F^,  those  at  n.  There  are  n  -f-  2  unknowns. 
From  Prin.  II  we  have  2Z  =  0,  and  2  T  =  0;  i.e.,  H0  —  Hn 
=  0,  and  [G,  -f  G,  +  ...]-  [  F0  +  Fw]  =  0.  While  from 
Prin.  III.,  taking  the  successive  knots,  1,  2,  etc.,  as  centres  of 
moments,  we  have 

=  0, 
f,  +  G&*  ~  «0  =  0, 

etc.,  for  n  knots. 

-Thus  we  have  n  -\-  2  independent  equations,  a  sufficient 
number,  and  they  are  all  of  the  first  degree  (with  reference  to 
the  unknowns),  and  easily  solved.  As  a  special  solution,  we 
maj>  by  §  42,  resolve  6ra  in  the  directions  of  the  first  and  sec- 
ond cord-segments,  and  obtain  their  tensions  by  a  parallelogram 
of  forces ;  then  at  the  second  knot,  knowing  the  tension  in  the 
second  segment,  we  may  find  that  in  the  third  and  £ra  in  like 
manner,  and  so  on.  Of  course  HQ  and  F0  are  components  of 
the  tension  in  the  first  segment,  Hn  and  Vn  of  that  in  the 
last. 

45.  The  converse  of  the  problem  in  §  44,  viz.,  given  the 
weights  Gr^  etc.,  xn  and  yn,  the  lengths  0,  5,  c,  etc.;  required 
HQ,  F0,  JTni  Fn,  and  the  co-ordinates  x^  y^  a?2,  ya,  etc.,  of  the 
fixed    knots   when    equilibrium   exists,    contains   2n  +  2   un- 
knowns.    Statics  furnishes  n  -f-  2  equations  (already  given  in 
§  44) ;  while  geometry  gives  the  other  n  equations,  one  for 
each  cord-segment,  viz.,  x*  -\-  y*  =  a3 ;  (a?,  —  x^f  -f-  (y,  —  y,)3 
—  t>* ;  etc. 


STATICS   OF   FLEXIBLE   CORDS. 


45 


!Y 


However,  most  of  these  %n-\-%  equations  are  of  the  second 
degree  ;  hence  in  the  general  case  they  cannot  be  solved. 

46.  Loaded  Cord  as  Parabola,  —  If  the  weights  are  equal  and 
infinitely  small,  and  are  intended  to  be  uniformly  spaced 
along  the  horizontal,  when  equilib- 
rium obtains,  the  cord  having  no 
weight,  it  will  form  a  parabola.  Let 
q  —  weight  of  loads  per  horizontal 
linear  unit,  0  be  the  vertex  of  the  «-^t 
curve  in  which  the  cord  hangs,  and 
m  any  point.  We  may  consider 
the  portion  Om  as  a  free  body,  if 
the  reactions  of  the  contiguous  portions  of  the  cord  are  put  in, 
HQ  and  T,  and  these  (from  Prin.  I.)  must  act  along  the  tangents 
to  the  curve  at  0  and  m,  respectively  ;  i.e.,  HQ  is  horizontal, 

and  T  makes  some  angle  cp  (whose  tangent  =  -/-,  etc.)  with 
the  axis  X.     Applying  Prin.  II., 

=  0  gives  T  cos  <p  -HQ=0',  i.e.,  T~  = 


2Y=  0  gives  Tsin  cp  —  qx  =  0  ;  i.e.,  1       =  qx. 


(1) 
(2) 


//7/ 


Dividing  (2)  by  (1),  member  by  member,  we  have   ~  =  ^  ; 

/.  dy  =  -~xdx,    the     differential    equation      of   the    curve ; 
•"o 

y  =  -jj  I    xdx  =  -—.-;  or  x*  = y,    the   equation   of  a 

parabola  whose  vertex  is  at  0  and  axis  vertical. 

NOTE.— The  same  result,  -/-  =  ^r,  mav  be  obtained  by  considering  that 
ax      7/0 

we  have  here  (Prin.  II.)  a  free  rigid  body  acted  on 
by  three  forces,  T,  Ho,  and  R  —  qx,  acting  verti- 
cally through  the  middle  of  the  abscissa  x\  the 
resultant  of  HQ  and  R  must  be  equal  and  oppo- 
site to  I7,  Fig.  52.  . '.  tan  <p  =  -^,  or  J-  =  ~. 

Evidently  also  the  tangent-line  bisects  the  ab- 
scissa x. 


46 


MECHANICS   OF   ENGINEERING. 


FIG.  53. 


47.  Problem  under  §  46.     [Case  of  a  suspension-bridge  in 
which  the  suspension-rods  are  vertical,  -the  weight  of  roadway 
is  uniform  per  horizontal  foot,  and  large  compared  with  that 
of  the  cable  and  rods.     Here  the  roadway  is  the  only  load  :  it 
is  generally  furnished  with  a  stiffening  truss  to  avoid  deforma- 
tion under  passing  loads.]  —  Given  the  span   —  26.  Fig.  53, 

Yi  v,f  71   the  deflection  =  a,  and  the  rate  of  loading 

=  q  Ibs.  per  horizontal  foot  ;  required  the 
tension  in  the  cable  at  0,  also  at  m  ;  and 
*^e  len£tn  °^  ca°le  needed.  From  the 
equation  of  the  parabola  qx*  —  ^HQy,  put- 
ting x  =  b  and  y  =  a,  we  have  HQ  =  qb*  -±-  %a  =  the  tension 
at  0.  From  2  Y  =  0  we  have  Vl  =  qb,  while  2X=  0  gives 

H,  =  #0;  >.  the  tension  at  m  =  VH?  +  V?=  ~[_qb  i/S^+Fj. 

aft 

The  semi-length,  Om  ,  of  cable  (from  p.  88,  Todhunter's  In^ 
tegral  Calculus)  is  (letting  n  denote  HQ  -=-  2<?,  =  tf  ~-  4a) 
Om  =   tfna  +  a*  +  n  .  loge  [(  V~a-\-  4/^T+"a)  -T-   Vn]. 

48.  The  Catenary.*  —  A  flexible,  inextensible  cord  or  chain,  of 
uniform  weight  per  unit  of  length,  hung  at  two  points,  and 
supporting  its  own  weight  alone,  forms  a   curve  called   the 
catenary.     Let  the  tension  HQ  at  the  lowest  point  or  vertex  be 
represented  (for  algebraic  convenience)  by  the  weight  of  an 
imaginary  length,  'c,  of  similar  cord  weighing  q  Ibs.  per  unit 
of  length,  i.e.,  HQ  —  qc  ;   an  actual  portion   of   the   cord,  of 
length  s,  weighs  qs  Ibs.     Fig.  54  shows  as  free  and  in  equilib- 

rium  a  portion  of  the  curve  of  any 
lengta  si  reckoning  from  0  the 
vertex.  Required  the  equation  of 
the  curve.  The  load  is  uniformly 
spaced  along  the  curve,  and  not 
'x  horizontally,  as  in'§§  46  and  47. 


H 


2T=  Ogives  2      = 


while 


=  0  gives  T~  =  qc.     Hence,  by  division,  cdy  =  sdx,  and 


squaring, 


=  s*dx* 


(1) 


For  the  "  transformed  catenary,"  see  p.  395. 


.  STATICS    OF   FLEXIBLE   COKDS.  47 

Put   dy1  =  ds*  —  dx*,  and  we  have,  after  solving  for  dx 
cds  />«      ds  r«  

and  x  =c  .  loge[(*+  V~s*  +  c*)  -r-  c],   ...     (2) 

a  relation  between  the  horizontal  abscissa  and  length  of  curve. 
Again,  in  eq.  (1)  put  dx*  =  ds*  —  dy*,  and  solve  for  dy. 


sds  1 

This  glves  dy  =,     _  =  ^  .  A_±_|     Theref  ore 


+  <T^  +  *')  =  *(*'  +  «%  and  finally 

y  =  VT^F?-^    •    •    •   •   •    •    (3) 

Clearing  of  radicals  and  solving  for  c,  we  have 

c  =  (s*-y*)  +  2y  ......     (4) 

Example.  —  A  40-foot  chain  weighs  240  Ibs.,  and  is  so  hung 
from  two  points  at  the  same  level  that  the  deflection  is  10 
feet.  Here,  for  s  =  20  ft.,  y  =  10  ;  hence  eq.  (4)  gives  the 
parameter,  c  =  (400  —  100)  -f-  20  =  15  feet,  q  =  240  -r-  40 
—  6  Ibs.  per  foot.  /.  the  tension  at  the  middle  is  HQ  =  qc 
=  6  X  15  =  90  Ibs.;  while  the  greatest  tension  is  at  either 
support  and  =  tW+T20s  =  150  Ibs. 

Knowing  c  =  15  feet,  and  putting  s  =  20  feet  =  half 
length  of  chain,  we  may  compute  the  corresponding  value  of 
x  from  eq.  (2);  this  will  be  the  half-span  [loge  m  —  2.30258 
X  (common  log  m)].  To  derive  s  in  terms  of  a?,  transform 
eq.  (2)  in  the  sense  in  which  n  =  loge  m  may  be  transformed 
into  £w  =  m,  clear  of  radicals,  and  solve  for  5,  which  gives  * 


Again,  eliminate  s  fro/n  (2)  by  substitution  from  (3),  trans- 
form as  above,  clear  of  radicals,  and  solve  for  y  +  c9  whence 


*  e  here  denotes  2.71828,  the  base  of  the  natural  system  of  logarithms. 


48  MECHANICS   OF  ENGINEERING. 

which  is  the  equation  of  a  catenary  with  axes  as  in  Fig.  54. 
If  the  horizontal  axis  be  taken  a  distance  =  c  below  the  ver- 
tex, the  new  ordinate  y'  =  y  -|-  c,  while  x  remains  the  same; 
the  last  equation  is  simplified. 

If  the  span  and  length  of  chain  are  given,  or  if  the  span 
and  deflection  are  given,  c  can  be  determined  from  (5)  or  (6) 
only  by  successive  assumptions  and  approximations. 

48a.  Addendum  to  §  55.  Mass. — In  PHYSICS,  the  fundamental  units  are 
those  of 

SPACE,  involving  a  unit  of  length  (and  thence  of  area  and  volume) ; 
TIME,          "         a  unit  of  time,  usually  the  second  ; 

MASS,  "  a  unit  of  mass,  which  (by  Government  decree)  may  be  the 
quantity  of  matter  in  a  specified  piece  of  platinum,  or  specified  volume  of  water, 
etc.  (a  beam-balance  being  used  to  determine  equal  quantities  of  mass) ;  while 
FORCE  involves  a  derived  unit,  being  measured  by  its  effect  in  accelerating 
the  velocity  of  a  moving  mass,  since  it  is  proportional  both  to  the  mass  and  the 
acceleration.  The  unit  force  (called  absolute  unit)  is  the  force  necessary  to  pro- 
duce unit  acceleration  in  a  unit  of  mass;  so  that  to  produce  an  acceleration  =  p  in 
a  mass  =  m  requires  a  force  =  F=  mp,  and  the  force  thus  obtained  is  in  absolute 
units.  This  is  called  the  dynamic  measure  of  a  force. 

JSxample. — In  the  C.G.S.  system  of  units,  required  the  constant  force  necessary 
to  cause  a  mass  of  400  grams  to  gain  200  velocity  units  in  2  sec ;  i.e.,  p  =  100 
centims.  per  sec.,  per  sec.  From  F  =  mp  we  have 

F=  400  x  100  =  40000  abs.  units  of  force  (or  dynes,  in  C.G.S.  system). 
In  the  ft.-lb.-sec.  system  the  absolute  unit  is  called  &poundal. 

In  MECHANICS  or  ENGINEERING,  however,  it  is  more  convenient  to  regard  the 
fundamental  units  to  be  those  of 

SPACE,  as  ft.,  metre,  etc.,  area  and  volume  corresponding  ; 

TIME,    as  seconds,  hours,  etc. ; 

Force,  as  Ibs.,  grams,  kilograms,  tons,  etc.,  indicated  by  a  spring  balance; 
while  for 

MASS  we  assume  a  derived  unit,  a  mode  of  measuring  it  being  developed  as 
follows: 

If  by  experiment  (block  on  smooth  table,  for  instance)  we  find  that  a  constant 
force  P  (Ibs.,  tons,  kilos.)  will  maintain  an  acceleration  =  p  in  the  rectilinear 
motion  (in  line  of  force)  of  a  body  whose  weight  (by  previous  trial  with  a  spring 
balance)  is  G  (Ibs.,  tons,  or  other  unit);  and  if  in  a  second  experiment,  by  allow- 
ing the  force  O  to  act  on  the  same  body  in  vacuo,  a  free  vertical  fall  with  accelera- 
tion =  g  is  the  result,— we  find  that  the  proportion  (Newton's  2d  Law)  P:O:\p:g 

is  verified.    This  may  be  written  P=  —  .  p,  and  may  then  be  read:  Force  =  mass 

X  acceleration,  if  we  call  the  quotient  G  +  g  the  MASS  of  the  body  whose  weight 
(by  spring  balance)  is  =  G  at  a  locality  where  the  acceleration  of  gravity  =  g  ;  for 
this  quotient  will  be  the  same  at  all  localities  on  the  earth's  surface. 

Example  (same  as  above). — If  a  body  whose  weight  O  =  400  grams  (force)  is  to 
have  its  velocity  increased,  in  2  sec.,  from  300  centims.  per  sec.  to  500  centims. 
per  sec.,  at  a  uniform  rate,  we  must  provide  a  constant  force 

400  40000 

P=  98l X  10°  =  ~98T  =  4°'77  Srams;  or  -040  kilos- 

This  is  called  the  gravitation  measure  of  a  force.  Hence  it  is  evident  that  to  re- 
duce absolute  units  (called  dynes  and  poundals)  in  the  C.G.S.  and  ft.-lb.-sec. 
systems,  respectively)  to  ordinary  practical  units  of  force  (Ibs.,  tons,  kilos.,  etc., 
of  a  spring  balance),  we  divide  by  the  value  of  g  proper  to  the  system  of  units  em- 
ployed; and  vice  versd. 


PART  II.-DYNAMICS. 


CHAPTEE  I. 

RECTILINEAR  MOTION  OF  A  MATERIAL  POINT, 

49.  Uniform  Motion  implies  that  the  moving  point  passes 
over  equal  distances  in  equal  times;  variable  motion,  that  un-. 
equal  distances  are  passed  over  in  equal  times.  In  uniform 
motion  the  distance  passed  over  in  a  unit  of  time,  as  one  sec- 
ond, is  called  the  velocity  (=  v),  which  may  also  be  obtained 
by  dividing  the  length  of  any  portion  (=  s)  of  the  path  by 
the  time  (=  t)  taken  to  describe  that  portion,  however  small  or 
great ;  in  variable  motion,  however,  the  velocity  varies  from 
point  to  point,  its  value  at  any  point  being  expressed  as  the 
quotient  of  ds  (an  infinitely  'small  distance  containing  the 
given  point)  by  dt  (the  infinitely  small  portion  of  time  in 
which  ds  is  described). 

49#.  By  acceleration  is  meant  the  rate  at  which  the  velocity 
of  a  variable  motion  is  changing  at  any  point,  and  may  be  a 
uniform  acceleration,  in  which  case  it  equals  the  total  change 
of  velocity  between  any  two  points,  however  far  apart,  divided 
by  the  time  of  passage ;  or  a  variable  acceleration,  having  a 
different  value  at  every  point,  this  value  then  being  obtained 
by  dividing  the  velocity-increment,  dv,  or  gain  of  velocity 
in  passing  from  the  given  point  to  one  infinitely  near  to  it,  by 
dt,  the  time  occupied  in  acquiring  the  gain.*  (Acceleration 
must  be  understood  in  an  algebraic  sense,  a  negative  accelera- 
tion implying  a  decreasing  velocity,  or  else  that  the  velocity  in 
a  negative  direction  is  increasing.)  The  foregoing  applies  to 
motion  in  a  path  or  line  of  any  form  whatever,  the  distances 
mentioned  being  portions  of  the  path,  and  therefore  measured 
along  the  path. 

*  See  addendum  on  p.  12. 


50  MECHANICS   OF   ENGINEERING. 

50.  Eectilinear  Motion,  or  motion  in  a  straight  line.  —  The 
general  relations  of  the  quantities  involved  may  be  thus  stated 
(see  Fig.  55)  :  Let  v  =  velocity  of  the  body  at  any  instant  ; 

~8  Q  _____  s  ____  »d8_da_     +S      tlien  <h>  =  gain  of  velocity 

~~|    T\     ?        in  an  instant  of  time  dt.    Let 

•  dt^dt\    1        t  =  time   elapsed    since   the 

body  left  a  given  fixed  point, 

which  will  be  taken  as  an  origin,  0.  Let  s  =  distance  (+  or 
—  )  of  the  body,  at  any  instant,  from  the  origin  O\  then  ds  = 
distance  traversed  in  a  time  dt.  Let  p  =  acceleration  =  rate 
at  which  v  is  increasing  at  any  instant.  All  these  may  be 
variable  ;  and  t  is  taken  as  the  independent  variable,  i.e.,  time 
is  conceived  to  elapse  by  equal  small  increments,  each  =  dt  ; 
hence  two  consecutive  ds's  will  not  in  general  be  equal,  their 
difference  being  called  d?s.  Evidently  d*t  is  =  zero,  i.e.,  dt  is 
constant. 

Since  -,-  =  number  of  instants  in  one  second,  the  velocity  at 
(tt 

any  instant  (i.e.,  the  distance  which  would  be  described  at  that 

1  ds 

rate  in  one  second)  is  v  =  ds  .    -  ;  .'.  v  =    -.     .     .     .     .     (I.) 


!  /  (ds\     c 

Similarly,  p  =  dv  .  -^-,  and  I  since  dv  =  d(^J  =  ~d 


dv 


Eliminating  dt,  we  have  also  vdv  —  pds.       ....     .(III.) 

These  are  the  fundamental  differential  formulae  of  rectilinear 
motion  (for  curvilinear  motion  we  have  these  and  some  in  ad- 
dition) as  far  as  kinematics,  i.e.,  as  far  as  space  and  time,  is 
concerned.  The  consideration  of  the  mass  of  the  material 
point  and  the  forces  acting  upon  it  will  give  still  another  rela- 
tion (see  §  55). 

51.  Rectilinear  Motion  due  to  Gravity.  —  If  a  material  point 
fall  freely  in  vacuo,  no  initial  direction  other  than  vertical 
having  been  given  to  its  motion,  many  experiments  have 


RECTILINEAR   MOTION    OF  A   MATERIAL   POINT.        51 

shown  that  this  is  a  uniformly  accelerated  rectilinear  motion 
in  a  vertical  line  having  an  acceleration  (called  the  accelera- 
tion of  gravity)  equal  to  32.2  feet  per  square  second,*  or  9.81 
metres  per  square  second  ;  i.e.,  the  velocity  increases  at  this 
constant  rate  in  a  downward  direction,  or  decreases  in  an  up- 
ward direction. 

[NOTE.  —  By  "  square  second  "  it  is  meant  to  lay  stress  on  the  fact  that  an 
acceleration  (being  =  d*s  -*-  dP)  is  in  quality  equal  to  one  dimension  of 
length  divided  by  two  dimensions  of  time.  E.g.,  if  instead  of  using  the 
foot  and  second  as  units  of  space  and  time  we  use  the  foot  and  the  minute, 
g  will  =  32.2  X  3600;  whereas  a  velocity  of  say  six  feet  per  second  would 
=  6  X  60  feet  per  minute.  The  value  of  g—  32.2  implies  the  units  foot 
and  second,  and  is  sufficiently  exact  for  practical  purposes.] 

52.  Free  Fall  in  Vacuo.  —  Fig.  56.     Let  the  body  start  at  0 

with  an  initial  downward  velocity  =  c.     The  accelera-  _s 
tion  is  constant  and  =  -f-  g.     Reckoning  both  time  and 

distance  (-J-  down  wards)  from  0,  required  the  values  of  j° 

the  variables  s  and  v  after  any  time  t.     From  eq.  (II.),  c 

§  50,  we  have  +  g  =  dv  -=-  dt  ;  .'.  dv  =  gdt,  in  which  the  v  s 

two  variables  are  separated.  I  i 

v  ~  c  = 


rr*        rv 
dv  —  ffJo  dt>  i<e->  \_0 


v 

v  =       or 


gt  —  0  ;  and  finally,  v  =  c  +  gt  ........     (1)  FIG.  5* 

(Notice  the  correspondence  of  the  limits  in  the  foregoing 

operation  ;  when  t  =  0,  v  =  +  c.) 

From  eq.  (I.),  §  50,  v  =  ds  -=-  dt  ;  /.  substituting  from  (1), 

ds  =  (G  -f-  gt)dt,  in  which  the  two  variables  s  and  t  are  sepa- 

rated. 


or  s=  ct+%gt\        ,     .    .    v  *  .-.     (2) 

Again,  eq.  (III.),  §  50,  vdv  =  yds,  in  which  the  variables  v 
and  s  are  already  separated. 

.-.  fjvdv  =  gjs'ds  ;  or  |J^3  = 

*  =  -W--         --- 
*  Or,  32.2  "  feet  per  second  per  second." 


52  MECHANICS    OF    ENGIN  EKUI  NCr. 

If  the  initial  velocity  =  zero,  i.e.,  if  the  body  falls  from  rest, 
eq.  (3)  gives  s=^udv=  tftyfa*  [From  the  frequent  re- 

v* 
currence  of  these  forms,  especially  in  hydraulics,  ^-is  called  the 

t/ 

"height  due  to  the  velocity  v,"  i.e.,  the  vertical  height  through 
which  the  body  must  fall  from  rest  to  acquire  the  velocity  v ; 
while,  conversely,  y%gh  js  called  the  velocity  due  to  the  height 
or  head  A.] 

By  eliminating  g  between  (1)  and  (3),  we  may  derive  another 
formula  between  three  variables,  s,  v,  and  £,  viz., 

s  =  ftc  _|_  vy (4) 

53.  Upward  Throw. — If  the  initial  velocity  were  in  an  up- 
ward direction  in  Fig.  56  we  might  call  it  —  c,  and  introduce  it 
with  a  negative  sign  in  equations  (1)  to  (4),  just  derived ;  but 
for  variety  let  us  call  the  upward  direction  -(-,  in  which  case 
an  upward  initial  velocity  would  =  -|-  <?,  while  the  acceleration 
=  —  g,  constant,  as  before.  (The  motion  is  supposed  confined 
within  such  a  small  range  that  g  does  not  sensibly  vary.)  Fig. 

57.     From  p  —  dv  -r-  dt  we  have  dv  =  —  gdt  and 

!     A 

v       /w  m 

_gjc          yjo 

From  v  =  ds  -r-  dt,  ds  —  cdt  —  gtdt, 


I 

-^S     vdv  =pds  gives  Je  vdv  =  —  gJQ  ds,  whence 

%(v*  —  <?')  =  —  gs,  or  finally,  s 
And  by  eliminating  g  from  (l)a  and  (3)a, 


FIG.  57. 

<?2  —  v9 
%(v*  —c')  =  —  gs,  or  finally,  s  =  — ^ — •      •     (3)0 


The  following  is  now  easily  verified  from  these  equations  : 
the  body  passes  the  origin  again  (s  =  0)  with  a  velocity  =  —  <?, 
after  a  lapse  of  time  =  %c  -r-  g.  The  body  comes  to  rest  (for 

*  In  Hydraulics  h  is  used  instead  of  s. 


RECTILINEAR    MOTION    OF   A    MATERIAL   POINT.         53 

an  instant)  (put  v  =  0)  after  a  time  =  o  -j-  ^,  and  at  a  distance 
s  =  c*  -~  Zg  (u  height  due  to  velocity  <?")  from  0.  For  t  > 
c  -T-  g,  v  is  negative,  showing  a  downward  motion  ;  for  t  > 
2c  -7-  g,  s  is  negative,  i.e.,  the  body  is  below  the  starting-point 
while  the  rate  of  change  of  v  is  constant  and  =  —  g  at  all 
points. 

54.  Newton's  Laws.  —  As  showing  the  relations  existing  in 
general  between  the  motion  of  a  material  point  and  the  actions 
(forces)  of  other  bodies  upon  it,  experience  furnishes  the  fol- 
lowing three  laws  or  statements  as  a  basis  for  dynamics  : 

(1)  A  material  point   under  no  forces,  or  under  balanced 
forces,  remains  in  a  state  of  rest  or  of  uniform  motion  in  a 
right  line.     (This  property  is  often  called  Inertia^ 

(2)  If  the  forces  acting  on  a  material  point  are  unbalanced, 
an  acceleration  of  motion  is  produced,  proportional  to  the  re- 
sultant force  and  in  its  direction. 

(3)  Every  action  (force)  of  one  body  on  another  is  always 
accompanied  by  an  equal,  opposite,  and  simultaneous  reaction, 
(This  was  interpreted  in  §  3.) 

As  all  bodies  are  made  up  of  material  points,  the  results  ob- 
tained in  Dynamics  of  a  Material  Point  serve  as  a  basis  for  the 
Dynamics  of  a  Rigid  Body,  of  Liquids,  and  of  Gases. 

55.  Mass.*  —  If  a  body  is  to  continue  moving  in  a  right  line, 
the  resultant  force  P  at  all  instants  must  be  directed  along  that 
line  (otherwise  it  would  have  a  component  deflecting  the  body 
from  its  straight  course). 

In  accordance  with  Newton's  second  law,  denoting  by^>  the 
acceleration  produced  by  the  resultant  force  (ft  being  the 
body's  weight),  we  must  have  the  proportion  P  :  G  :  :  p  :  g  ; 
i.e., 


Eq.  IY.  and  (I.),  (II.),  (III.)  of  §  50  are  the  fundamental 
equations  of  Dynamics.     Since  the  quotient  G  -f-  g  is  invaria- 


*  See  Addendum  on  p.  48. 


54  MECHANICS    OF   ENGINEERING. 

ble,  wherever  the  body  be  moved  on  the  earth's  surface  (G  and 
g  changing  in  the  same  ratio),  it  will  be  used  as  the  measure 
of  the  mass  _flfor  quantity  of  matter  in  the  body.  In  this  way 
it  will  frequently  happen  that  the  quantities  G-  and  g  will  ap- 
pear in  problems  where  the  weight  of  the  body,  i.e.,  the  force 
of  the  earth's  attraction  upon  it,  and  the  acceleration  of  gravity 
have  no  direct  connection  witli  the  circumstances.  No  name 
will  be  given  to  the  unit  of  mass,  it  being  always  understood 
that  the  fraction  G  -f-  g  will  be  put  for  M  before  any  numeri- 
cal substitution  is  made.  From  (IY.)  we  have,  in  words, 
(  accelerating  force  =  mass  X  acceleration/ 

\  also,       acceleration  -—  accelerating  force  -f-  mass. 

56.  Uniformly  Accelerated  Motion.  —  If  the  resultant  force  is 
constant  as  time  elapses,  the  acceleration  must  be  constant  (from 
eq.  (IY.),  since  of  course  Jkf  is  constant)  and  =  P  -r-  M.  The 
motion  therefore  will  be  uniformly  accelerated,  and  we  have 
only  to  substitute  -\-  p  (constant)  for  g  in  eqs.  (1)  to  (4)  of 
§  52  for  the  equations  of  this  motion,  the  initial  velocity  being 
=  c  (in  the  line  of  the  force). 

.     .     .     (1)  ;          s  —  ct 


If  the  force  is  in  a  negative  direction,  the  acceleration  will 
be  negative,  and  may  be  called  a  retardation'  the  initial  veloc- 
ity should  be  made  negative  if  its  direction  requires  it. 

57.  Examples  of  Unif.  Ace.  Motion.  —  Example  1.     Fig.  58. 

A  small  block  whose  weight  is  -J-  Ib.  has  already  described  a 

—  S  c       P     M    v  B        distance  Ao  =  48  inches  over  a 

A    SMOOTH    -3z±.     !T!£fa^  ~j    smooth  portion   of   a   horizontal 

FIG.  58.  table  in  two  seconds  ;  at  0  it  en* 

counters  a  rough  portion,  and  a  con  sequent  constant  friction  of 

2  oz.     Required  the  distance  described  beyond  0,  and  the  time 

occupied  in  coming  to  rest.     Since  we  shall  use  32.2  for  <?, 

times  must  be  in  seconds,  and  distances  in  feet  ;  as  to  the  unit 


RECTILINEAR  MOTION   OF   A   MATERIAL   POINT.         55 

uf  force,  as  that  is  still  arbitrary,  say  ounces.  Since  AO  was 
smooth,  it  must  have  been  described  with  a  uniform  motion 
(the  resistance  of  the  air  being  neglected);  hence  with  a  veloc- 
ity =  4  ft.  ^  2  sec.  =  2  ft.  per  sec.  The  initial  velocity  for 
the  retarded  motion,  then,  is  c  =  -\-  2  at  0.  At  any  point  be- 
yond 0  the  acceleration  =  force  -f-  mass  =  (—  2  oz.)  -r-  (8  oz. 
-^  32.2)  =  —  8.05  ft.  per  square  second,  i.e.,  p  =  —  8.05  = 
constant ;  hence  the  motion  is  uniformly  accelerated  (retarded 
here),  and  we  may  use  the  formulae  of  §  56  with  c  =  -\-  2,^>  = 
—  8.05.  At  the  end  of  the  motion  v  must  be  zero,  and  the 
corresponding  values  of  s  and  t  may  be  found  by  putting  v  =  0 
in  equations  (3)  and  (1),  and  solving  for  s  and  t  respectively : 
thus  from  (3),*  =  i(-4)-r-  (—  8.05),  i.e.,«  =  0.248  +,  which 
must  be  feet ;  while  from  (1),  t  =  (—  2)  -r-  (—  8.05)  =  0.248  -f, 
which  must  be  seconds. 

Example  2.  (Algebraic.) — Fig.  59.  The  two  masses  Ml  = 
G1  -r-  g  and  M  =  G  -r-  g  are  connected  by  a  flexible,  inexten- 
sible  cord.  Table  smooth.  Required  the  acceleration  common 
to  the  two  rectilinear  motions,  and  the  tension  in  the  string  S, 


FIG.  59.  FIG.  60. 

there  being  no  friction  under  G»  none  at  the  pulley,  and  no 
mass  in  the  latter  or  in  the  cord.  At  any  instant  of  the  mo- 
tion consider  #,  free  (Fig.  60),  N'  being  the  pressure  of  the 
table  against  6>  Since  the  motion  is  in  a  horizontal  right  line 
2(vert.  compons.)—  0,  i.e., -ZV—  Gl  =  0,  which  determines  N. 
S,  the  only  horizontal  force  (and  resultant  of  all  the  forces)  = 
M^j  i.e., 

O  /7  ,_  f-t\ 

At  the  same  instant  of  the  motion  consider  G  free  (Fig.  61); 
the  tension  in  the  cord  is  the  same  value  as  above  =  8.  The 
accelerating  force  is  G  —  S,  and 

.-.  =  mass  X  ace.,  or  G  —  8  =  (G  -s-  g)p.     .     (2) 


56  MECHANICS    OF   ENGINEERING. 


1° 


s      From  equations  (1)  and  (2)  we  obtain  p  =  (Gg)  -~ 
(G  4-  6ra)  =  a  constant;  hence  each  motion  is  uniformly 
accelerated,  and  we  may  employ  equations  (1)  to  (4)  of 
*,  H-"  §  56  to  find  the  velocity  and  distance  from  the  starting- 
points,  at  the  end  of  any  assigned  time  t,  or  vice  versa. 
'  The  initial  velocity  must  be  known,  and  may  be  zero. 
Also,  from  (1)  and  (2)  of  this  article, 

8  =  (GG,}  +  (G  +  G,}  =  constant. 

Example  3. — A  body  of  2f  (short)  tons  weight  is  acted  on 
during  -J  minute  by  a  constant  force  P.  It  had  previously  de- 
scribed 316-f  yards  in- 180  seconds  under  no  force';  and  subse- 
quently, under  no  force,  describes  9900  inches  in  ^  of  an  hour. 
Required  the  value  of  P.  Ans.  P  =  22.1  Ibs. 

Example  4. — A  mass  of  1  ton  weight,  having  an  initial 
velocity  of  48  inches  per  second,  is  acted  on  for  J  minute  by  a 
force  of  400  avoirdupois  ounces.  Kequired  the  final  velocity. 

Ans.     10.037  ft.  per  sec. 

Example  5. — Initial  velocity,  60  feet  pei  second  ;  mass  weighs 
0.30  of  a  ton.  A  resistance  of  112£  Ibs.  retards  it  for  -^  of 
a  minute.  Required  the  distance  passed  over  during  this  time. 

Ans.     286.8  feet. 

Example  6. — Required  the  time  in  which  a  force  of  600  avoir- 
dupois ounces  will  increase  the  velocity  of  a  mass  weighing  1-J 
tons  from  480  feet  per  minute  to  240  inches  per  second. 

Ans.     30  seconds. 

Example  7. — What  distance  is  passed  over  by  a  mass  of  (0.6) 
tons  weight  during  the  overcoming  of  a  constant  resistance 
(friction),  if  its  velocity,  initially  144  inches  per  sec.,  is  reduced 
to  zero  in  8  seconds.  Required,  also,  the  friction. 

Ans.     48  ft.  and  55  Ibs. 

Example  8. — Before  the  action  of  a  force  (value  required)  a 
body  of  11  tons  had  described  uniformly  950  ft.  in  12  minutes. 
Afterwards  it  describes  1650  feet  uniformly  in  180  seconds. 
The  force  acts  30  seconds.  P  —  ?  Ans.  P  —  178  Ibs. 


RECTILINEAR   MOTION    OF   A    MATERIAL    POINT.         57 

58.  Graphic  Representations.     Unif.  Ace.  Motion. — With  the 
initial  velocity  =  0,  the  equations  of  §  56  become 


v  =  ptj (1)  s  ==  %ptf, (2) 

s  =  v9  -r-  2p,  .     .     .     (3)      and      5  =  %vt (4) 

Eqs.  (1),  (2),  and  (3)  contain  each  two  variables,  which  may 
graphically  be  laid  off  to  scale  as  co-ordinates  and  thus  give  a 
curve  corresponding  to  the  equation.  Thus,  Fig.  62,  in  (I.),  we 


have  a  right  line  representing  eq.  (I.)  ;  in  (II.),  a  parabola  with 
axis  parallel  to  s,  and  vertex  at  the  origin  for  eq.  (2)  ;  also  a 
parabola  similarly  situated  for  eq.  (3).  Eq.  (4)  contains  three 
variables,  #,  v,  and  t.  This  relation  can  be  shown  in  (I.),  s  be- 
ing represented  by  the  area  of  the  shaded  triangle  =  %vt. 
(II.)  and  (III.)  have  this  advantage,  that  the  axis  OS  may  be 
made  the  actual  path  of  the  body.  [Let  the  student  determine 
how  the  origin  shall  be  moved  in  each  case  to  meet  the  supposi- 
tion of  an  initial  velocity  =  -f-  c  .or  —  <?.] 

59.  Variably  Accelerated  Motions.  —  We  here  restate  the  equa- 

tions 

ds        /T.  dv      tfs 

v  =  dl  •  •  0.);j>  =  -g8rj£  .  .  (II.);vdv=pdS..  (III.); 

and  resultant  force 


which  are  the  only  ones  for  general  use  in  rectilinear  motion. 

PROBLEM  1.  —  In  pulling  a  mass  M  along  a  smooth,  horizon- 
tal table,  by  a  horizontal  cord,  the  tension  is  so  varied  that 
,?  =  4£8  (not  a  homogeneous  equation  ;  the  units  are,  say,  the 
foot  and  second).  Required  by  what  law  the  tension  varies. 


58  MECHANICS    OF   ENGINEERING. 

ds       d(±t*) 
From  (I.)  v  =       =  =  12*  ;  from  (IL),  p  = 


;  and  (IY.)  the  tension  —  P  —  Mp  =  24J/fc,  i.e.,  varies 
directly  as  the  time. 

PROBLEM  2.  "  Harmonic  Motion,"  Fig.  63.  —  A  small  block 


FIG.  63 

on  a  smooth  horizontal  table  is  attached  to  two  horizontal 
elastic  cords  (arid  they  to  pegs)  in  such  a  way  that  when  the 
block  is  at  0,  each  cord  is  straight  but  not  tense  ;  in  any  other 
position,  as  m,  one  cord  is  tense,  the  other  slack.  The  cords 
are  alike  in  every  respect,  and,  as  with  springs,  the  tension 
varies  directly  with  the  elongation  (=  s  in  figure).  If  for  an 
elongation  sl  the  tension  is  jTl9  then  for  any  elongation  s  it  is 
T  =  T^  -T-  sr  The  acceleration  at  any  point  m,  then,  is 
p  =  —  (T  -=-  M)  =  —  (7>  -T-  Ms,\  which  for  brevity  put 
p  =  —  as,  a  being  a  constant.  Required  the  equations  of 
motion,  the  initial  velocity  being  =  -(-  c,  at  0.  From  eq.  (III.) 

vdv  =  —  asds  ;  /.    /   vdv  =  —  a   I  sds> 
ie.,  i(va  -  c3)  =  -  %as*  ;     or,  tf  =  c*  -  as\     .     (1) 


From  (I.),  dt  =  ds-±-v\  hence  from  (1), 


or 


t  =  ==  =  i  Psin 


RECTILINEAR   MOTION    OF   A    MATERIAL   POINT.         59 

Inverting  (2),  we  have  s  =  (c  -^  Va)  sin  (t  Va),  ...  (3) 
Again,  by  differentiating  (3),  see  (I.),  v  —  c  cos  (t  Va)  (4) 
Differentiating  (4),  see  (II.),  p  =  —  c  Va  sin  (t  Va). .  .  (5) 

These  are  the  relations  required,  but  the  peculiar  property 
of  the  motion  is  made  apparent  by  inquiring  the  time  of  pass- 
ing from  0  to  a  state  of  rest ;  i.e.,  put  v  =  0  in  equation  (4), 
we  obtain  t  =  \n  -T-  Va,  or  J-TT  -r-  Va,  or  f  7t  -+-  Va,  and  so  on, 
while  the  corresponding  values  of  s  (from  equation  (3)),  are 
-f-  (o  -7-  Va),  —  (c  -r-  Va),  -\-  (c  -f-  Va),  arid  so  on.  This  shows 
that  the  body  vibrates  equally  on  both  sides  of  0  in  a  cycle  or 
period  whose  duration  =  2?r  -f-  Va,  and  is  independent  of  the 
initial  velocity  given  it  at  0.  Each  time  it  passes  0  the 
velocity  is  either  -|-  c,  or  —  c,  the  acceleration  =  0,  and  the 
time  since  the  start  is  =  mt  -$-  Va,  in  which  n  is  any  whole 
number.  At  the  extreme  point  p  =  =p  c  Va,  from  eq.  (5). 
If  then  a  different  amplitude  be  given  to  the  oscillation  by 
changing  c,  the  duration  of  the  period  is  still  the  same,  i.e., 
the  vibration  is  isochronal.  The  motion  of  an  ordinary  pen- 
dulum is  nearly,  that  of  a  cycloidal  pendulum  exactly,  harmonic. 

If  the  crank-pin  of  a  reciprocating  engine  moved  uniformly 
in  its  circular  path,  the  piston  would  have  a  harmonic  motion 
if  the  connecting-rod  were  infinitely  long,  or  if  the  design  in 


FIG.  64. 

Fig.  64  were  used.  (Let  the  student  prove  this  from  eq.  (3).) 
Let  2/1  =  length  of  stroke,  and  o  =  the  uniform  velocity  of  the 
crank-pin,  and  M  =  mass  of  the  piston  and  rod  AB.  Then 
the  velocity  of  M  at  mid-stroke  must  =  <?,  at  the  dead-points, 
zero;  its  acceleration  at  mid-stroke  zero;  at  the  dead-points 
the  ace.  —  c  Va,  and  s  =  r  =  c  -r-  Va  (from  eq.  (3))  ;  .'.  Va 
=  c  +  r,  and  the  ace.  at  a  dead  point  (the  maximum  ace.) 


60 


MECHANICS    OF   ENGINEERING 


FIG.  65. 


=  ca  -T-  r.  Hence  on  account  of  the  acceleration  (or  retarda- 
tion) of  JJ^in  the  neighborhood  of  a  dead-point  a  pressure  will 
be  exerted  on  the  crank-pin,  equal  to  mass  X  ace.  =  Me*  -f-  r 
at  those  points,  independently  of  the  force  transmitted  due  to 
steam-pressure  on  the  piston -head,  and  makes  the  resultant 
pressure  on  the  pin  at  C  smaller,  and  at  D  larger  than  it  would 
be  if  the  "inertia"  of  the  piston  and  rod  were  not  thus  taken 
into  account.  We  may  prove  this  also  by  the  free-body  method, 
considering  AB  free  immediately  after  passing  the  dead-point 
G  Q  ;  (7,  neglecting  all  friction.  See  Fig. 
65.  The  forces  acting  are :  G,  the 
weight ;  J¥,  the  pressures  of  the 
guides ;  P,  the  known  effective  steam- 
pressure  on  piston-head ;  and  P',  the  .unknown  pressure  of 
crank-pin  on  side  of  slot.  There  is  no  change  of  motion  ver- 
tically ; .-.  N'  +  N  —  G  =  0,  and  the  resultant  force  is  P  —  Pf 
=  mass  X  accel.  =  Me*  -~-  r,  hence  P'  =  P  —  Me*  -=-  r. 
Similarly  at  the  other  dead-point  we  would  obtain  P'  =  P  + 
Me*  -T-  r.  In  high-speed  engines  with  heavy  pistons,  etc., 
Me*  -T-  /•  is  no  small  item.  [The  upper  half-revol.,  alone,  IB 
here  considered.] 

PROBLEM  3. — Supposing  the  earth  at  rest  and  the  resistance 
of  the  air  to  be  null,  a  body  is  given  an  initial  upward  vertical 
velocity  —  c.  Required  the  velocity  at  any  distance  s  from 
the  centre  of  the  earth,  whose  attraction  varies  in- 
versely as  the  square  of  the  distance  s. 

See  Fig.  66. — The  attraction  on  the  body  at  the 
surface  of  the  earth  where  s  =  r,  the  radius,  is  its 
weight  G ;  at  any  point  m  it  will  be  P  =  G(r*  -r-  s2), 
while  its  mass  =  G  -r-  g. 

Hence  the  acceleration  at  ra  =_£>  =  (  —  P) -r-  M 
=  —  g(r*  -T-  s*).  Take  equation  III.,  vdv  =  pds, 
and  we  have 


+s 


Fio.  66. 


vdv  =  - 


or 


=  -  gr       -  - 


, 


(1) 


RECTILINEAR  MOTION   OF   A   MATERIAL   POINT.        61 


Evidently  v  decreases,  as  it  should.  Now  inquire  how  small 
a  value  c  may  have  that  the  body  shall  never  return;  i.e., 
that  v  shall  not  =  0  until  s  =  co.  Put  v  =  0  and  s  =  oo  in 
(1)  and  eolve  for  c  ;  and  we  have 


c  = 


=      2  X  32.2  X  21000000, 


=  about  36800  ft.  per  sec.  or  nearly  7  miles  per  sec.  Con- 
versely, if  a  body  be  allowed  to  fall,  from  rest,  toward  the 
earth,  the  velocity  with  which  it  would  strike  the  surface 
would  be  less  than  seven  miles  per  second  through  whatever 
distance  it  may  have  fallen. 

If  a  body  were  allowed  to  fall  through  a  straight  opening  in 
the  earth  passing  through  the  centre,  the  motion  would  be  har- 
monic, since  the  attraction  and  consequent  acceleration  now 
vary  directly  with  the  distance  from  the  centre.  See  Prob.  2. 
This  supposes  the  earth  homogeneous. 

PROBLEM  4. — Steam  working  expansively  and  raising  a  weight, 
Fig.  67. — A  piston  works  without 
friction  in  a  vertical  cylinder.  Let 
8  =  total  steam-pressure  on  the 
underside  of  the  piston ;  the  weight 
G,  of  the  mass  G  -s-  g  (which  in- 
cludes the  piston  itself)  and  an 
atmospheric  pressure  =  A,  con- 
stitute a  constant  back-pressure. 
Through  the  portion  OB  =  s,,  of  FIO.  67. 

the  stroke,  S  is  constant  =  /S,,  while  beyond  -#,  boiler  com- 
munication being  "  cut  off,"  S  diminishes  with  Boyle's  law,  i.e., 
in  this  case,  for  any  point  above  Bt  we  have,  neglecting  the 
"clearance",  F being  the  cross-section  of  the  cylinder, 

Full  length  of  stroke  =  ON  =  sn.  Given,  then,  the  forces 
Sl  and  Ay  the  distances  st  and  sw  and  the  velocities  at  0  and 
nt  ^Vboth  =  0  (i.e.,  the  mass  M  =  (r-^-gisto  start  from  rest  at 
O.  and  to  come  to  rest  at  N\  required  the  proper  weight  G  to 


62  MECHANICS   OF  ENGINEERING. 

fulfil  these  conditions,  S  varying  as  already  stated.     The  accel- 
eration at  any  point  will  be 

p=[S-A-0]  +  M.     .     .     .    .     (1) 

Hence   (eq.  III.)  Mvdv  =  [S  —  A  —  6P]d8,  and  .-.  for  the 

whole  stroke 

_S-A  -  €F\d8\  i.e., 

/>»i  f*Siids  /»Sn  /*Sn 

0  =  S,  /   ds  +  tf  A  /     -  -  A  I    ds  -  G  I    ds, 

V/O  *  1t/sl      S  Jo  t/0 

or  SA  [l  +  log.  J]  -  ^  +  £*n.  ...     (2) 

Since  S  =  S,  =  constant,  from  0  to  j£,  and  variable,  = 
$X  -r-  «,  from  ^  to  ^y,  we  have  had  to  write  the  summation 


in  two  parts. 

From  (2),  G  becomes  known,  and  /.  Jfalso  (=  G  -+•  g}. 

Required,  further,  the  time  occupied  in  this  upward  stroke. 
From  0  to  B  (the  point  of  cut-off)  the  motion  is  uniformly 
accelerated,  since  p  is  constant  (S  being  =  8^  in  eq.  (1)  ), 
with  the  initial  velocity  zero;  hence',  from  eq.  (3),  §  56, 
the  velocity  at  B  =  v,  —  1/2  [^  —  A~—  €F\sl  -r-  Jf  is  known  ; 
/.  the  time  tl  =  2^  -r-  vl  becomes  known  (eq.  (4),  §  56)  of  de- 
scribing OB.  At  any  point  beyond  B  the  velocity  v  may  be  ob- 
tained thus :  From  (III.)  vdv  =  pds,  and  eq.  (1)  we  have, 
summing  between  B  and  any  point  above, 


This  gives  the  relation  between  the  two  variables  v  and  s 
anywhere  between  B  and  W\  if  we  solve  for  v  and  insert  its 
value  in  dt  =  ds  -f-  v,  we  shall  have  dt  —  a  function  of  s  and 
ds,  which  is  not  integrable.  Hence  we  may  resort  to  approxi- 


RECTILINEAR   MOTION    OF   A    MATERIAL   POINT.         63 

mate  methods  for  the  time  from  B  to  N.  Divide  the  space 
BN  into  an  uneven  number  of  equal  parts,  say  five;  the  dis- 
tances of  the  points  of  division  from  0  will  be  s1?  sa,  s8,  s4,  s» 
and  sn.  For  these  values  of  s  compute  (from  above  equation) 
^  (already  known),  v»  v»  v4,  VM  and  vn  (known  to  be  zero).  To 
the  first  four  spaces  apply  Simpson's  Rule,  and  we  have  the 
time  from  B  to  the  end  of  «9 


r6,      r  ds 
\    t  =  /      -  ; 
LI       •/,-«» 


approx.  = 


12 


while  regarding  the  motion  from  5  to  ^V  as  uniformly  retarded 
(approximately)  with  initial  velocity  =  v6  and  the  final  =  zero, 
we  have  (eq.  (4),  §  56), 

t  =  2s   -  s    -s-  v. 


c 


By  adding  the  three  times  now  found  we  have  the  whole  time 
of  ascent.  In  Fig.  67  the  dotted  curve  on  the  left  shows  by 
horizontal  ordinates  the  variation  in  the  velocity  as  the  distance 
s  increases ;  similarly  on  the  right  are  ordinates  showing  the 
variation  of  S.  The  point  E,  where  the  velocity  is  a  maximum 
—  vm,  may  be  found  by  putting  p  =  0,  i.e.,  for  S  =  A  -f-  G, 
the  acelerating  force  being  —  0,  see  eq.  (1).  Below  E  the  ac- 
celerating force,  and  consequently  the  acceleration,  is  positive ; 
above,  negative  (i.e.,  the  back-pressure  exceeds  the  steam- 
pressure).  The  horizontal  ordinates  between  the  line  HE'KL 
and  the  right  line  RI 'are  proportional  to  the  accelerating  force. 
If  by  condensation  of  the  steam  a  vacuum  is  produced  be- 
low the  piston  on  its  arrival  at  N",  the  accelerating  force  is 
downward  and  =  A  -f-  G.  [Let  the  student  determine  how 
the  detail  of  this  problem  would  be  changed,  if  the  cylinder 
were  horizontal  instead  of  vertical.] 

60.  Direct  Central  Impact. — Suppose  two  masses  Ml  and  J/, 
to  be  moving  in  the  same  right  line  so  that  their  distance  apart 
continually  diminishes,  and  that  when  the  collision  or  impact 
takes  place  the  line  of  action  of  the  mutual  pressure  coincides 
with  the  line  joining  their  centres  of  gravity,  or  centres  of 


64  MECHANICS    OF   ENGINEERING. 

mass,  as  they  may  be  called  in  this  connection.  This  is  called 
a  direct  central  impact,  and  the  motion  of  each  mass  is  varia- 
bly accelerated  and  rectilinear  during  their  contact,  the  only 
force  being  the  pressure  of  the  other  body.  The  whole  mass 
of  each  body  will  be  considered  concentrated  in  the  centre  of 
mass,  on  the  supposition  that  all  its  particles  undergo  simul- 
taneously the  same  change  of  motion  in  parallel  directions. 
(This  is  not  strictly  true ;  the  effect  of  the  pressure  being 
gradually  felt,  and  transmitted  in  vibrations.  These  vibrations 
endure  to  some  extent  after  the  impact.)  When  the  centres 
of  mass  cease  to  approach  each  other  the  pressure  between  the 
bodies  is  a  maximum  and  the  bodies  have  a  common  velocity ; 
after  this,  if  any  capacity  for  restitution  of  form  (elasticity) 
exists  in  either  body,  the  pressure  still  continues,  but  dimin- 
ishes in  value  gradually  to  zero,  when  contact  ceases  and  the 
bodies  separate  with  different  velocities.  Reckoning  the  time 
from  the  first  instant  of  contact,  let  t'  =  duration  of  the  first 
period,  just  mentioned  ;  t"  that  of  the  first  -f-  the  second  (resti- 
tution). Fig.  68.  Let  Ml  and  J/~2  be  the  masses,  and  at  any 
J?j_  | — ^~|.Jk>  instant  during  the  contact  let  ^  and  0, 
be  simultaneous  values  of  the  velocities 
of  the  mass-centres  respectively  (reckon- 
Fl°-  M-  ing  velocities  positive  toward  the  right), 

and  P  the  pressure  (variable).  At  any  instant  the  acceleration 
of  Jfcf,  is  pl  =  —  (P  -T-  -3/j),  while  at  the  same  instant  that  of 
My  is  PI  =  +  (P  -r-  -3Q ;  M^  being  retarded,  J/"a  accelerated, 
in  velocity.  Hence  (eq.  II.,  p  =  dv  -r-  dt)  we  have 

M1dv1=-Pdt:>    and     Mtdv,  =  +  Pdt.   .    .    (!)• 

Summing  all  similar  terms  for  the  first  period  of  the  impact, 
we  have  (calling  the  velocities  before  impact  <?,  and  <?3,  and  the 
common  velocity  at  instant  of  maximum  pressure  C) 

f-c1)  =  -t£tpdt;    (2) 

-*,)  =  • 


RECTILINEAR   MOTION    OF   A    MATERIAL   POINT.         65 

The  two  integrals  are  identical,  numerically,  term  by  term, 
since  the  pressure  which  at  any  instant  accelerates  J/,  is  nu- 
merically equal  to  that  which  retards  J/~t;  hence,  though  we  do 
not  know  how  P  varies  with  the  time,  we  can  eliminate  the 
definite  integral  between  (2)  and  (3)  and  solve  for  C.  If 
the  impact  is  inelastic  (i.e.,  no  power  of  restitution  in  either 
body,  either  on  account  of  their  total  inelasticity  or  damaging 
effect  of  the  pressure  at  the  surfaces  of  contact),  they  continue 
to  move  with  this  common  velocity,  which  is  therefore  their 
final  velocity.  Solving,  we  have 


. 


Next,  supposing  that  the  impact  is  partially  elastic,  that  the 
bodies  are  of   the  same   material,  and    that   the   summation 

/     Pdt  for  the  second  period  of  the  impact  bears  a  ratio,  0, 

/v 
Pdt,  already  used,  a  ratio  peculiar  to  the  material, 

if  the  impact  is  not  too  severe,  we  have,  summing  equations 
(1)  for  the  second  period  (letting  Vl  and  Fa  =  the  velocities 
after  impact), 

Jf,  fj'dv,  =  -  £'pdt,  i.e,  J/,(  F,-  C)  =  -  eJ*Pdt  ;  (5) 


M,        dv,  =  +        pdt,  i.e.,  M,(  V,-  G)  =  +  ePdt.  (6) 

e  is  called  the  coefficient  of  restitution. 

Having  determined  the  value  of  J   Pdt  from  (2)  and  (3)  in 

terms  of  the  masses  and  initial  velocities,  substitute  it  and  that 
of  (7,  from  (4),  in  (5),  and  we  have  (for  the  final  velocities) 

F,  =  [MlCl  +  Mj>-  etffa-cj]  -r-  [IT,  +  JQ;    •  CO 
and  similarly 

F,  =  [^,c,  +  J(fA+^(c,-^)]-[-^  +  ^].    •    (8) 
For  e  =  0,  i.e.,  for  inelastic  impact,  F,=  F,=  C  in  eq.  (4)  ;  for 


66  MECHANICS    OF    ENGINEERING. 

e=  1,  or  elastic  impact,  (7)  and  (8)  become  somewhat  simpli- 
fied. 

To  determine  e  experimentally,  let  a  ball  (J/i)  of  the  sub- 
stance fall  upon  a  very  large  slab  (J/"a)  of  the  same  substance, 
noting  both  the  height  of  fall  h»  and  the  height  of  rebound  Hv. 
Considering  Mt  as  —  oo,  with 


eq.  (7)  gives 

Let  the  student  prove  the  following  from  equations  (2),  (3), 
(5),  and  (6): 

(a)  For  any  direct  central  impact  whatever, 

[The  product  of  a  mass  by  its  velocity  being  sometimes 
called  its  momentum,  this  result  may  be  stated  thus : 

In  any  direct  central  impact  the  sum  of  the  momenta  before 
impact  is  equal  to  that  after  impact  (or  at  any  instant  during 
impact).  This  principle  is  called  the  Conservation  of  Momen- 
tum. The  present  is  only  a  particular  case  of  a  more  general 
proposition. 

It  can  be  proved  that  O,  eq.  (4),  is  the  velocity  of  the  centre 
of  gravity  of  the  two  masses  before  impact ;  the  conservation 
of  momentum,  then,  asserts  that  this  velocity  is  unchanged  by 
the  impact,  i.e.,  by  the  mutual  actions  of  the  two  bodies.] 

(b)  The  loss  of  velocity  of  Mr  and  the  gain  of  velocity  of 
J/a,  are  twice  as  great  when  the  impact  is  elastic  as  when  in- 
elastic. 

(c)  If  e  =  1,  and  M,  =  M»  then  Vl  =  +  c,,  and  F2  =  c,. 

Example. — Let  Ml  and  M*  be  perfectly  elastic,  having  weights  =  4  and 
6  Ibs.  respectively,  and  let  Ci  =  10  ft.  per  sec.  and  c2  =  —  6  ft.  per  sec. 
(i.  e.,  before  impact  Jf2  is  moving  in  a  direction  contrary  to  that  of  MI). 
By  substituting  in  eqs.  (7)  and  (8),  with  e  =  1,  Jtfi  =  4  -s-  g,  and  Jf2  =  5  -f-  g, 
we  have 

F,  =  .1 1~4  x  10  +  5  x  (-  6)  -  5  (lO  -  (-  6))]=  -  7.7  ft.  per  sec. 

Fa  =  if~4  x  10  +  5  x  (-  6)  +  4  (lO  -  (-  6))]=  +  8.2  ft.  per  sec. 

as  the  velocities  after  impact.     Notice  their  directions, -as  indicated  by  their 
signs 


"  VIRTUAL   VELOCITIES."  67 


CHAPTER  II. 

"VIRTUAL  VELOCITIES." 

61.  Definitions.  —  If  a  material  point  is  moving  in  a  direction 
not  coincident  with  that  of  the  resultant  force  acting  (as  in 
curvilinear  motion  in  the  next  chapter),  and  any  element  of  its 
path,  ds,  projected  upon  this  force;*  the  length  of  this  projec- 
tion, du,  Fig.  69,  is  called  the  "VIRTUAL  VELOCITY"  of  the 
force,  since  du  -r-  dt  may  be  considered  the  veloc- 

ity of  the  force  at  this  instant,  just  as  ds  -r-  dt  i 

that  of  the  point.     The  product  of  the  force  by 

its  du  will  be  called  its  virtual  moment,  reckoned 

-j-  or  —  according  as  the  direction  from  0  to  D  is        IG'  6  ' 

the  same  as  that  of  the  force  or  opposite. 

62.  Prop.  I.  —  The  virtual  moment  of  a  force  equals  the 
algebraic  sum  of  those  of  its  components.     Fig.  YO.     Take  the 

direction  of  ds  as  an  axis  JT;  let  jPJ  and  P^ 
^  be  components  of  P  ';  &„  a»  and  a  their 
angles  with  X.  Then  (§  16)  P  cos  a  = 
-A  cos  ^1+^2  cos  aa-  Hence  P(ds  cos  a)= 
P^(ds  cos  <*,)  +  PJ(ds  cos  a^).  But  ds  cos  a 
=  the  projection  of  ds  upon  P,  i.e.,  =  du  ; 


FIG.  70. 


cog 


P^duy  If  in  Fig.  70  <xl  were  >  90°,  evidently  we  would 
have  Pdu  =  —  P^du^  +  P^du^  i.e.,  P^du^  would  then  be 
negative,  and  ODJ  would  fall  behind  0\  hence  the  definition 
of  -f-  and  —  in  §  61.  For  any  number  of  components  the 
proof  would  be  similar,  and  is  equally  applicable  whether  they 
are  in  one  plane  or  not. 

63.  Prop.  II.  —  The  sum  of  the  virtual  moments  equals  zero, 

for  concurrent  forces  in  equilibrium. 

*  We  should  rather  say  "  projected  on  the  line  of  action  of  the  force;" 
but  the  phrase  used  may  be  allowed,  for  brevity. 


68  MECHANICS   OF   ENGINEERING. 

(If  the  forces  are  balanced,  the  material  point  is  moving  in 
a  straight  line  if  moving  at  all.)  The  resultant  force  is  zero. 
Hence,  from  §  62,  PjLu^  -\-  P$u^  +  etc.  =  0,  having  proper 
regard  to  sign,  i.e.,  2(Pdu)  =  0. 

64.  Prop.  III. — The  sum  of  the  virtual  moments  equals  zero 
for  any  small  displacement  or  motion  of  a  rigid  ~body  in  equi- 
librium under  non-concurrent  forces  in  a  plane;  all  points  of 
the  body  moving  parallel  to  this  plane.  (Although  the  kinds 
of  motion  of  a  given  rigid  body  which  are  consistent  with 
balanced  non-concurrent  forces  have  not  yet  been  investigated, 
we  may  imagine  any  slight  motion  for  the  sake  of  the  alge- 
braic  relations  between  the  different  du's  and  forces.) 

First,  let  the  motion  be  a  translation,  all 
points  of  the  body  describing  equal  parallel 
lengths  =  ds.    Take  X parallel  to  ds ;  let  <*1T 
$r  etc.,  be  the  angles  of   the  forces  with  Z 


Then  (§  35)  2(P  cos  *)  =  0  ;  .-.  ds2(Pcos  a) 
—  0 ;  but  ds  cos  al  =  dul ;  ds  cos  at  =  du9 ; 
etc. ;  .-.  2(Pdu)  =  0.  Q.  E.  D. 

Secondly,  let  the  motion  be  a   rotation 
FIO.  71.  through  a  small  angle  d6  in  the  plane  of  the 

forces  about  any  point  0  in  that  plane,  Fig.  72.    With  0  as  a  pole 

let  p,  be  the  radius-vector  of  the  point  of  application  of  P,,  and 

0j  its  lever-arm  from  0\  similarly  for  the 

other  forces.     In  the  rotation  each  point  of 

application  describes  a  small  arc,  ds^  ds^ 

etc.,  proportional  to  />„  /o2,  etc.,  since  ds1 

^  ftlde,  ds,  =  P,d0,   etc.      From    §  36,  ^ 

P^  +  etc.  —  0  ;  but  from  similar  triangles    ' 

dsl  :  dul  ::  P!  :  »j  ;      /.  a1  =  pjLu^  -r-  dst 

=  du,  -T-  dO ;  similarly  a^  =  du^  +  dV,  etc. 

Pence  we  must  have  [P^d^  +  Pzdu9  + .  .  .]  ~  dO  =  0,  i.e., 

2(Pdu)  =  0.     Q.  E.  D. 
Now  since  any  slight  displacement  or  motion  of  a  body  may 

be  conceived  to  be  accomplished  by  a  small  translation  fol- 
lowed by  a  rotation  through  a  small  angle,  and  since  the  fore- 


"VIRTUAL  VELOCITIES."  69 

going  deals  only  with  projections  of  paths,  the  proposition  is 
established  and  is  called  the  Principle  of  Virtual  Velocities. 
[A  similar  proof  may  be  used  for  any  slight  motion  what- 
ever in  space  when  a  system  of  non-concurrent  forces  is  bal- 
anced.] Evidently  if  the  path  (ds)  of  a  point  of  application  is 
perpendicular  to  the  force,  the  virtual  velocity  (du\  and  con- 
sequently the  virtual  moment  (Pdu)  of  the  force  are  zero. 
Hence  we  may  frequently  make  the  displacement  of  such  a 
character  in  a  problem  that  one  or  more  of  the  forces  may  be 
excluded  from  the  summation  of  virtual  moments. 

65.  Connecting-Rod  by  Virtual  Velocities. — Let  the  effective 
steam-pressure  P  be  the  means,  through  the  connecting-rod 
and  crank  (i.e.,  two  links),  of  raising  the  weight  G  very  slowly; 
neglect  friction  and  the  weight  of  the  links  themselves.  Cdn- 
sider  AB  as  free  (see  (o)  in  Fig.  73),  BC  also,  at  (c);  let  the 


(«•) 

FIG.  73. 

"small  displacements"  of  both  be  simultaneous  small  portions 
of  their  ordinary  motion  in  the  apparatus.  A  has  moved  to  Al 
through  dx ;  B  to  B^  through  ds,  a  small  arc ;  C  has  not 
moved.  The  forces  acting  on  AB  are  P  (steam-pressure),  N 
(vertical  reaction  of  guide),  and  N'  and  T  (the  tangential  and 
normal  components  of  the  crank-pin  pressure).  Those  on  BC 
are  N'  and  T  (reversed),  the  weight  G,  and  the  oblique  pressure 
of  bearing  P' .  The  motion  being  slow  (or  rather  the  accelera- 
tion being  small),  each  of  these  two  systems  will  be  considered  as 
balanced.  Now  put  2(Pdu)  =  0  for  AB,  and  we  have 

Pdx  _|_  &  x  0  +  N'  X  0  -  Tds  =  0.     .     .     (1) 

For  the  simultaneous  and   corresponding   motion    of 
2(Pdu)  =  0  irives 


70  MECHANICS    OF   ENGINEERING. 

N*  X  0  -f  Tds  -  Gdh  +  P'  X  0  =  0,     .     .     (2) 

dh  being  the  vertical  projection  of  (7's  motion. 

From  (1)  and  (2)  we  have,  easily,  Pdx  —  Gdh  =  0,  .  (3) 
which  is  the  same  as  we  might  have 
obtained  by  putting  2(Pdu)  =  Ofor 
the  two  links  together,  regarded  col- 
lectively as  a  free  body,  and  describ- 
»•  74-  ing  a  small  portion  of  the  motion 

they  really  have  in  the  mechanism,  viz.,  (Fig.  74,) 

Pdx  +  Nx  0-  Gdh-\-P'  X  0  =  0.    .    .    (4) 
We  may  therefore  announce  the — 

66.  Generality  of  the  Principle  of  Virtual  Velocities. — If  any 

mechanism  of  flexible  inextensihle  cords,  or  of  rigid  bodies 
jointed  together,  or  both,  at  rest^  or  in  motion  with  very  small 
accelerations,  he  considered  free  collectively  (or  any  portion  of 
it\  and  all  the  external  forces  put  in  •  then  (Disregarding 
mutual  frictions)  for  a  small  portion  of  its  prescribed  motion, 
2(Pdu)  must  =  0,  in  which  the  du,  or  virtual  velocity,  of 
each  force,  P,  is  the  projection  of  the  path  of  the  point  of 
application  upon  the  force  (the  product,  Pdu,  being  -\-  or  — 
according  to  §  61). 

67.  Example. — In  the  problem  of  §  65,  having  given  the 
weight  G,  required  the  proper  steam-pressure  (effective)  P  to 
hold  G  in  equilibrium,  or  to  raise  it  uniformly,  if  already  in 
motion,  for  a  given  position  of  the  links.     That  is.  Fig.  75, 
given  a,  r,  c,  a,  and  ft,  re- 
quired the  ratio  dh  :  dx ;  for, 

from  equation  (3),  §  65,  P 
=  G(dh  :  dx).    The  projec- 
tions of  dx  and  ds  upon  AB       dx  A, 
will  be  equal,  since  AB  =  FlGK  75- 

A^B^,  and  makes  an  (infinitely)  small  angle  with  A^B^,  i.e., 
dx  cos  a  =  ds  cos  (ft  —  a).  Also,  dh  =  (c  :  r)ds  sin  ft. 


"VIRTUAL  VELOCITIES."  71 

Eliminating  ds,  we  have, 

dh    _  c  sin  fi  cos  OL  c  sin  fi  cos  a 

dx  ~~  r  cos  (ft  —  a)  '  ^  cos  (/?  —  a)' 

68.  When  the  acceleration  of  the  parts  of  the  mechanism  is 
not  practically  zero,  2(Pdu)  will  not  ==  0,  but  a  function  of 
the  masses  and  velocities  to  be  explained  in  the  chapter  on 
Work,  Energy,  and  Power.    If  friction  occurs  at  moving  joints, 
enough  "  free  bodies"  should  be  considered  that  no  free  body 
extend  beyond  such  a  joint  ;  it  will  be  found  that  this  friction 
cannot  be  eliminated  in  the  way  in  which  T  and  N'  were,  in 
§65. 

69.  Additional  Problems  ;  to  be  solved  by  "  virtual  velocities." 
PROBLEM  1.  —  Find  relations  between  the  forces  acting  on  a 
straight  lever  in  equilibrium  ;  also,  on  a  bent  lever. 

PROBLEM  2.  —  When  an  ordinary  copying-press  is  in  equilib- 
rium, find  the  relation  between  the  force  applied  horizontally 
and  tangential  ly  at  the  circumference  of  the  wheel,  and  the 
vertical  resistance  under  the  screw-shaft. 

Solution.—  Considering  free  the  rigid  body  consisting  of  the  wheel  and 
screw-shaft,  let  R  be  the  resistance  at  the  point  of  the  shaft  (pointing 
along  the  axis  of  the  shaft),  and  Pthe  required  horizontal  tangential  force 
at  edge  of  wheel.  Let  radius  of  wheel  be  r.  Besides  R  and  P  there  are 
also  acting  on  this  body  certain  pressures,  or  "  supporting  forces,"  consist- 
ing of  the  reactions  of  the  collars,  and  reactions  of  the  threads  of  nut  against 
the  threads  of  screw.  Denote  by  s  the  "pitch  "  of  the  screw,  i.e.,  the  dis- 
tance the  shaft  would  advance  for  a  full  turn  of  the  wheel.  Then  if  we 
imagine  the  wheel  to  turn  through  a  small  angle  dQ,  the  corresponding 

advance,  ds,  of  the  shaft  would  be  ^—  ,  from  the  proportion  s  :  ds  ::  2n  :  dQ. 

alt 

The  path  of  the  point  of  application  of  P  would  be  a  small  portion  of 
a  helix,  the  projection  of  which  on  the  line  of  P  is  rdQ,  while  ds  projects 
in  its  full  length  on  the  line  of  the  force  R.  In  the  case  of  each  of  the 
other  forces,  however,  the  path  of  the  point  of  application  is  perpendicular 
to  the  line  of  the  force  (which  is  normal  to  the  rubbing  surfaces,  friction 
being  disregarded).  Hence,  substituting  in  2(Pdu)  =  0,  we  have 


+  P.rdO-R.  ds  +  0  +  0  =  0; 
whence 


MECHANICS  OF  K 


CHAPTER  III. 

CURVILINEAR  MOTION  OF  A  MATERIAL  POINT. 

[Motion  in  a  plane,  only,  will  be  considered  in  this  chapter.] 

70.  Parallelogram  of  Motions. — It  is  convenient  to  regard 
the  curvilinear  motion  of  a  point  in  a  plane  as  compounded,  or 
made  up,  of  two  independent  rectilinear  motions  parallel 
respectively  to  two  co-ordinate  axes  X.  and  Y,  as  may  be  ex- 
plained thus :  Fig.  76.  Consider  the 
drawing-board  CD  as  fixed,  and  let  the 
head  of  a  T-square  move  from  A 
toward  B  along  the  edge  according  to 
any  law  whatever,  while  a  pencil  moves 
from  M  toward  Q  along  the  blade.  The 
result  is  a  curved  line  on  the  board,  whose 
form  depends  on  the  character  of  the 
two  JTand  Y  component  motions,  as  they  may  be  called.  If 


FIG.  76. 


in  a  time  tf,  the  T'-square  head  has  moved  an  ^distance  — 
and  the  pencil  simultaneously  a  Y  distance  =  MP,  by  com- 
pleting the  parallelogram  on  these  lines,  we  obtain  7?,  the 
position  of  the  point  on  the  board  at  the  end  of  the  time  tv. 
Similarly,  at  the  end  of  the  time  £/  we  find  the  point  at  JK '. 

71.  Parallelogram  of  Velocities. — Let  the  X  and  Emotions 
be  uniform,  required  the  resulting  motion.  Fig.  77.  Let  cx 
and  cy  be  the  constant  uniform  JTand  Y  velocities.  Then  in 
any  time,  t,  we  have  x  =  cxt  and  y  = 
Cyt\  whence  we  have,  eliminating  t, 
as  -j-  y  =  c,,,  ~  cv  =  constant,  i.e.,  x  is 
proportional  to  y,  i.e.,  the  path  is  a  0^' 
straight  line.  Laying  off  OA  =  cm  V- 
and  AB  =  cv,  B  is  a  point  of  the  path, 
and  OB  is  the  distance  described  by  the  point  in  the  first 


FIQ.  77. 


CUKVILINEAli   MOTION    OF   A   MATERIAL   POINT.         73 

second.  Since  by  similar  triangles  OR  :  x  :  :  OB  :  c^  we 
have  also  OR  —  OB  .  t  ;  hence  the  resultant  motion  is  uniform, 
and  its  velocity,  OB  =  c,  is  the  diagonal  of  the  parallelogram 
formed  on  the  two  component  velocities. 

Corollary.  —  If  the  resultant  motion  is  curved,  the  direction 
and  velocity  of  the  motion  at  any  point  will  be  given  by  the 
diagonal  formed  on  the  component  velocities  at  that  instant. 
The  direction  of  motion  is,  of  course,  a  tangent  to  the  curve. 

72.  Uniformly  Accelerated  X  and  Y  Motions.  —  The  initial 
velocities  of  hoth  ~being  zero.  Required  the  resultant  motion. 
Fig.  78.  From  §  56,  eq.  (2)  (both  cx  andctf 
being  =  0),  we  have  x  —  -<zpxt*  and  y  = 
ipy£\  whence  x  -^  y  =  px^  py—  constant, 
and  the  resultant  motion  is  in  a  straight 
line.  Conceive  lines  laid  off  from  0  on  IK 
and  Y  to  represent^  and^,  to  scale,.  and 
form  a  parallelogram  on  them.  From  similar  triangles  (OR 
being  the  distance  described  in  the  resultant  motion  in  any 

time  t\~OR\x\\  ~OB  \px;.'.  OlT=  $OJBf.  Hence,  from  the. 
form  of  this  last  equation,  the  resultant  motion  is  uniformly 
accelerated,  and  its  acceleration  is  OB  =  p,  (on  the  same  scale 


This  might  be  called  the  parallelogram  of  accelerations,  but 
is  really  a  parallelogram  of  forces,  if  we  consider  that  a  free 
material  point,  initially  at  rest  at  (9,  and  acted  on  simulta- 
neously by  constant  forces  Px  and  Py  (so  that  px  =  Px  -^  M 
and  py  —  Pv  -r-  M\  must  begin  a  uniformly  accelerated  recti- 
linear motion  in  the  direction  of  the  resultant  force,  having  no 
initial  velocity  in  any  direction. 

73.  In  general,  considering  the  point  hitherto  spoken  of  as  a 
free  material  point,  under  the  action  of  one  or  more  forces,  in 
view  of  the  foregoing,  and  of  Newton's  second  law,  given  the 
initial  velocity  in  amount  and  direction,  the  starting-point, 
the  initial  amounts  and  directions  of  the  acting  forces  and  the 


74 


MECHANICS    OF   ENGINEERING. 


laws  of  their  variation  if  they  are  not  constant,  we  can  resolve 
them  into  a  single  X  and  a  single  Y  force  at  any  instant, 
determine  the  X and  Emotions  independently,  and  afterwards 
the  resultant  motion.  The  resultant  force  is  never  in  the  direc- 
tion of  the  tangent  to  the  path  (except  at  a  point  of  inflection). 
The  relations  which  its  amount  and  direction  at  any  instant 
bear  to  the  velocity,  the  rate  of  change  of  that  velocity,  and 
the  radius  of  curvature  of  the  path  will  appear  in  the  next 
paragraph. 

74.  General  Equations  for  the  curvilinear  motion  of  a  ma- 
terial point  in  a  plane. — The  motion  will  be  considered  result- 
ing  from    the   composition    of 
\  independent  X  and  Y  motions, 

"  •  ^  JTand  Y  being  perpendicular  to 
each  other.  Fig.  79.  In  two 
consecutive  equal  times  (each 
=  dt\  let  dx  and  dx'  =  small 
spaces  due  to  the  X  motion ; 
and  dy  and  CK '=  dy' ,  due  to 
the  Y  motion.  Then  ds  and 

dsf  are  two  consecutive  elements 

FIG.  79.  of  the  curvilinear  motion.     Pro- 

long ds,  making  BE  —  ds\  then  EF  =  d*x,  CF=  d*y,  and 
CO  =  d*s  (EO  being  perpendicular  to  BE).  Also  draw  CL 
perpendicular  to  BG  and  call  CL  d*n.  Call  the  velocity  of 
the  X  motion  vm  its  acceleration  px ;  those  of  the  Y  motion, 
vy  and  py.  Then, 

dx  dy  dvx  _  d*x          ,       _  dvv 

V*  =  df'>    Vy  =  ~dt'>  Px  =  ~dt^  dtf*          Py  =  ~di 

For  the  velocity  along  the  curve  (i.e.,  tangent) 
0  =  ds  ~  dt,  we  shall  have,  since  ds*  =  dx*  +  dy*, 
fds\        (dx' 

V  *  =      -T7-        =      I  -T7 


Hence  v  is  the  diagonal   formed  on  vx  and  vv  (as  in  §  71). 
Let  pt  =  the  acceleration  of  v,  i.e.,  the  tangential  acceleration, 


CURVILINEAR    MOTION    OF   A    MATERIAL    POINT.         75 

then  pt  =.  d*s  -~  dtf,  and,  since  d*s  =  the  sum  of  the  projec- 
tions of  JST^and  CF  on  EC,  i.e.,  d?s  —  d*x  cos  a  -\-  d*y  sin  a, 
we  have 


3S  =  ~J#  cos  a  ~r  ~jf*  sm  a9  1-e-j  Pt  =Px  cos  a  -{-pyBin  a.  (2) 

By  Normal  Acceleration  we  mean  the  rate  of  change  of  the 
velocity  in  the  direction  of  the  normal.  In  describing  the  ele- 
ment AB  =  ds,  no  progress  has  been  made  in  the  direction  of 
the  normal  B II  i.e.,  there  is  no  velocity  \\\  the  direction  of  the 
normal;  but  in  describing  BG  (on  account  of  the  new  direc- 
tion of  path)  the  point  has  progressed  a  distance  CL  (call  it 
d?ri)  in  the  direction  of  the  old  normal  BH  (though  none  in 
that  of  the  new  normal  CI).  Hence,  just  as  the  tang.  ace. 

ds'  —  ds       d*s                                         CL  —  zero       d*n 
=  — -75 —  =  -TTJ,  so  the  normal  accel.  =  -73 =  -j-f. 

(tu  Civ  Cut  Cvt 

It  now  remains  to  express  this  normal  acceleration  (=pn)  in 
terms  of  the  X  and  Y  accelerations.  From  the  figure,  CL 
=  CM-  ML,  i.e., 

dFn  =  d?y  cos  a  —  d?x  sin  a  {since  EF  =  d?x\ ; 

•*•  ~7~a     — ~  77"?  C°S  &   —   "TT  SHI  (X. 

Hence  pn=pycos  a —pxsin  a.  .     .     .     .     .     (3) 

The  norm.  ace.  may  also  be  expressed  in  terms  of  the  tang, 
velocity  v,  and  the  radius  of  curvature  r,  as  follows : 
da'  =•  rda,  or  da  =  da'  -4-  r ;     also      d?n  =  ds'dot,  =  dsf*  •—  r, 

d'n       (dsV  1                        tf 
i.e.,  -rr,  =  l-rr     -,     or     #M  =  r W 


If  now,  Fig.  80,  we  resolve  the  forces  X  =  Mpx  and  T 
=  Mpy,  which  at  this  instant  account  for  the 
X  and  Y  accelerations  (M  =  mass  of  the 


***  material  point),  into  components  along  the 
tangent  and  normal  to  the  curved  path,  we 
shall'  have,  as  their  equivalent,  a  tangential 

teTw,  force 

T  =  Mpx  cos  a  -\-  Mpv  sin  a, 


76  MECHANICS   OF   ENGINEERING. 

and  a  normal  force 

N  =  Mpy  cos  a  —  Mpx  sin  a. 
But  [see  equations  (2),  (3),  and  (4)]  we  may  also  write 

Cl  *?)  Oj 

T=Mpt  =  MTt;    and    N  =  Mpn=  M°~.    .    (5) 

Hence,  if  a  free  material  point  is  moving  in  a  curved  path, 
the  sum  of  the  tangential  components  of  the  acting  forces  must 
equal  (the  mass)  X  tang,  accel.;  that  of  the  normal  components, 
=  (the  mass)  X  normal  accel.  =  (mass)  X  (square  of  veloc.  in 
path)  -r-  (rad.  curv.). 

It  is  evident,  therefore,  that  the  resultant  force  (=  diagonal 
on  T  and  N  or  on  ^Tand  Y,  Fig.  80)  does  not  act  along  the  tan- 
gent at  any  point,  but  toward  the  concave  side  of  the  path  ;  un- 
less r  =  oo. 

Radius  of  curvature.  —  From  the  line  above  eq.  (4)  ^we 
have  d*n  =  ds"  -r-  r\  hence  (line  above  eq.  (3)),  dsf*  -7-r  — 
d?y  cos  a  —  d*x  sin  a  ;  but  cos  a  —  dx  -f-  ds,  and  sin  a  =  dy  -r-  ds, 


i.e., =  ax'a  \  -^  \  =  ax  a  (tan  <*), 

fds"ds\       r/dx\*dt3M 


d  tan 

-.3       .        I      ...    a 

or. 


which  is  equally  true   if,  for  vx  and  tan  a,  we  put  vy  and 

tan  (90°  —  a),  respectively. 

Change  in  the  velocity  square.  —  Since  the  tangential  accelera- 

dv  .          ,  dv 

tion  -3-  =pt)  we  have  ds-j-  =ptds\  i.e., 

-j-dv=ptds,     or    vdv=ptds    and     /.  —  -  —  =  J  ptds.  (T) 

having  integrated  between  any  initial  point  of  the  curve  where 
v  =  c,  and  any  other  point  where  v  =  v.  This  is  nothing 
more  than  equation  (III.),  of  §  50. 


^CURVILINEAR   MOTION    OF    A    MATERIAL    POINT.         77 

75.  Normal  Acceleration.     Second  Method. — Fig.  81.      Let 
C  be  the  centre  of  curvature  and   OD  —  2r.     Let   OB'  be  a 
portion  of  the  oscillatory  parabola  (vertex  at  ^jv. 

0 ;  any  oscillatory  curve  will  serve).  When  \\ 
ds  is  described,  the  distance  passed  over  in 
the  direction  of  the  normal  is  AB ;  for  2<&, 
It  would  be  AB'  =  ±AB  (i.e.,  as  the 
square  of  OB'\  property  of  a  parabola), 
and  so  on.  Hence  the  motion  along  the  normal  is  uniformly 
accelerated  with  initial  velocity  =  0,  since  the  distance  AB^ 
varies  as  the  square  of  the  time  (considering  the  motion  along 
the  curve  of  uniform  velocity,  so  that  the  distance  OB  is  di- 
rectly as  the  time).  If  pn  denote  the  accel.  of  this  uniformly 
accelerated  motion,  its  initial  velocity  being  =  0,  we  have  (eq. 
2,  §  56)  AB  =  \pndt\  i.e.,  ^n  =  2JJ?  -7-  df.__But  from  the 
similar  triangles  ODB  and  GAB  we  have,  AB  :ds::ds:2r, 
hence  2 AB  =  ds*  -r-  r,  .'.  pn  =  ds*  -r-  rdf  =v*~-r. 

76.  Uniform  Circular  Motion.     Centripetal  Force. — The  ve- 
locity being  constant,  j^  must  be  =  0,  and  .*.  T(or^Tif  there 
are  several  forces)  must  =  0.     The  resultant  of  all  the  forces, 
therefore,   must  be  a   normal  force  =  (Me*  -=-  r)  =  a  con- 
stant (eq.  5,  §  Y4).      This  is   called  the  "  deviating  force," 
or  "  centripetal  force ;"  without  it  the  body  would  continue 
in  a  straight  line.     Since  forces    always  occur  in  pairs  (§  3), 
a|  "centrifugal  force,"  equal    and   opposite  to    the    "centri- 
petal" (one  being  the  reaction  of    the  other),  will  be  found 
among  the  forces  acting  on  the  body  to  whose  constraint  the 
deviation  of  the  first  body  from  its  natural  straight  course  is 
due.     For  example,  the  attraction  of.  the  earth  on  the  moon 
acts  as  a  centripetal  or  deviating  force  on  the  latter,  while  the 
equal  and  opposite  force  acting  on  the  earth  may  be  called 

Cvx  the  centrifugal.     If  a  small  block  moving  on  a 

^  smooth  horizontal  table  is  gradually  turned  from 
its  straight  course  AB  by  a  fixed  circular  guide, 
tangent  to  AB  at  B,  the  pressure  of  the  guide 
against  the  block  is  the  centripetal  force  J/b'-r-  r 
FIG.  82.        directed  toward  the  centre  of  curvature,  while 


> 

CENTRIP.  It- 


78 


MECHANICS    OF   ENGINEERING. 


FIG.  83. 


the  centrifugal  force  J/b2  -f-  r  is  the  pressure  of  the  block 
against  the  guide,  directed  away  from  that  centre.  The  cen- 
trifugal force,  then,  is  never  found  among  the  forces  acting  on 
the  body  whose  circular  motion  we  are  dealing  with. 

The  Conical  Pendulum,  or  governor-ball. — Fig.  83.  If  a 
material  point  of  mass  ==  M  =  G  -r-  g,  suspended  on  a  cord  of 
length  =  I,  is  to  maintain  a  uniform  cir- 
cular motion  in  a  horizontal  plane,  with  a 
given  radius  r,  under  the  action  of  gravity 
and  the  cord,  required  the  velocity  c  to  be 
given  it.  At  B  we  have  the  body  free. 
The  only  forces  acting  are  G  and  the  cord- 
tension  P.  The  sum  of  their  normal  com- 
ponents, i.e.,  21V,  must  =  Me*  -5-  /*,  i.e.,  P  sin  a  =  Me*  -f-  r ; 
but,  since  2  (vert,  comps.)  =  0,  P  cos  a  =  G.  Hence 
£rtan  a  =  Gc*-±-  gr\  .*.  c=  Vgrt&n  a.  Let  u  =  number  of 
revolutions  per  unit  of  time,  then  u-=.c-^-  %7tr  =  Vg  -r-  2?r  Vh ; 
i.e.,  is  inversely  proportional  to  the  (vertical  projection)*  of 
the  cord-length.  The  time  of  one  revolution  is  =  1  -j-  u. 

Elevation  of  the  outer  rail  on  railroad  curves  (considera- 
tions of  traction  disregarded). — Consider  a  single  car  as  a 
material  point,  and  free,  having  a  given  p\ 

velocity   =   c.      P   is   the   rail-pressure  \u 

against  the  wheels.     So  long  as  the  car  £ ~r — ^  \ 

follows  the  track  the  resultant  R  of  P 

and  G  must  point  toward  the  centre  of 

curvature  and  have  a  value  =  J^b2  -j-  r. 

But  7?—  G  tan  a,  whence  tan  a  —  c2-=-  gr. 

If  therefore  the  ties  are,  placed   at  this 

angle  a  with  the  horizontal,  the  pressure 

will  come  upon  the  tread  and  not  on  the  flanges  of  the  wheels  ; 

in  other  words,  the  car  will  not  leave  the  track.    (This  is  really 

the  same  problem  as  the  preceding.) 

Apparent  weight  of  a  body  at  the  equator. — This  is  less  than 
the  true  weight  or  attraction  of  the  earth,  on  account  of  the 
uniform  circular  motion  of  the  body  with  the  earth  in  its 
diurnal  rotation.  If  the  body  hangs  from  a  spring-balance. 


FIG.  84. 


CURVILINEAR   MOTION    OF   A    MATERIAL    POINT.         79 

whose  indication  is  G  Ibs.  (apparent  weight),  while  the  true 
attraction  is  G'  Ibs.,  we  have  G'  —  G  =  Me*  -=-  r.  For  M 
we  may  use  G  -r-  g  (apparent  values);  for  r  about  20,000,000 
ft.;  for  c,  25,000  miles  in  24  hrs.,  reduced  to  feet  per  second. 
It  results  from  this  that  G  is  <  G'  by  -^^G'  nearly,  and 
(since  173  =  289)  hence  if  the  earth  revolved  on  its  axis  seven- 
teen times  as  fast  as  at  present,  G  would  =  0,  i.e.,  bodies 
would  apparently  have  no  weight,  the  earth's  attraction  on 
them  being  just  equal  to  the  necessary  centripetal  or  deviating 
force  necessary  to  keep  the  body  in  its  orbit. 

Centripetal  force  at  any  latitude. — If  the  earth  were  a  ho- 
mogeneous liquid,  and  at  rest,  its  form  would  be  spherical ;  but 
when  revolving  uniformly  about  the  polar  diameter,  its  form 
of  relative  equilibrium  (i.e.,  no  motion  of  the  particles  relatively 
to  each  other)  is  nearly  ellipsoidal,  the  polar  diameter  being  an 
axis  of  symmetry. 

Lines  of  attraction  on  bodies  at  its  surface  do  not  intersect 
in  a  common  point,  and  the  centripetal  force  requisite  to  keep 
a  suspended  body  in  its  orbit  (a  small  circle  of  the  ellipsoid), 
at  any  latitude  ft  is  the  resultant,  -Z\T,  of  the  attraction  or  true 
weight  G'  directed  (nearly)  toward  the  centre,  and  of  G  the 
tension  of  the  string.  Fig.  85.  G  =  the  apparent  weight,  in- 
dicated by  a  spring-balance  and  MA  is  its  _ ,^G 

line  of  action  (plumb-line)  normal  to  the  /<?^~  """P^*^ 
ocean  surface.  Evidently  the  apparent  / 
weight,  and  consequently  g,  are  less  than  (~ — 
the  true  values,  since  N  must  be  perpen- 
dicular to  the  polar  axis,  while  the  true 
values  themselves,  varying  inversely  as 
the  square  of  MC,  decrease  toward  the  equator,  hence  the  ap- 
parent values  decrease  still  more  rapidly  as  the  latitude  dimin* 
ishes.  The  following  equation  gives  the  apparent  g  tor  any 
latitude  /3,  very  nearly  (units,  foot  and  second): 

g  =  32.1808  -  0.0821  cos  2/3. 

(The  value  32.2  is  accurate  enough  for  practical  purposes.', 
Since  the  earth's  axis  is  really  not  at  rest,  but  moving  about 


80 


MECHANICS    OF   ENGINEERING. 


the  sun,  and  also  about  the  centre  of  gravity  of  the  moon  and 
earth,  the  form  of  the  ocean  surface  is  periodically  varied,  i.e., 
the  phenomena  of  the  tides  are  produced. 

77.  Cycloidal  Pendulum. — This  consists  of  a  material  point 
at  the  extremity  of  an  imponderable,  flexible,  and  inextensible 
cord  of  length  =  I,  confined  to  the  arc  of  a  cycloid  in  a  ver- 
tical plane  by  the  cycloidal  evolutes  shown  in  Fig.  86.  Let 
the  oscillation  begin  (from  rest)  at  J.,  a  height  =  h  above  0 


*he  vertex.  On  reaching  any  lower  point,  as  E  (height  ==  & 
above  0\  the  point  has  acquired  some  velocity  v9  which  is  at 
this  instant  increasing  at  some  rate  =  pt.  Now  consider  the 
point  free,  Fig.  87;  the  forces  acting  are  P  the  cord-  tension, 
normal  to  path,  and  G  the  weight,  at  an  angle  cp  with  the 
path.  From  §  74,  eq.  (5),  2T  =  Mpt  gives 

G  cos  cp  +  P  cos  90°  —  (G  -r-  g)pt\  .\pt  —  g  cos  (p 
Hence  (eq.  (7),  §  74),  vdv  —  ptds  gives 

<vdv  =  g  cos  cpds  ;  but  ds  cos  <p  =  —  dz  ;  /.  vd/v  =  —  gdz. 
Summing  between  A  and  B,  we  have 


or  v  = 


the  same  as  if  it  had  fallen  freely  from  rest  through  the  height 
h  _^  2.  (This  result  evidently  applies  to  any  form  of  path 
when^  besides  the  weight  G,  there  is  but  one  other  force,  and 
that  always  normal  to  the  path.} 

From  'SN  =  Mv*  -=-  n,  we  have  P  —  G  sin  <p  =  Mv*  -r-r^ 


CURVILINEAR    MOTION    OF   A    MATERIAL    POINT.         81 

whence  P,  the  cord-tension  at  any  point,  may  be  found  (here 
TI—  the  radius  of  curvature  at  any  point  =  length  of  straight 
portion  of  the  cord). 

To  find  the  time  of  passing  from  A  to  0,  a  half-oscillation, 
substitute  the  above  value  of  va  in  v  =  ds  -~  dt,  putting  ds* 
=  dx*  +  dz\  and  we  have  df  =  (dx*  +  da*)  ~  \Zg(h  —  s)~]. 
To  find  dx  in  terms  of  dz,  differentiate  the  equation  of  the 
curve,  which  in  this  position  is 


x  =  r  ver.  sin."1  (2  -r-  r)  -\-  V%rz  —  z*  ; 
whence 

rdz  (r  —  z]dz     _  (%r  —  z) 

VZrz  —  z*       V2rz  —  z*  .'  VZrz  — 


(r  =  radius  of  the  generating  circle).     Substituting,  we  have 


_ 

hz-z*' 

r°*      F  (*   d*        /rvh       i  «       F 

.-.      t  =  \  /  -  I       ._  =  \  /  -\  ver.  sin.-1  -rj=  n\  /  -. 

LA         V  gJ*     Vhz-z*      V  g  L0  tA        \g 

Hence  the  whole  oscillation  occupie^  a  time  =  n  VI  -r-  g 
(since  I  =  4r).  This  is  independent  of  A,  i.e.,  the  oscillations 
are  isochronal.  This  might  have  been  proved  by  showing  that 
pt  is  proportional  to  OB  measured  along  the  curve  /  i.e.,  that 
the  motion  is  harmonic.  (§  59,  Prob.  2.) 

78.  Simple  Circular  Pendulum.  —  If  the  material  point  oscil- 
lates in  the  arc  of  a  circle,  Fig.  88,  proceeding 
as  in  the  preceding  problem,  we  have  finally,     fs^-  —  ^     j 
after  integration  by  series,  as  the  time  of  a  full        \         yf 

oscillation  ,  j  x  v.\B/  H 

,r°.        fir.     i    A       9   A'.          -i  '~dr      -*-* 

*\J=*VffL  ^8  '  1  +  256  17'  .?** 

Hence  for  a  small  A  the  time  is  nearly  7t  VI  -^  g,  and  the  os- 


82  MECHANICS    OF  ENGINEERING. 

cillations  nearly  isochronal.  (For  the  Compound  Pendulum, 
see  §117.) 

79.  Change  in  the  Velocity  Square. — From  eq.  (7),  §  74,  we 
have  %(v*  —  c2)  -=fptds.  But,  from  similar  triangles,  du  be- 
ing the  projection  of  any  ds  of  the  path  upon  the  resultant 
force  R  at  that  instant,  Rdu  =  Tds  (or,  Prin.  of  Yirt.  Yels. 
§  62,  Rdu  =  Tds  +  N  X  0).  T  and  Nwe  the  tangential  and 
normal  components  of  R.  Fig.  89.  Hence,  finally, 

£Mv*  —  \  Me*  =fRdu, (a) 

for  all  elements  of  the  curve  between  any  two  points.     In  .gen- 
eneral  R  is  different  in  amount  and  direc- 
t,v      tion  for  each  ds  of  the  path,  but  du  is  the 
x,   distance  through  which  R  acts,  in  its  own 
FIG.  89.  direction,  while  the  body  describes  any  ds ; 

Rdu  is  called  the  work  done  by  R  when  ds  is  described  by  the 
body.  The  above  equation  is  read  :  The  difference  between  the 
initial  and  final  kinetic  energy  of  a  body  =  the  work  done  lyy 
the  resultant  force  in  that  portion  of  the  path. 

(These  phrases  will  be  further  spoken'  of  in  Chap.  YI.) 
Application  of  equation  (a)  to  a  planet  in  its  orbit  about 
the  sun. — Fig.  90.  Here  the  only  force  at  any  instant  is  the  at 
traction  of  the  sun  R  =  C  -=-  u*  (see  Prob.  3,  §  59), 
where  C  is  a  constant  and  u  the  variable  radius    hs. 
vector.      As  u  diminishes,  v  increases,  therefore 
dv  and  du  have  contrary  signs ;  hence  equation   |  tf^V 
(a)  gives   (c  being  the  velocity  at  some  initial  " 
point  0) 

~V<te-£l;.fc 


>|—  1  -1    — i 

|_^"  ' '  —J,  which  is  independ- 
ent of  the  direction  of  the  initial  velocity  c. 

NOTE. — If  u0  were  =  infinity,  the  last  member  of  equation  (5)  would  re- 
duce to  C  ±  «lt  and  is  numerically  the  quantity  called  potential  in  the 
theory  of  electricity. 


CURVILINEAR   MOTION   OF   A   MATERIAL   POINT.        83 


Application  of  eq.   (a)  to  a  projectile  in  vacuo. — <?,  the. 
body's  weight,  is  the  only  force  acting,  and  0 
therefore  =  R,  while  M  =  G  -f-  g*     There- 
fore equation  (a)  gives 


ff  * 

0*.  v.  =  Vc*  +  2<72/a,  which  is  independent  of  FlG  91 

the  angle,  <*,  of  projection. 

Application  of  equation  (a)  to  a  body  sliding,  without  fric* 
tion,  on  a  fixed  curved  guide  in  a  vertical  plane ;  initial  velo- 
city =  c  at  0. — Since  there  is  some  pressure  at  each  point  be- 
tween the  body  and  the  guide,  to  consider  the  body  free  in 
space,  we  must  consider  the  guide  removed  and  that  the  body 

describes  the  given  curve  as  a  re- 
sult of  the  action  of  the  two  forces, 
its  weight  6r,  and  the  pressure  P, 
of  the  guide  against  the  body.  G 
is  constant,  while  P  varies  from 
point  to  point,  though  always  (since 
there  is  no  friction)  normal  to  curve. 
At  any  point,  R  being  the  resultant 
of  G  and  P,  project  ds  upon  R,  thus  obtaining  du;  on  #, 
thus  obtaining  dy  ;  on  P,  thus  obtaining  zero.  But  by  the 
principle  of  virtual  velocities  (see  §  62)  we  have  Rdu  =  Gdy 
-f-  P  X  zero  =  Gdy,  which  substituted  in  eq.  (a)  gives 


and  therefore  depends  only  on  the  vertical  distance  fallen 
through  and  the  initial  velocity,  i.e.,  is  independent  of  the 
form  of  the  guide. 

As  to  the  value  of  P,  the  mutual  pressure  between  the  guide 
and  body  at  any  point,  since  ^N  must  equal  Mv*  ~  r,  r  being 
the  variable  radius  of  curvature,  we  have,  as  in  §  77, 

P  —  G  sin  cp  =  Mv*  H-  r ;    .-.  P  =  #[sin  g>+  (v*  ~  gr)]. 
As,  in  general,  <p  and  r  are  different  from  point  to  point  of 


84  MECHANICS    OF    ENGINEERING. 

the  path,  P  is  not  constant.  (The  student  will  explain  how  P 
may  be  negative  on  parts  of  the  curve,  and.  the  meaning  of 
this  circumstance.) 

80.  Projectiles  in  Vacuo. — A  ball  is  projected  into  the  air 
Y  (whose  resistance   is  neglected,  hence  the 

phrase  in  vacuo)  at  an  angle  =  aQ  with  the 
horizontal ;  required  its  path ;  assuming  it 
confined  to  a  vertical  plane.  Resolve  the 
motion  into  independent  horizontal  (X) 
and  vertical  (Y)  motions,  6r,  the  weight, 
the  only  force  acting,  being  correspondingly  replaced  by  its 
horizontal  component  —  zero,  and  its  vertical  component 
=  —  6r.  Similarly  the  initial  velocity  along  X '=  cx  =•  c  cos  #0, 
along  Y,  —  cy  =  csin  or0.  The  JT  acceleration  =px  —  0  -h  M 
=  0,  i.e.,  the  X  motion  is  uniform,  the  velocity  vx  remains 
=  cx  =  c  cos  a0  at  all  points,  hence,  reckoning  the  time  from  0, 
at  the  end  of  any  time  t  we  have 

x  =  c(cos  a0)t (1) 

In  the  Y  motion,  pv  =  (—  @)  -r-  M  =  —  g,  i.e.,  it  is  uniformly 
retarded,  the  initial  velocity  being  cy  =  c  sin  a0 ;  hence,  after 
any  time  £,  the  Y  velocity  will  be  (see  §  56)  vy  =  c  sin  a0  —  gt, 
while  the  distance 

Between  (1)  and  (2)  we  may  eliminate  tf,  and  obtain  as  the 
equation  of  the  trajectory  or  path 

gx* 

y  =  x  tan  a.  —  T— ^ — 5 — . 
2<r  cos  <x0 

For  brevity  put  e*  =  %gh,  h  being  the  ideal  height  due  to  the 
velocity  c,  i.e.,  c2  -f-  2</  (see  §  53  ;  if  the  ball  were  directed  ver- 
tically upward,  a  height  h  =  c*  -r-  2<7  would  be  actually  at- 
tained, of0being  =  90°),  and  we  have 

y  =  x  tan  a0  —  --j 5 (3) 

4A  cos    a0 

This  is  easily  shown  to  be  the  equation  of  a  parabola,  with  its 
axis  vertical. 


CURVILINEAR   MOTION    OF   A   MATERIAL   POINT.         85 

The  horizontal  range. — Fig.  94.  Putting  y  =  0  in  equa- 
tion (3),  we  obtain 

r  x     -\ 

x    tan  a.  —  Tl 5 —     =0.  Y      c 

L  4A  cos2  tf0J  JL 

which  is  satisfied  both  by  x  =  0  (i.e.,  at  the 

origin),  and  by  x  =  4A  cos  or0  sin  <*0.     Hence 

the  horizontal  range  for  a  given  c  and  a0  is  Fl°-  94- 

xr  =  4A  cos  aQ  sin  a0  =  2A  sin  2a0. 

For  a0  =  45°  this  is  a  maximum  (<?  remaining  the  same), 
being  then  =  2A.  Also,  since  sin  2<*0  =  sin  (180°  —  2<*0)  = 
sin  2(90°  —  arc),  therefore  any  two  complementary  angles  of 
projection  give  the  same  horizontal  range. 

Greatest  height  of  ascent ;  that  is,  the  value  of  y  maximum, 
=  ym. — Fig.  94.  Differentiate  (3),  obtaining 

dy_  • x__ 

dx  °      2A  cos2  of0' 

which,  put  =  0,  gives  a?  =  2A  sin  a0  cos  or0,  and  this  value  of 
x  in  (3)  gives  ym  —  h  sin2  ac. 

(Let  the  student  obtain  this  more  simply  by  considering  the 
T  motion  separately.) 

To  strike  a  given  point ;  c  being  given  and  a0  required. — 
Let  x'  and  y'  be  the  co-ordinates  of  the  given  point,  and  or/ 
the  unknown  angle  of  projection.  Substitute  these  in  equa- 
tion (3),  h  being  known  =  c2  -=-  2^,  and  we  have 

y'  =  xf  tan  &'  —  -rj 5 — 7.     Put  cos2  <*/  =  ^ — : 5—7, 

4A  cos2  «.'  1  +  tan2  or/' 

and  solve  for  tan  or/,  whence 


tan  or'0  =    2A  ±  1/4A*  -  a/a  -  4%r    -5-  x'  .  . 


Evidently,  if  the  quantity  under  the  radical  in  (4)  is  negative, 
tan  or/  is  imaginary,  i.e.,  the  given  point  is  out  of  range  with 
the  given  velocity  of  projection  c  =  V2ghj  if  positive,  tan  a0' 
has  two  values,  i.e.,  two  trajectories  may  be  passed  through 
the  point  ;  while  if  it  is  zero,  tan  or/  has  but  one  value. 

The  envelope,  for  all  possible  trajectories  having  the  same 


86 


MECHANICS    OF    ENGINEERING. 


FIG.  95. 


initial  velocity  c  (and  hence  the  same'A);  i.e.,  the  curve  tan- 
gent to  them  all,  has  but  one  point  of  contact  with  any  one  of 
them;  hence  each  point  of  the  envelope,  Fig.  95,  must  have 

co-ordinates  satisfying  the  con- 
dition, 4A2  —  a/2  —  4:hy'  =  0  ;  i.e. 
(see  equation  (4) ),  that  there  is 
but  one  trajectory  belonging  to  it. 
Hence,  dropping  primes,  the 
equation  of  the  envelope  is  4A2  — 
a?2  —  4Ay  =  0.  Now  take  0"  as  a 
new  origin,  a  new  horizontal  axis  X" ,  and  reckon  y"  positive 
downwards ;  i.e.,  substitute  x  =  x"  and  y  =  h  —  y" .  The 
equation  now  becomes  a?//a  —  ±hy" ;  evidently  the  equation  of 
a  parabola  whose  axis  is  vertical,  whose  vertex  is  at  O'1 ',  and 
whose  parameter  =  4A  =  double  the  maximum  horizontal 
range.  0  is  therefore  its  focus. 

The  range  on  an  inclined  plane. — Fig,  96.      Let   OC  be 
the  trace  of  the  inclined  plane  ;  its  equation    y 
is  y  =  x  tan  /?,  which,  combined  with  the 
equation  of  the  trajectory  (eq.  3),  will  give 
the   co-ordinates   of    their   intersection    C. 
That  is,  substitute  y  =  x  tan  ft  in  (3)  and 
solve  for  a?,  which  will  be  the  abscissa  ajn  of  C.     This  gives 

sin  acn      sin  0      sin  (a0  —  ft) 

cos  CKO  cos  ft ' 
sin 


=  tan  a0  —  tan  ft  = 


j   7  o 

4A  cos   «  cos  aQ      cos 

.*.  xl  =  4Acos  tf0  sift  («0  —  fi)  -r-  cos  /?,  and  the  range 
which  =  a?,  -^  cos  /?,  is  =  (4A  -r-  cos2  /?)  cos  orc  sin  (<*0  —  /?).  (5) 
The  maximum  range  on  a  given  inclined  plane,  /?,  c  (and 
.'.  A),  remaining  constant,  while  orc  varies. — That  is,  required 
the  value  of  a0  which  renders  OC  a  maximum.  Differential 
ing  (5)  with  respect  to  <*0,  putting  this  derivative  —  0,  we  have 
[4A  -r-  cosa  ft]  [cos  a0  cos  (at  —  ft)  —  sin  a0  sin  (<r0  —  /5)]  =  0 ; 
whence  cos  [>0  +  (a,  —  /?)]  =  0;  i.e.,  2«0  —  /?  =  90°;  or, 
a0  =  45°  -f-  |y5,  for  a  maximum  range.  By  substitution  this 
maximum  becomes  known. 


The  velocity  at  any  point  of  the  path  is  v  =  Vvx*  -j-  vv*  = 


CURVILINEAR   MOTION    OF   A    MATERIAL   POINT.         87 


r  c2  —  2ctg  sin  <*Q  -\-y*t*  (see  the  first  part  of  this  §  80)  ;  while 
the  time  of  passage  from  0  to  any  point  whose  abscissa  is  x  is 
t  =  x  -r-  c  cos  ^0  ;  obtained  from  equation  (1).  E.g.,  to  reach 
the  point  B,  Fig.  94,  we  pat  x  =  &r  =  4A  sin  a  cos  a,  and  ob- 
-f- ^.  This  will  give  the  velocity  at  ^  = 


tain  tr  =  2c  sin 

t7?  =  G. 


81.  Actual  Path  of  Projectiles.  —  Small  jets  of  water,  so  long  as 
they  remain  unbroken,  give  close  approximations  to  parabolic 
paths,  as  also  any  small  dense  object,  e.g.,  a  ball  of  metal,  hav- 
ing a  moderate  initial  velocity.  The  course  of  a  cannon-ball, 
however,  with  a  velocity  of  1200  to  1400  feet  per  second  is 
much  affected  by  the  resistance  of  the  air,  the  descending 
branch  of  the  curve  being  much  steeper  than  the  ascending; 
see  Fig.  96$.  The  equation  of  this  curve  has  not  yet  been 
determined,  but  only  the  expression  for  the  slope  (i.e., 
dy  :  dx)  at  any  point.  See  Professor  Bart- 
lett's  Mechanics,  §  151  (in  which  the  body 
is  a  sphere  having  no  motion  of  rotation). 
Swift  rotation  about  an  axis,  as  well  as  an 
unsymmetrical  form  with  reference  to  the 
direction  of  motion,  alters  the  trajectory 
still  further,  and  may  deviate  it  from  a  vertical  plane. 
presence  of  wind  would  occasion  increased  irregularity. 
Johnson's  Encyclopaedia,  article  "  Gunnery." 


fi      ...--T- 


FIG.  96a. 


The 
See 


82.  Special  Problem  (imaginary ;  from  Weisbach's  Mechan- 
ics. The  equations  are  not  homogeneous). — Suppose  a  ma- 
terial point,  mass  =  J^,  to  start 
from  the  point  0,  Fig.  97,  with 
a  velocity  =  9  feet  per  second 
along  the  —  Y  axis,  being  sub- 
•  jected  thereafter  to  a  constant 
attractive  X  force,  of  a  value  X 
=  1%M,  and  to  a  variable  Y 
force  increasing  with  the  time 
(in  seconds,  reckoned  from  O), 


. 


y     x 


Fio.  97. 


88  MECHANICS    OF   ENGINEERING. 

viz.,  Y=SMt.     Eequired  the  path,  etc.     For  the  X  motion 
we  have  px  =  X  ~  M  =  12,  and  hence 


and 


/»Vx  s*t  s>t 

I     dvx  =  /  pxdt  =  12  /   ^  ;  i.e.,  ^  =  12*: 

t'o  t/o 

/*»  ^.f  s>t 

[  J^    dx  =J^  vxdt ;  i.e.,  x  ----  12 ^  ^  =  6*a.  .     (1) 

/^y  /^^ 

6^yu=8  /  &ft: 
9  i/« 

i.e.,  vv  +  9  =  4*3,     and    J    dy  =    f  vydt; 

r*  r* 

.-.  y  —  4  /  z?<^  —  9  /  ^*,     or    y  =  4*8  —  9*.  .     .     (2) 

e/o  t/« 

Eliminate  t  between  (1)  and  (2),  and  we  have,  as  the  equa- 
tion of  the  path. 


which  indicates  a  curve  of  the  third  order. 
The  velocity  at  any  point  is  (see  §  74,  eq.  (1)  ) 


v  =     vx*  +  vv*  =  4£3  +  9  ......     (4) 

length  of  curve  measured  from  0  will  be  (since  ?)  = 


The  «^6,  tan  or,  at  any  point  =vy-irvx=  (4f  —  9)  -f-  12£, 
6?  tan  a       4^  +  9 


radius  of  curvature  at  any  point  (§74,  eq.  (6)),  sub- 
stituting ^  =  12£,  a)so  from  (4)  and  (6),  is 

,      .     .     (7) 


and  the  normal  acceleration  =  v*  -i-  r  (eq.  (4),  §  74),  becomes 
from  (4)  and  (7)  pn  =  12  (ft.  per  square  second),  a  constant. 
Hence  the  centripetal  or  deviating  force  at  any  point,  i.e.,  the 


CURVILINEAR    MOTION    OF   A    MATERIAL   POINT.         89 

the  forces  X  and  I7,  is  the  same  at  all  points,  and  = 


From  equation  (3)  it  is  evident  that  the  curve  is  symmetrical 
about  the  axis  X.  Negative  values  of  t  and  s  would  apply  to 
points  on  the  dotted  portion  in  Fig.  97,  since  the  body  may  be 
considered  as  having  started  at  any  point  whatever,  so  long  as 
all  the  variables  have  their  proper  values  for  that  point. 

(Let  the  student  determine  how  the  conditions  of  this  motion 
could  be  approximated  to  experimentally.) 

83/  Relative  and  Absolute  Velocities.—  Fig.  98.     Let  J^Tbe  a 

material  point  having  a  uniform  motion  of  velocity  vt  along  a 

straight  groove  cut  in  the  deck  of  a  steamer,  which  itself  has 

a  uniform  motion  of  translation,  of  velocity  v»  over  the  bed  of 

a  river.     In  one  second  M  ad- 

vances  a   distance   v9  along   the 

groove,  which  simultaneously  has 

moved  a  distance  v1  —  AB  with 

the  vessel.     The  absolute  path  of 

M  during  the  second  is  evidently  FIG.  98. 

w  (the  diagonal  formed  on  vl  and  v^,  which  may  therefore  be 

called  the  absolute  velocity  of  the  body  (considering  the  bed 

of  the  river  as  fixed)  ;  while  v9  is  its  relative  velocity,  i.e.,  rela- 

tive to  the  vessel.     If  the  motion  of  the  vessel  is  not  one  of 

translation,  the  construction  still  holds  good  for  an  instant  of 

time,  but  v1  is  then  the  velocity  of  that  point  of  the  deck  over 

which  M  is  passing  at  this  instant,  and  v9  is  M'  s  velocity  rela- 

tively to  that  point  alone. 

Conversely,  if  M  be  moving  over  the  deck  with  a  given 
absolute  velocity  =  w,  VT  being  that  of  the  vessel,  the  relative 
velocity  ^  may  be  found  by  resolving  w  into  two  components, 
one  of  which  shall  be  vt  ;  the  other  will  be  vy 

If  w  is  the  absolute  velocity  and  direction  of  the  wind,  the 
vane  on  the  mast-head  will  be  parallel  to  MT,  i.e.,  to  v9  the 
relative  velocity  ;  while  if  the  vessel  be  rolling  and  the  mast- 
head therefore  describing  a  sinuous  path,  the  direction  of  the 
vane  varies  periodically. 


90  MECHANICS    OF    ENGINEERING. 

Evidently  the  effect  of  the  wind  on  the  sails,  if  any,  will 
depend  on  vt  the  relative,  and  not  directly  on  w  the  absolute, 
velocity.  Similarly,  if  w  is  the  velocity  of  a  jet  of  water,  and 
vl  that  of  a  water-wheel  channel,  which  the  water  is  to  enter 
without  sudden  deviation,  or  impact,  the  channel-partition 
should  be  made  tangent  to  vt  and  not  to  w. 

Again,  the  aberration  of  light  of  the  stars  depends  on  the 
same  construction  ;  v1  is  the  absolute  velocity  of  a  locality  of  the 
earth's  surface  (being  practically  equal  to  that  of  the  centre) ; 
w  is  the  absolute  direction  and  velocity  of  the  light  from  a 
certain  star.  To  see  the  star,  a  telescope  must  be  directed 
along  MT,  i.e.,  parallel  to  vt  the  relative  velocity ;  just  as  in 
the  case  of  the  moving  vessel,  the  groove  must  have  the  direc- 
tion MT,  if  the  moving  material  point,  having  an  absolute 
velocity  w,  is  to  pass  down  the  groove  without  touching  its 
sides.  Since  the  velocity  of  light  =  192,000  miles  per  second 
=  w,  and  that  of  the  eartli  in  its  orbit  =  19  miles  per  second 
=  Vv  the  angle  of  aberration  SMT,  Fig.  98,  will  not  exceed 
•20  seconds  of  arc ;  while  it  is  zero  when  w  and  v1  are  parallel. 

Returning  to  the  wind  and  sail-boat,*  it  will  be  seen  from 
Fig.  98  that  when  v1  =  or  even  >  w,  it  is  still  possible  for  v^ 
to  be  of  such  an  amount  and  direction  as  to  give,  on  a  sail 
properly  placed,  a  small  wind-pressure,  having  a  small  fore-and 
aft  component,  which  in  the  case  of  an  ice-boat  may  exceed 
the  small  fore-and-aft  resistance  of  such  a  craft,  and  thus  vl  will 
be  still  further  increased  ;  i.e.,  an  ice-boat  may  sometimes  travel 
faster  than  the  wind  which  drives  it.  This  has  often  been 
proved  experimentally  on  the  Hudson  liiver. 

*  See  §  571  for  the  mechanics  of  the  sail-boat. 


MOMENT   OF   INEKTIA.  91 


CHAPTEK  IY. 

MOMENT  OF  INERTIA. 

[NOTE. — For  the  propriety  of  this  term  and  its  use  in  Mechanics,  see 
§§  114,  216,  and  229 ;  for  the  present  we  deal  only  with  the  geometrical 
nature  of  these  two  kinds  of  quantity.] 

85.  Plane  Figures. — Just  as  in  dealing  with  the  centre  of 
gravity  of  a  plane  figure  (§  23),  we  had  occasion  to  sum  the 
series  fzdF,  z  being  the  distance  of  any  element  of  area,  dF, 
from  an  axis ;  so  in  subsequent  chapters  it  will  be  necessary  to 
know  the  value  of  the  series  fz*dF  for  plane  figures  of  various 
shapes  referred  to  various  axes.      This  summation  fz*dF  of 
the  products  arising  from  multiplying  each  elementary  area  of 
the  figure  by  the  square  of  its  distance  from  an  axis  is  called 
the  moment  of  inertia  of  the  plane  figure  with  respect  to  the 
axis  in  question  •  its  symbol  will  be  7.     If  the  axis  is  perpen- 
.dicular  to  the  plane  of  the  figure,  it  may  be  named  the  polar 
mom.  of  inertia  (§  94)  ;  if  the  axis  lies  in  the  plane,  the  rec- 
tangular mom.  of  inertia  (§§  90-93).     Since  the  7  of  a  plane 
figure  evidently  consists  of  four  dimensions  of  length,  it  may 
always  be  resolved  into  two  factors,  thus  I  =  Fl£,  in  which 
F=  total  area  of  the  figure,  while  &  =   1/7 -f-  F,  is  called  the 
radius  of  gyration,  because  if  all  the  elements  of  area  were 
situated  at  the  same    radial    distance,  &,  from   the  axis,  the 
moment  of  inertia  would  still  be  the  same,  viz., 

7 = JTtdF  =  WfdF  =  F~k\ 

86.  Rigid  Bodies. — Similarly,  in  dealing  with   the   rotary 
motion  of  a  rigid  body,  we  shall  need  the  sum  of  the  series 
>fffdM<  meaning  the  summation  of  the  products  arising  from 
multiplying  the  mass  dM oi  each  elementary  volume  dV  of  a 


92 


MECHANICS    OF   ENGINEERING. 


rigid  body  by  the  square  of  its  distance  from  a  specified  axis. 
This  will  be  called  the  moment  of  inertia  of  the  lody  with 
respect  to  the  particular  axis  mentioned  (often  indicated  by  a 
subscript),  and  will  be  denoted  by  /.  As  before,  it  can  often 
be  conveniently  written  JST^a,  in  which  3f  is  the  whole  mass, 
and  k  its  "radius  of  gyration"  for  the  axis  used,  k  being 
=  VI '-r-  M.  If  the  body  is  homogeneous,  the  heaviness,  y,  of 
all  its  particles  will  be  the  same,  and  we  may  write 

i=/fttur=  (r  •*  g)ftfdv=  (y  -  g)  m. 

87.  If  the  body  is  a  homogeneous  plate  of  an  infinitely  small 
thickness  =  r,  and  of  area  =  F,  we  have  I  =  (y  -j-  g)fp*d  V 
=  (y  -r-  g)rfp*dF\  i.e.,  =  (y  -~  g)  X  thickness  X  mom.  iner- 
tia of  the  plane  figure. 

88.  Two  Parallel  Axes.     Reduction  Formula.— Fig.  99.    Let 

Z  and  Z'  be  two  parallel  axes.  Then  Iz 
=fp*d-M,8LndIz,=tfp'*dM.  But  d  being 
the  distance  between  the  axes,  so  that  <za 
-]-  5a=  £?3,  we  have  p'a=  (x  —  &)a+(y— &)a 

=fp*dM+  tffdM  -  ZafxdM 

-ZbfydM.    .     (1) 

But/pWJf  =  Iz,fdM=  M,  and  from  the 
theory  of  the  centre  of  gravity  (see  §  23,  eq.  (1),  knowing  that 
dM=yd  V-r-g,  and  /.  that  [fyd  V~\  -j-  g=M)  we  have/a%Of 
=  M.X  SLudJydM  =  -3/y  5  hence  (1)  becomes 

in  which  a  and  5  are  the  x  and  £/  of  the  axis  Z'\  x  and  y  refer 
to  the  centre  of  gravity  of  the  body.  If  Z  is  a  gravity-axis 
(call  it  <?),  both  x  and  y  =  0,  and  (2)  becomes 

2Z,  =  lg-\-Mcl?.  ...     or    kz>*  —  kg  -{-d*.    .     .     (3) 

It  is  therefore  evident  that  the  mom.  of  inertia  about  a  grav- 
ity-axis is  smaller  than  about  any  other  parallel  axis. 

Eq.  (3)  includes  the  particular  case  of  a  plane  figure,  by 


FIG.  99. 


MOMENT   OF   INERTIA. 


93 


writing  area  instead  of  mass,  i.e.,  when  Z  (now  g)  is  a  gravity- 
axis, 

f,,=Ig  +  Fd* (4) 

89.  Other  Reduction  Formulae;  for  Plane  Figures. — (The  axes 
here  mentioned  lie  in  the  plane  of  the  figure.)  For  two  sets 
of  rectangular  axes,  having  the  same  origin,  the  following  holds 
good.  Fig.  100.  Since 

•Ix=f!fdF,     and     IT=/afdF9 
we  have  Ix  +  IY  =f(a?  +  y*)dF. 

Similarly,  Iv  +  Iv  =f(v*  +  u*)dF. 

But  since  the  a?  and  y  of  any  dFliwe  the  same  hypothennse  as 
the  u  and  v,  we  have  tf  +  u*  =  a?a+  y2;  .-.  fx+JY~ 


FIG.  100. 


FIG.  lOOa. 


Let  XT)e  an  axis  of  symmetry  •  then,  given  Ix  and  IT  (0  is 
anywhere  on  X\  required  7^,  U  being  an  axis  through  0  and 
making  any  angle  a  with  X. 

lu  =tfv*dl<J=tf(y  cos  a  —  x  sin  otfdF;  i.e., 

In  =  cosa  afy*dF—  2  sin  a  cos  afxydF-\-  sina  afx*dF. 

But  since  the  area  is  symmetrical  about  JT,  in  summing  up  the 
products  xydF,  for  every  term  x(  +  y)dF,  there  is  also  a  term 
a?(  —  y)<£F  to  cancel  it ;  which  gives  fxy dF  =  0.  Hence 

Jn  =  cos3  alx  -f-  sina  or/r. 

The  student  may  easily  prove  that  if  two  distances  a  and  J 
be  set  off  from  0  on  ^  and  Y  respectively,  made  inversely 
proportional  to  V Ix  and  VlY,  and  an  ellipse  described  on  a  and 
1)  as  semi-axes  ;  then  the  moments  of  inertia  of  the  figure  about 


94 


MECHANICS   OF   ENGINEERING. 


any  axes  through  O  are  inversely  proportional  to  the  squares 
of  the  corresponding  semi-diameters  of  this  ellipse ;  called 
therefore  the  Ellipse  of  Inertia.  It  follows  therefore  that  the 
moments  of  inertia  about  all  gravity-axes  of  a  circle,  or  a 
regular  polygon,  are  equal ;  since  their  ellipse  of  inertia  must 
be  a  circle.  Even  if  the  plane  figure  is  not  symmetrical,  an 
"  ellipse  of  inertia"  can  be  located  at  any  point,  arid  has  the 
properties  already  mentioned  ;  its  axes  are  called  the  principal 
axes  for  that  point. 


90.  The  Rectangle. — First,  about  its 


Fig.  101.     Since 


dz 


all  points  of  a  strip  parallel  to  the  base 
'""*       have  the  same  co-ordinate,  z,  we  may  take 
dz  the  area  of  such  a  strip  for  dF=  bdz; 


j* 

Jo 


FIG.  101. 


FIG.  102. 


Secondly,  about  a  gravity-axis  parallel  to  base. 


dF=  Uz  .-.  Ig  = 


Thirdly,  about  any  other  axis  in  its  plane.     Use  the  results 
already  obtained  in  connection  with  the  reduction-formulae  of 

§§88,  89. 

90a.  The  Triangle.—  First,  about  an  axis  through  the  vertex 

and  parallel  to  the  base  ;  i.e.,  lv      «. ....&.....  ... .^       « .&.. ^  ^ 

in  Fig.   103.     Here  the   length     1 V y. ~y    ~]j-fcV~~        / 

of  the  strip  is  variable  ;  call  it  y.    ^  \  j       /  Y"*---"/y--a 

From  similar  triangles  \\  /  \  /   %h 

_^     ^     y        ...\L..      [ y 

y  =  (b  -T-  Tl)Z  ;  FIG.  103.  FIG.  104. 


.-.  I  = 


=  fz*ydz  =  (&-=- 


Secondly,  about  g,  a  gravity-axis  parallel  to  the  base.     Fig. 
104.     From  §  88,  eq.  (4),  we  have,  since  F=  %bh  and 

d  =    A  I  =  IV 


MOMENT    OF    INERTIA. 


Thirdly,  Fig.  104,  about  the  base  ;  IB  =  ? 
(4),  IB  =  Ig  +  Fd\  with  d  =  %h  ;  hence 


From  §  88,  eq. 


91.  The  Circle.  —  About  any  diameter,  as  g,  Fig.  105.  Polar 
co-ordinates,  lg  =  fz*dF.  Here  we  take<fJT==  area  of  an  ele- 
mentary rectangle  =  pdcp  .dp,  while  z  =  psin  ^. 


f 

A 

H 

E 

.       1 

:±!r.: 

—  at— 

:±^ 

s 

N 

1 

i—  i 

-\ 

FIG.  105. 


FIG.  106. 


=  J  J(P  sin  <p)*pd<pdp  =J     I  sin3  ^9?y     pWp  I 

^,4      /^27r  ^,4      ,^2n- 

=  —  /     sin2  <p<#<p  =  ~  /     ^(1  —  c 


92.    Compound  Plane  Figures.  —  Since  I  —  f£dF  is  an  in- 

finite series,  it  may  be  considered  as  made  up  of  separate 
groups  or  subordinate  series,  combined  by  algebraic  addition, 
corresponding  to  the  subdivision  of  the  compound  figure  into 
component  figures,  each  subordinate  series  being  the  moment  of 
inertia  of  one  of  these  component  figures;  but  these  separate 
moments  must  all  he  referred  to  the  same  axis.  It  is  con- 
venient to  remember  that  the  (rectangular)  I  of  a  plane 
figure  remains  unchanged  if  we  conceive  some  or  all  of  its 
elements  shifted  any  distance  parallel  to  the  axis  of  refer- 
ence. E.g.,  in  Fig.  106,  the  sum  of  the  IB  of  the  rectangle  CE, 
and  that  of  FD  is  =  to  the  IB  of  the  imaginary  rectangle 


96 


MECHANICS    OF    ENGINEERING. 


formed  by  shifting  one  of  them  parallel  to  B,  until  it  touches 
the  other ;  i.e.,  IB  of  CE+  IB  of  FD  =  #A'  (§  90).  Hence 
the  IB  of  the  T  shape  in  Fig.  106  will  be  =  IB  of  rectangle 
AD  —  IB  of  rect.  CE  —  IB  of  rect.  FD. 

That  is,          IB  of  T  =  t[M'  -  &A'].  •  •  .  (§  90).     .     .     (1) 

About  the  gravity-axis,  g,  Fig.  106.  To  find  the  distance  d 
from  the  base  to  the  centre  of  gravity,  we  may  make  use  of 
eq.  (3)  of  §  23,  writing  areas  instead  of  volumes,  or,  experi- 
mentally, having  cut  the  given  shape  out  of  sheet-metal  or 
card-board,  we  may  balance  it  on  a  knife-edge.  Supposing  d 
to  be  known  by  some  such  method,  we  have,  from  eq.  (4)  of 
§  88,  since  the  area  F=bh  —  J  A>  Ig  =  IB—  Fd1 ; 


— -O- 


The  double-^  (on),  and  the  box  forms  of   Fig.  106&,  if 
- -••"&• *      « *> -     symmetrical  about  the  gravity- 
axis  g,  have  moments  of  inertia 
alike  in  form.     Here  the  grav- 
ity-axis (parallel  to  base)  of  the 
compound  figure  is  also  a  grav- 
FIG.  io6a.  ity  axis  (parallel  to  base)  of  each 

of  the  two  component  rectangles,  of  dimensions  b  and  A,  bl  and 
At,  respectively. 

Hence  by  algebraic  addition  we  have  (§  90),  for  either  com- 
pound figure, 

(If  there  is  no  axis  of  symmetry  parallel  to  the  base  we  must 
proceed  as  in  dealing  with  the  T-form.)  Similarly  for  the  ring, 


FIG.  107.  FIG.  108. 

Fig.  107,  or  space  between  two  concentric  circumferences,  we 
have,  about  any  diameter  or  g  (§  91), 


MOMENT   OF   INERTIA.  97 

The  rhombus  about  a  gravity-axis,  g,  perpendicular  to  a 
diagonal,  Fig.  108.  —  This  axis  divides  the  figure  into  two 
equal  triangles,  symmetrically  placed,  hence  the  Ig  of  the 
rhombus  equals  double  the  moment  of  inertia  of  one  triangle 
about  its  base  ;  hence  (§ 


....  (5) 

(The  result  is  the  same,  if  either  vertex,  or  both,  be  shifted 
any  distance  parallel  to  AB.) 

For  practice,  the  student  may  derive  results  for  the  trapezoid  ; 
for  the  forms  in  Fig.  106,  when  the  inner  corners  are  rounded 
into  equal  quadrants  of  circles;  for  the  double-T,  when  the 
lower  flanges  are  shorter  than  the  upper;  for  the  regular 
polygons,  etc. 

93.  If  the  plane  figure  be  bounded,  wholly  or  partially,  by 
curves,  it  may  be  subdivided  into  an  infinite  number  of  strips, 
and  the  moments  of  inertia  of  these  (referred  to  the  desired 
axis)  added  by  integration,  if  the  equations  of  the  curves  are 
known  ;  if  not,  Simpson's  Rule,  for  a  finite  even  number  of 
strips,  of  equal  width,  may  be  employed  for  an  approximate 
result.     If  these  strips  are  parallel  to  the  axis,  the  1  of  any  one 
strip  —  its  length  X  its  width  X  square  of  distance  from  axis; 
while  if  perpendicular  to,  and   terminating  in,  the  axis,  its 
I  —  J-  its  width  X  cube  of  its  length  (see  §  90). 

A  graphic  method  of  determining  the  moment  of  inertia  of 
any  irregular  figure  will  be  given  in  a  subsequent  chapter. 

94.  Polar  Moment  of  Inertia  of  Plane  Figures  (§  85).—  Since 
the  axis  is  now  perpendicular  to  the  plane  of  the  figure,  inter- 
secting it  in  a  point,  0,  the  distances  of  the  ele- 

ments of  area  will  all  radiate  from  this  point, 

and  would  better  be  denoted  by  p  instead  of  z  ; 

hence,  Fig.  109,yp2^7^is  the  polar  moment  of 

inertia   of   any  plane  figure   about   a  specified 

point  0  ;  this  may  be   denoted  by  Ip.     But  /oa         FIO.  109. 

=  a?a  -f-  y3,  for  each  dF\  hence 

=  I       7. 


98  MECHANICS    OF   ENGINEERING. 

i.e.,  the  polar  moment  of  inertia  about  any  given  point  in 
the  plane  equals  the  sum  of  the  rectangular  moments  of  iner- 
tia about  any  two  axes  of  the  plane  figure,  which  intersect  at 
right  angles  in  the  given  point.  We  have  therefore  for  the 
circle  about  its  centre 


For  a  ring  of  radii  T\  and  7\, 
For  the  rectangle  about  its  centre, 
For  the  square,  this  reduces  to 

(See  §§90  and  91.) 

95.  Slender,  Prismatic,  Homogeneous  Rod. — Returning  to  the 
moment  of  inertia  qf  rigid  bodies,  or  solids,  we  begin  with  that 
of  a  material  line,  as  it  might  be  called,  about 
an  axis  through  its  extremity  making  some  an- 
gle a  with  the  rod.     Let  I  =  length  of  the  rod, 
.2^  its  cross-section  (very  small,  the  result  being 
strictly  true  only  when  F  =  0).     Subdivide 
FIG.  no.  the  rod  into  an  infinite  number  of  small  prisms, 

each  having  F  as  a  base,  and  an  altitude  —  ds.  Let  y  =  the 
heaviness  of  the  material ;  then  the  mass  of  an  elementary 
prism,  or  dM,  =-(y  -r-  g)Fds,  while  its  distance  from  the  axis 
Z  is  p  =  s  sin  a.  Hence  the  moment  of  inertia  of  the  rod 
with  respect  to  Z  as  an  axis  is 


Iz  =       p*dM  =  (y  -f- 

But  yFl  -r-  g  =  mass  of  rod  and  I  sin  a  =  a,  the  distance  of 
the  further  extremity  from  the  axis  ;  hence  Iz  —  \Ma?  and 
the  radius  of  gyration,  or  Ic,  is  found  by  writing  ^Ma'2=  Mk*  ; 
.-.  tf  =  i^8,  or  Tc  =  V%a  (see  §  86).  If  a  =  90°.  a  =  I. 

96.  Thin  Plates.     Axis  in  the  Plate.—  Let  the  plates  be  homo- 
geneous and  of  small  constant  thickness  —  T.     If  the  surface  of 


MOMENT    OF    INERTIA.  99 

the  plate  be  =  F,  and  its  heaviness  y,  then  its  mass  =  yFr  ~-  g. 
From  §  87  we  have  for  the  plate,  about  any  axis, 
/  —  (y  -T-  g)t  X  mom.  of  inertia  of  the  plane  figure  formed  by 
the  shape  of  the  plate  .............     (1) 

Rectangular  plate.     Gravity-axis  parallel  to  base.  —  Dimen- 
sions b  and  h.     From  eq.  (1)  and  §  90  we  have 

If=(y  •*-  fr  •  A^s  =  (rMr  -5-  g)tf?=  ^Mh";  .:  If  =  ^h'. 

Similarly,  if  the  base  is  the  axis,  IB  =  -JJ/A2,  /.  tf  =  -JA3. 

Triangular  plate.     Axis  through  vertex  parallel  to  base.  — 
From  eq.  (1)  and  §  90#,  dimensions  being  b  and  A, 


Circular  plate,  with  any  diameter  as  axis.  —  From  eq.  (1) 
and  §  91  we  have 


97.  Plates  or  Right  Prisms  of  any  Thickness  (or  Altitude). 
Axis  Perpendicular  to  Surface  (or  Base).  —  As  before,  the  solid  is 
homogeneous,  i.e.,  of  constant  heaviness  y\  |z 

let  the  altitude  =  A.     Consider  an  elementary 
prism,  Fig.  Ill,  whose  length  is  parallel  to  the 
axis  of  reference  Z.     Its  altitude  =  A  =  that 
of  the  whole  solid  ;  its  base  =  dF  =  an  element  f 
of  T^the  area  of  the  base  of  solid  ;  and  each  * 
point  of  it  has  the  same  p.     Hence  we  may         FIG.  m. 
take  its  mass,  =  yJidF  ~  g,  as  the  dM  in  summing  the  series 


=  (y  h  -f-  g)  X  polar  mom.  of  inertia  of  base.  .    .    (2) 

By  the  use  of  eq.  (2)  and  the  results  in  §  94  we  obtain  the 
following: 

Circular  plate,  or  right  circular  cylinder,  about  the  geo- 

metrical axis,     r  =  radius,  A  =  altitude. 

f.  =  (yh  -  0*w  =  (yk™*  •*  gW  =  *^°;  •'•  *"  =  &• 

Right  parallelopiped  or  rectangular  plate.  —  Fig.  112, 
Ig  =  (Yh  + 


100  MECHANICS   OF  ENGINEERING. 

For  a  hollow  cylinder,  about  its  geometric  axis, 


FIG.  112.  FIG.  113. 

98.  Circular  Wire.  —  Fig.  113  (perspective).  Let  Z  be  a 
gravity-axis  perpendicular  to  the  plane  of  the  wire  ;  X  and  Y 
lie  in  this  plane,  intersecting  at  right  angles  in  the  centre  0. 
The  wire  is  homogeneous  and  of  constant  (small)  cross-section. 
Since,  referred  to  Z,  each  dM  has  the  same  p  =  /*,  we  have 
7^  =ffdM=  Mr\  Now  7^  must  equal  7r,  and  (§  94)  'their 
sum  =  7Z, 

/.  Ix,  or  7r,  =  JJff*,         and         &x3,  or  &F2  =  Jr8. 


99.  Homogeneous  Solid  Cylinder,  about  a  diameter  of  its  base. 
—  Fig.  114.  Ix  =  ?  Divide  the  cylinder  into  an  infinite  num- 
ber of  laminae,  or  thin  plates,  parallel  to  the 
base.  Each  is  some  distance  z  from  X,  of 
thickness  dz,  and  of  radius  r  (constant).  In 
^—  each  draw  a  gravity-axis  (of  its  own)  parallel  to 
FIG.  ii4.  X.  We  may  now  obtain  the  Ix  of  the  whole 
cylinder  by  adding  the  7x's  of  all  the  laminae.  The  Ig  of  any 
one  lamina  (§96,  circular  plate)  =  its  mass  Xi^2;  hence  its 
/x  (eq-  (3),  §  88)  =  its  Ig+  (its  mass)  X  s2.  Hence  for  the 
whole  cylinder 


=  / 


i.e.,  7X  = 

100.  Let  the  student  prove  (1)  that  if  Fig.  114  represent 
any  right  prism,  and  kF  denote  the  radius  of  gyration  of  any 
one  lamina,  referred  to  its  gravity-axis  parallel  to  X,  then  the 
Ix  of  whole  prism  =  M(kf  -\-  £Aa)  ;  and  (2)  that  the  moment 


MOMENT   OF   INEKTIA. 


101 


of  inertia  of  the  cylinder  about  a  gravity-axis  parallel  to  the 
baseis  = 


101.  Homogeneous  Right  Cone.  —  Fig.  115.  First,  about  an 
axis  V,  through  the  vertex  and  parallel  to  the  base.  As  before, 
divide  into  laminae  parallel  to  the  base.  Each  is  a 
circular  thin  plate,  but  its  radius,  a?,  is  not  =  r,  but, 
from  proportion,  is  x  =  (r  -;-  h)z. 

The  /of  any  lamina  referred  to  its  own  gravity- 
axis  parallel  to  Fis  (§96)  =  (its  mass)  X  i#a,  and 
its  Iv  (eq.  (3),  §88)  is  .'.   =  its   mass   X   i#8  +     FIG.  115. 
its  mass  X  £*• 

Hence  for  the  whole  cone, 


Secondly,  about  a  gravity-axis  parallel  to  the  "base.  —  From 
cq.  (3),  §  88,  with  d  =  %h  (see  Prob.  7,  §  2(5),  and  the  result 
just  obtained,  we  have  /=  M-/$[r*  -f"  i^*]- 

Thirdly,  about  its  geometric  axis,  Z.  —  Fig.  116.  Since  the 
axis  is  perpendicular  to  each  circular  lamina  through  the  centre, 
its/z(§97)is 

=  its  mass  X  J(rad.)a  =  (yntfdz  -r-  g)^*?. 
Now  x  =  (r  -i-  h)z,  and  hence  for  the  whole  cone 

I  =       «r>  H-   K        z'dz  =       r'h 


FIG.  116. 


FIG.  118. 


102.  Homogeneous   Right  Pyramid  of  Rectangular  Base. — 

About  its  geometrical  axis.     Proceeding  as   in  the  last  para- 


102  MECHANICS    OF   ENGINEERING. 

graph,  we  derive  Iz  —  MfacP,  in  which  d  is  the  diagonal  of  the 
base. 

103.  Homogeneous  Sphere.  —  About  any  diameter.  Fig.  118. 
Iz  —  ?  Divide  into  laminae  perpendicular  to  Z.  By  §  97,  and 
noting  that  a?2  =  r*—  z*,  we  have  finally,  for  the  whole  sphere,, 


For  a  segment,  of  one  or  two  bases,  put  proper  limits  for  z. 
in  the  foregoing,  instead  of  +  r  and  —  r. 

104.  Other  Cases.  —  Parabolic  plate,  Fig.  119,  homogeneous 
and  of  (any)  constant  thickness,  about 
an  axis  through  0,  the  middle  of  the 
chord,  and  perpendicular  to  the  plate, 
This  is 

FIG.  119.  FIG.  120.  /  =  Jf|(J«9  +  f  A'). 

The  area  of  the  segment  is  —  %hs. 

For  an  elliptic  plate,  Fig.  120,  homogeneous  and  of  any 
constant  thickness,  semi-axes  a  and  5,  we  have  about  an  axis 
through  0,  normal  to  surface  I0  =  M^jf  -\-  £a]  ;  while  for  a 
very  small  constant  thickness 

and     I 


The  area  of  the  ellipse  =  nab. 

Considering  Figs.  119  and  120  as  plane  figures,  let  the 
student  determine  their  polar  and  rectangular  moments  of 
inertia  about  various  axes. 

(For  still  other  cases,  see  p.  518  of  Rankine's  Applied 
Mechanics,  and  pp.  593  and  594  of  Coxe's  Weisbach.) 

105.  Numerical  Substitution.  —  The  moments  of  inertia  of 
plane  figures  involve  dimensions  of  length  alone,  and  will  be 
utilized  in  the  problems  involving  flexure  and  torsion  of  beams, 
where  the  inch  is  the  most  convenient  linear  unit.  E.g.,  the 


MOMENT    OF   1JSEKTIA.  103 

polar  moment  of  inertia  of  a  circle  of  two  inches  radius  about 
its  centre  is  \7tr*  —  25.13  -|-  biquadratic,  or  four-dimension, 
inches,  as  it  may  be  called.  Since  this  quantity  contains  four 
dimensions  of  length,  the  use  of  the  foot  instead  of  the  inch 
would  diminish  its  numerical  value  in  the  ratio  of  the  fourth 
power  of  twelve  to  unity. 

The  moment  of  inertia  of  a  rigid  "body,  or  solid,  however, 
=  Ml£  =  (G-  -r-  g)k*,  in  which  G,  the  weight,  is  expressed  in 
units  of  force,  g  involves  both  time  and  space  (length),  while  &a 
involves  length  (two  dimensions).  Hence  in  any  homogeneous 
formula  in  which  the  /  of  a  solid  occurs,  we  must  be  careful  to 
employ  units  consistently  ;  e.g.,  if  in  substituting  G  ~  g  for  M 
(as  will  always  be  done  numerically)  we  put  g  =  32.2,  we 
should  use  the  second  as  unit  of  time,  and  the  foot  as  linear 
unit. 

106.  Example.  —  Eequired  the  moment  of  inertia,  about  the 
axis  of  rotation,  of  a  pulley  consisting  of  a  rim,  four  parallelo- 
pipedical  arms,  and  a  cylindrical  hub  which  may  be  considered 
solid,  being  filled  by  a  portion  of  the  shaft. 
Fig.  121.  Call  the  weight  of  the  hub  G, 
its  radius  r\  similarly,  for  the  rim,  G^ 
and  7*2  ;  the  weight  of  one  arm  being  =  £?,. 
The  total  /  will  be  the  sum  of  the  /'s  of 
the  component  parts,  referred  to  the  same 
axis,  viz.  :  Those  of  the  hub  and  rim  will 
be  (G  +  gfr?  and  «?,  -  ff^r'  +  r.'), 
respectively  (§  97),  while  if  the  arms  are  Fl°-  121- 

not  very  thick  compared  with  their  length,  we  have  for  them 
(§§  95  and  88) 


as  an  approximation  (obtained  by  reduction  from  the  axis  at 
the  extremity  of  an  arm  to  a  parallel  gravity-axis,  then  to  the 
required  axis,  then  multiplying  by  four).  In  most  fly-wheels, 
the  rim  is  proportionally  so  heavy,  besides  being  the  farthest 
removed  from  the  axis  of  rotation,  that  the  moment  of  inertia 
of  the  other  parts  may  be  for  practical  purposes  neglected. 


104  MECHANICS    OF   ENGINEERING. 

107.  Ellipsoid  of  Inertia. — The  moments  of  inertia  about 
all  axes  passing  through  any  given  point  of  any  rigid  body 
whatever  may  be  proved  to  be  inversely  proportional  to  the 
squares  of  the  diameters  which  they  intercept  in  an  imaginary 
ellipsoid,  whose  centre  is  the  given  point,  and  whose  position 
in  the  body  depends  on  the  distribution  of  its  mass  and  the 
location  of  the  given  point.  The  three  axes  which  contain  the 
three  principal  diameters  of  the  ellipsoid  are  called  the  Princi- 
pal Axes  of  the  "body  for  the  given  point.  This  is  called  the 
ellipsoid  of  inertia.  (Compare  §  89.)  Hence  the  moments  of 
inertia  of  any  homogeneous  regular  polyedron  about  all  gravity- 
axes  are  equal,  since  then  the  ellipsoid  becomes  a  sphere.  It 
can  also  be  proved  that  for  any  rigid  body,  if  the  co-ordinate 
axes  X,  Y,  and  Z,  are  taken  coincident  with  the  three  principal 
axes  at  any  point,  we  shall  have 

=0;    and 


DYNAMICS   OF   A    RIGID   BODY.  105 


CHAPTEE  Y. 

DYNAMICS  OF  A  RIGID  BODY. 

108.  General  Method. — Among  the  possible  motions  of  a 
rigid  body  the  most  important  for  practical  purposes  (and  for- 
tunately the  most  simple  to  treat)  are  :  a  motion  of  translation, 
in  which  the  particles  move  in  parallel  right  lines  with  equal 
accelerations  and  velocities  at  any  given  instant;  and  rotation 
about  a  fixed  axis,  in  which  the  particles  describe  circles  in 
parallel  planes  with  velocities  and  accelerations  proportional 
(at  any  given  instant)  to  their  distances  from  the  axis.  Other 
motions  will  be  mentioned  later.  To  determine  relations,  or 
equations,  between  the  elements  of  the  motion,  the  mass  and 
form  of  the  body,  and  the  forces  acting  (which  do  not  neces- 
sarily form  an  unbalanced  system),  the  most  direct  method  to 
be  employed  is  that  of  two  equivalent  systems  of  forces  (§  15), 
one  consisting  of  the  actual  forces  acting  on  the  body,  con- 
sidered free,  the  other  imaginary,  consisting  of  the  infinite 
number  of  forces  which,  applied  to  the  separate  material  points 
composing  the  body,  would  account  for  their  individual  mo- 
tions, as  if  they  were  an  assemblage  of  particles  without  mutual 
actions  or  coherence.  If  the  body  were  at  rest,  then  considered 
free,  and  the  forces  referred  to  three  co-ordinate  axes,  they 
would  constitute  a  balanced  system,  for  which  the  six  summa- 
tions ^X,  ^Y,  *2Z,  ^(mom.)x,  2(mom.)Ti  and  ^(mom.)z, 
would  each  =  0 ;  but  in  most  cases  of  motion  some  or  all  of 
these  sums  are  equal  (at  any  given  instant),  not  to  zero,  but  to 
the  corresponding  summation  of  the  imaginary  equivalent 
system,  i.e.,  to  expressions  involving  the  masses  of  the  particles 
(or  material  points),  their  distribution  in  the  body,  and  the 


106  MECHANICS    OF   ENGINEERING. 

elements  of  the  motion.  That  is,  we  obtain  six  equations  by 
putting  the  ^2X  si  the  actual  system  equal  to  the  2X  of  the 
imaginary,  and  so  on  ;  for  a  definite  instant  of  time  (since  some 
of  the  quantities  may  be  variable). 

109.  Translation. — Fig.  122.  At  a  given  instant  all  the  par- 
ticles have  the  same  velocity  =  v,  in  parallel  right  lines  (par- 
allel to  the  axis  JT,  say),  and  the 
same  acceleration  p.  Required 
the  ^>X.  of  the  acting  forces, 
^M  shown  at  (I.).  (II.)  shows  the 
imaginary  equivalent  system,  con- 
sisting of  a  force  =  mass  X  ace. 
=  dMp  applied  parallel  to  JTto 
each  particle,  since  such  a  force 
would  be  necessary  (from  eq.  (IY.) 
§  55)  to  account  for  the  accelerated  rectilinear  motion  of  the 
particle,  independently  of  the  others.  Putting  (2X  )I=(2X)II, 

we  have 

v' 

(2X)I=fpdM=pfdM^Mp:  .    .    .    (Y.) 

It  is  evident  that  the  resultant  of  system  (II.)  must  be  paral- 
lel toX;  hence*  that  of  (I.),  which  =  (-2X)7  and  may  be  de- 
noted by  7?,  must  also  be  parallel  to  X\  let  a  =  perpendicular 
distance  from  R  to  the  plane  YX\  a  will  be  parallel  to  Z. 
Now  put  [^(mom.^Jj  =  [J£(mpni.F)]xn  (Y is  an  axis  perj^n^f ^/~ 
dicular  to  paper  through  0)  and  we  have  —  Ra  =  —fdMpz^ 
=  —pfdMz  =  —  pMz  (§88),  i.e.,  a  —  z.  A  similar  result 
may  be  proved  as  regards  y.  Hence,  if  a  rigid  body  has  a 
motion  of  translation,  the  resultant  force  must  act  in  a  line 
through  the  centre  of  gravity  (here  more  properly  called  the 
centre  of  mass),  and  parallel  to  the  direction  of  motion.  Or, 
practically,  in  dealing  with  a  rigid  body  having  a  motion  of 
translation,  we  may  consider  it  concentrated  at  its  centre  of 
mass.  If  the  velocity  of  translation  is  uniform,  R  =  ^M  X  0 
=  0,  i.e.,  the  forces  are  balanced. 

*  The  forces  of  system  (I. )  cannot  form  a  couple  ;  since  those  of  system  (II. ) 
do  not  reduce  to  a  couple,  all  pointing  one  way. 


DYNAMICS    OF  A   RIGID   BODY.  107 

110.  Rotation  about  a  Fixed  Axis.—  First,  as  to  the  elements 
of  space  and  time  involved.  Fig.   123.     Let  0  be  the  axis  of 
rotation  (perpendicular  to  paper),  OY  a  fixed 
line  of  reference,  and  OA  a  convenient  line  of 
the  rotating  body,  passing  through  the  axis  and 
perpendicular  to  it,  accompanying  the  body  in 
its  angular  motion,  which  is  the  same  as  that  of 
OA..     Just  as  in  linear  motion  we  dealt  with        Fl°-  123- 
linear  space  ($),  linear  velocity  (V),  and  linear  acceleration  (p), 
so  here  we  distinguish  at  any  instant  ; 
a,  the  angular  space  between  0  Y  and  OA  ; 

GO  —    ,   ,  the  angular  velocity,  or  rate  at  which  a  is  changing  ; 

and 

0  =  -TT  =  -3-5-,  the  angular  acceleration,  or  rate  at  which  GO 

is  changing. 

These  are  all  reckoned  in  jr-measure  and  may  be  +  or  —  , 
according  to  their  direction  against  or  with  the  hands  of  a 
watch. 

(Let  the  student  interpret  the  following  cases  :  (1)  at  a  cer- 
tain instant  GO  is  +,  and  6  —  ;  (2)  GO  is  —  ,  and  6  -J-;  (3)  a  is 
—  ,  GO  and  9  both  +;  (4)  a  -J-,  GO  and  6  both  —  .)  For  rotary 
motion  we  have  therefore,  in  general, 

da  dao 

"  =  -&>    .....     0^)      6=dt=- 

and  .-.  oodGo  =  Bda  ; 


corresponding  to  eqs.  (I.),  (II.),  and  (III.)  in  §  50,  for  rectilinear 
motion. 

Hence,  for  uniform  rotary  motion,  GO  being  constant  and 
6  =  0,  we  have  a  =  cot,  t  being  reckoned  from  the  instant 
when  a  =  0. 

For  uniformly  accelerated  rotary  motion  6  is  constant,  and 


108  MECHANICS    OF    ENGINEERING. 

if  <s?0  denote  the  initial  angular  velocity  (when  a  and  t  =3  0), 

we  may  derive,  precisely  as  in  §56, 

GO  =  H  +  6t ;      .     .     (1)  «  =  atf +  *6tf ;   .     .     (2) 


a  = 


(3)     and     «  =  *(«.  +  «)*• 


If  in  any  problem  in  rotary  motion  6,  GO,  and  <*  have  been 
determined  for  any  instant,  the  corresponding  linear  values  for 
any  point  of  the  body  whose  radial  distance  from  the  axis  is  p, 
will  be  s=  ap  (=  distance  described  by  the  point  measured 
along  its  circular  path  from  its  initial  position),  v  =  oop  =  its 
velocity,  and  pt  =  6p  its  tangential  acceleration,  at  the  instant 
in  question. 

Examples.  —  (1)  What  value  of  G?,  the  angular  velocity,  is 
implied  in  the  statement  that  a  pulley  is  revolving  at  the  rate 
of  100  revolutions  per  minute  ? 

100  revolutions  per  minute  is  at  the  rate  of  2?r  X  100 
=  628.32  (^-measure  units)  of  angular  space  per  minute 
=  10.472  per  second  ;/.(»  =  628.32  per  minute  or  10.472 
per  second. 

(2)  A  grindstone  whose  initial  speed  of  rotation  is  90  revo- 
lutions per  minute  is  brought  to  rest  in  30  seconds,  the  an- 
gular retardation  (or  negative  angular  acceleration)  being  con- 
stant ;  required  the  angular  acceleration,  0,  and  the  angular 
space  a  described.  Use  the  second  as  unit  of  time. 
oov  =  27r|~§-  =  9.4248  per  second  ;  /.  from  eq.  (1) 

0  -  -.IL^L  =  _  9^24  -s-  30  =  —  0.3141  (^-measure  units) 
t 

per  "  square  second."     The  angular  space,  from  eq.  (2)  is 

a  =  a>f  +  ±0ff  =  30  X  9.42  —  4(0.314)900  =  141.3 
(jr-measure  units),  i.e.,  the  stone  has  made  22.4  revolutions  in 
coming  to  rest  and  a  point  2  ft.  from  the  axis  has  described  a 
distance  s  =  ap  —  141.3  X  2  =  282.6  ft.  in  its  circular  path. 

111.  Rotation.  Preliminary  Problem.  Axis  Fixed.—  For 
clearness  in  subsequent  .matter  we  now  consider  the  following 


DYNAMICS    OF   A    RIGID    BODY.  109 

simple  case.  Fig.  124  shows  a  rigid  body,  consisting  of  a 
drum,  an  axle,  a  projecting  arm,  all 
of  which  are  imponderable,  and  a 
single  material  point,  whose  weight 
is  G  and  mass  M.  An  imponderable 
flexible  cord,  in  which  the  tension  is 
kept  constant  and  =  P,  unwinds 
from  the  drum.  The  axle  coincides 
with  the  vertical  axis  Z,  while  the  cord  '  Fl<3>  124> 

is  always  parallel  to  Y.  Initially  (i.e.,  when  t  =  0)  Jflies  at 
rest  in  the  plane  ZY.  Required  its  position  at  the  end  of  any 
time  t  (i.e.,  at  any  instant)  and  also  the  reactions  of  the  bearings 
at  0  and  0^  supposing  no  vertical  pressure  to  exist  at  0^  and 
that  P  and  M  are  at  the  same  level.  No  friction.  At  any  in- 
stant the  eight  unknowns,  a,  GO,  0,  JT0,  Y^ZW  JL,,  and  Y^  may 
be  found  from  the  six  equations  formed  by  putting  ^X,  etc., 
of  the  system  of  forces  in  Fig.  124,  equal,  respectively,  to  the 
2X,  etc.,  of  the  imaginary  equivalent  system  in  Fig.  125,  and 
two  others  to  be  mentioned  subsequently.  Since,  at  this  in- 
stant, the  velocity 'of  M  must  be  v  =  cop  and  its  tangential  ac- 
celeration pt  =  Op,  its  circular  motion 
could  be  produced,  considering  it  free  (eq. 
(5),  §  74),  by  a  tangential  force  T  =  mass 
X  Pt  =  M6p,  and  a  normal  centripetal 
force  N=Mv*  -7-  p=M(oop)*  ^  p—tfMp. 
~~FIG.  125.  Hence  the  system  in  Fig.  125  is  equivalent 

to  that  of  Fig.  124,  and  from  putting  the  ^  (mom.)z  of  one 
=  that  of  the  other,  we  derive 

Pa  =  Tp ;  i.e.,  Pa  =  QMp\     ....    (1) 

whence  0  becomes  known,  and  is  evidently  constant,  since  P, 
a,  M,  and  p  are  such.  .'.  the  angular  motion  is  uniformly  ac- 
celerated, and  from  eqs.  (1)  and  (2),  §  110,  GO  and  a  become 
known ; 

i.e.,  GO  =  Bt,      .     .     .     (2)     and     a  =  £6t* (3) 

Putting  (2Z  of  124)  =  (2Z  of  125).  gives 

Z.  -  G  =  0 ;  i.e..  Z0  =  0 (4) 


110  MECHANICS   OF   ENGINEERING. 


Proceeding  similarly  with  the  ^J5T  of  each  system, 
X0  -f-  Xv  =  Tcos  OL  —  JVsin  a  =  6Mp  cos  a  —  afMp&m  or,  (5) 
and  with  the  2  Y  of  each, 
P  +  Y0  +  Y,  =  -  Tsm  a-JVcosa=-  OMpsiu  a 

—  oo*Mp  cos  ex  ;  (6) 

while  with  the  2  (mom^j-  we  have,  conceiving  all  the  forces  in 
each  system  projected  on  the  plane  ZI^(see  §38),  and  noting 
that  y  =  p  cos  a  and  x  =  p  sin  ar, 

+  Op  cos  a  +  YJ  +  P£  =  -  (6Mp  sin  *)&-(<»•  Jf/o  cosa)5,(Y) 
and  with  the  ^  (mom.)F, 
—  #/>  sin  or  —  X,l  =  —  (OMp  cos  a)l  +  (oo*Mp  sin  a)J.  .     (8) 

From  (Y)  we  may  find  Yj>  from  (8),  JT,;  then  X0  and  J^ 
from  (5)  and  (6).  It  will  be  noted  that  as  the  motion  proceeds 
6  remains  constant  ;  GO  increases  with  the  time,  a  with  the 
square  of  the  time  ;  ZQ  is  constant,  =  G  ;  while  X.^  Y09  -Xj, 
and  Yl  have  variable  values  dependent  on  p  cos  a  and  p  sin  #, 
i.e.,  on  the  co-ordinates  y  and  x  of  the  moving  material  point. 

112.  Particular  Supposition  in  the  Preceding  Problem  with 
Numerical  Substitution.  —  Suppose  we  have  given  (using  the 
foot-pound-second  system  of  units  in  which  g  =  32.2)  G  =  64.4 
Ibs.,  whence 

M  =(Q  +  g)  =  2;  P  =  41bs.,  Z  =  4ft.,  5  =  2  ft,  a  =  2  ft, 
and  p  —  4  ft.;  and  that  J/  is  just  passing  through  the  plane 
ZJf,  i.e.,  that  «  =  -JTT.     We  obtain,  first,  the  angular  accelera- 
tion, eq.  (1), 

6  =  Pa-+  Mp°  =S  +  3Z=  0.25  =  J. 

From  eqs.  (2)  and  (3)  we  have  «£  the  instant  mentioned  (not- 
ing that  when  a  was  =  0,  t  was  =  0) 

o>a  =  2«#  =  Jw  =  0.7854  +, 
while  (2)  gives,  for  the  time  of  describing  the  quadrant, 

t  =  GO  -i-  6  =  3.544.  .  .  .  seconds. 
Since  at  this  instant  cos  a  —  0  and  sin  a  —  1,  we  have,  from 

(7), 

-f  0+r,X4  +  4x2^  -JX2X4X2;  .-.  T>  -  3  Ibs. 


DYNAMICS    OF   A    RIGID    BODY. 


Ill 


The  minus  sign  shows  it  should  point  in  a  direction  contrary 
to  that  in  which  it  is  drawn  in  Fig.  124.     Eq.  (8)  gives 

-64.4  X  ^-Xl  X  4=-0-hi7TX2x4x2;.-.X1  =  -  67.54  Ibs. 
And  similarly,  knowing  YI  and  X^  we  have  from  (5)  and  (6), 

.  Z0  =  +  61.26  Ibs.,  and  Y0  =  -  3.00  Ibs. 
The  resultant  of  X,  and  Y»  also  that  of  JT0,  Y0,  and  Z0,  caii 
now  be  found  by  the  parallelogram  (and  parallelopipedon)  of 
forces,  both  in  amount  and  position,  noting  carefully  the  direc- 
tions of  the  components.  These  resultants  are  the  actions  of 
the  supports  upon  the  ends  of  the  axle ;  jfyeir  equals  and 
opposites  would  be  the  actions  or  pressures  of  the  axle  against 
the  supports,  at  the  instant  considered  (when  M  is  passing 
through  the  plane  ZX\  i.e.,  with  A  =  \7t).  (At  the  same  in- 
stant, suppose  the  string  to  break  ;  what  would  be  the  effect  on 
the  eight  quantities  mentioned?) 

113.  Centre  of  Percussion  of  a  Rod  suspended  from  one  End.— 
Fig.  126.  The  rod  is  initially  at  rest  (see  (I.)  in  figure),  is  straight, 
homogeneous,  and  of  constant 
(small)  cross-section.  Neglect  its 
weight.  A  horizontal  force  or 
pressure,  jP,  due  to  a  blow  (and 
varying  in  amount  during  the  p 
blow),  now  acts  upon  it  from  the 
left,  perpendicularly  to  the  axis, 
Z,  of  suspension.  An  accelerated 
rotary  motion  begins  about  the  fixed  axis  Z.  (II.)  shows  the  rod 
free,  at  a  certain  instant,  with  the  reactions  X0  and  Y0  put  in 
at  O0.  (III.)  shows  an  imaginary  system  which  would  produce 
the  same  effect  at  this  instant,  and  consisting  of  a  dT  =  dMOp, 
and  a  dN  =  <*?dMf>  applied  to  each  dM9  the  rod  being  composed 
of  an  infinite  number  of  dM's,  each  at  some  distance  p  from 
the  axis.  Considering  that  the  rotation  has  just  ~begun,  <&,  the 
angular  velocity  is  as  yet  small,  and  will  be  neglected.  Ke- 
quired  Y0  the  horizontal  reaction  of  the  support  at  0  in  term? 
of  P.  By  putting  2  Yu  =  2  Yin,  we  have 

P  -  Y0  =fdT=  6/pdM  =  8M~p. 


0 


(n.) 

FIG.  126. 


(in.) 


112 


MECHANICS    OF    ENGINEERING. 


/.  Yo  =  P  —  BM p ;  p  is  the  distance  of  the  centre  of  gravity 
from  the  axis  (KB.  J'pdM  =  M~p  is  only  true  when  all  the 
p's  are  parallel  to  each  other).  But  the.  value  of  the  angular 
acceleration  6  at  this  instant  depends  on  P  and  «,  for  2  (mom.)^ 
in  (II.)  =  2  (mom.)z  in  (III.),  whence  Pa  =  8fp*dM=z  OIZ, 
where  Iz  is  the  moment  of  inertia  of  the  rod  about  Z,  and  from 
§  95  =  \Ml\  Now  p  =  il ;  hence,  finally, 


If  now  Y0  is  •=  0,  i.e.,  if  there  is  to  be  no  shock 
the  rod  and  axis,  we  need  only  apply  P  at  a  point  whose  dis- 
tance a  =  \l  from  the  axis ;  for  then  Y0  =  0.  This  point  is 
called  the  centre  of  percussion  for  the  given  rod  and  axis.  It 
and  the  point  of  suspension  0  are  interchangeable  (see  §  118). 
(Lay  a  pencil  on  a  table;  tap  it  at  a  point  distant  one  third  of 
the  length  from  one  end  ;  it  will  begin  to  rotate  about  a  vertical 
axis  through  the  farther  end.  Tap  it  at  one  end  ;  it  will  begin 
to  rotate  about  a  vertical  axis  through  the  point  first  mentioned. 
Such  an  axis  of  rotation  is  called  an  axis  of  instantaneous  rota- 
tion^ and  is  different  for  each  point  of  impact — just  as  the 
point  of  contact  of  a  wheel  and  rail  is  the  one  point  of  the 
wheel  which  is  momentarily  at  rest,  and  about  which,  therefore, 
all  the  others  are  turning  for  the  instant.  Tap  the  pencil  at 
its  centre  of  gravity,  and  a  motion  of  translation  begins ;  see 
§  109.) 

114.  Rotation.     Axis  Fixed.     General-  Formulae. — Consider 

I 


FIG.  127.  FIG.  128. 

Jug  now  a  rigid  body  of  any  shape  whatever,  let  Fig.  127  indi* 
cate  the  system  of  forces  acting  at  any  given  instant^  Z  being 


DYNAMICS    OF    A    RIGID   BODY.  118 

the  fixed  axis  of  rotation,  GO  and  6  the  angular  velocity  and 
angular  acceleration,  at  the  given  instant.  X  and  Y  are  two 
axes,  at  right  angles  to  each  other  and  to  Z,  fixed  in  space.  At 
this  instant  each  dM  of  the  body  has  a  definite  a?,  y,  and  <p 
(see  Fig.  128),  which  will  change,  and  also  a  p,  and  z,  which  will 
not  change,  as  the  motion  progresses,  and  is  pursuing  a  circu- 
lar path  with  a  velocity  =  cop  and  a  tangential  acceleration 
=  dp.  Hence,  if  to  each  dM  of  the  body  (see  Fig.  128)  we 
imagine  a  tangential  force  dT  —  dMOp  and  a  normal  force 
=  dM(copf  -T-  p  =  Go*dMp  to  be  applied  (eq.  (5),  §  74),  and 
these  alone,  we  have  a  system  comprising  an  infinite  number  of 
forces,  all  parallel  to  XY,  and  equivalent  to  the  actual  system 
in  Fig.  127.  Let  ^X,  etc.,  represent  the  sums  (six)  for  Fig. 
127,  whatever  they  may  be  in  any  particular  case,  while  for 
128  we  shall  write  the  corresponding  sums  in  detail.  Noting 
that 

fdN  cos  (p  =  affdMp  cos  (p  =  offdMy  =  G? My,($  88); 
that /<1ZV  sin  cp  =  cffdMp  sin  (p  =  ctffdMx  =  a?Mx', 
and  similarly,  that  fdT  cos  cp  =  OfdMp  cos  cp  =  6 My,  and 
/^jTsin  <p  =  6Mx\  while  in  the  moment  sums  (the  moment 
of  dT  cos  cp  about  Y,  for  example,  being  —  dT  cos  cp  .  z  — 
—  OdMp  (cos  cp)2=  —  8dMy2)  the  sum  of  the  moms.  r  of  all  the 
(dT cos  0>)'s  =  -  8/dMyz) 

fdTcos  cpz  =  6fdMyz,  fdN  sin  cpz  =  affdMxz,  etc., 
we  have,  since  the  systems  are  equivalent, 

.    .    .    .      (IX.) 
.    .    .    .       (X.) 

=      0; (XL) 

2  moms.x  =  -  OfdMxz  -  tffdMyz ;     .    (XII.) 

2  moms.r  =  -  OfdMyz  +  tffdMxz  ;      .  (XIII.) 

2  moms.z  =      e/dMp*  =  6IZ.       .     •     .  (XIV.) 

These  hold  good  for  any  instant.     As  the  motion  proceeds  x 

and  y  change,  as  also  the  sums  fdMxz  and  fdMyz.     If  the 

body,  however,   is   homogeneous,  and  symmetrical  about  the 

plane  XY,  fdMxz  and  fdMyz  would  always  =  zero  ;  since 

9 


114 


MECHANICS    OF   ENGINEERING. 


the  z  of  any  dMdoes  not  change,  and  for  every  term  dMy(-\-z), 
there  would  be  a  term  dMy( —  z)  to  cancel  it ;  similarly  for 
fdMxz.  The  eq.  (XIY.),  2  (moms,  about  axis  of  rotat.)  = 
fdTp  —  OfdMp*  =  (angular  accel.)  X  (mom.  of  inertia  of 
body  about  axis  of  rotat.),  shows  how  the  sum/^J/p2  arises  in 
problems  of  this  chapter.  That  a  force  dT  =  dM6p  should 
be  necessary  to  account  for  the  acceleration  (tangential)  Op  of 
the  mass  dM,  is  due  to  the  so-called  inertia  of  the  mass  (§  54)5 
and  its  moment  dTp,  or  BdMp*,  might,  with  some  reason,  be 
called  the  moment  of  'inertia  of  the  dM,  "Ai\&fOdMp*=  OfdMp* 
that  of  the  whole  body.  But  custom  has  restricted  the  name 
to  the  sumfdMp*,  which,  being  without  the  0,  has  no  term  to 
suggest  the  idea  of  inertia.  For  want  of  a  better  the  name  is 
still  retained,  however,  and  is  generally  denoted  by  /.  (See 
§§  86,  etc.) 

115.  Example  of  the  Preceding. — A  homogeneous  right  par- 
allelopiped  is  mounted  on  a  vertical 
axle  (no  friction),  as  in  figure.  0  is 
at  its  centre  of  gravity,  hence  both 
x  and  y  are  zero.  Let  its  heaviness 
be  7,  its  dimensions  A,  J,,  and  b  (see 
§  97).  XY  is  a  plane  of  symmetry, 
hence  both  fdMxz  and  fdMyz  are 
zero  at  all  times  (see  above).  The 
tension  P  in  the  (iriextensible)  cord 
is  caused  by  the  hanging  weight  Pt 
(but  is  not  =  PV  unless  the  rotation  is  uniform).  The  figure 
shows  both  rigid  bodies  free.  P1  will  have  a  motion  of  trans- 
lation ;  the  parallelepiped,  one  of  rotation  about  a  fixed  axis. 
No  masses  are  considered  except  Pt  -=-  </,  and  bhb}y  -f-  g.  The 
Iz  =  Mkz  of  the  latter  =  its  mass  X  TV(^,a  +  V},  §  97.  At 
any  instant,  the  cord  being  taut,  if  p  =  linear  acceleration  of 
P^  we  have 


FIG.  129. 


p—  da 

From  (XIY.),      Pa  =  6IZ  ;    .'.  P  =  8IZ  -:-  a.    . 


eq.  (a) 
.      (1) 


DYNAMICS   OF  A   RIGID  BODY.  116 

For    the  free  mass  Pl  +  g  we   have    (§109)  Pl  —  p  = 
mass  X  ace., 


Equate  these  two  values  of  P  and  solve  for  6,  whence 


All  the  terms  here  are  constant,  hence  0  is  constant  ;  there- 
fore the  rotary  motion  is  uniformly  accelerated,  as  also  the 
translation  of  Pr  The  formulae  of  §  56,  and  (1),  (2),  (3),  and 
(4)  of  §  110,  are  applicable.  The  tension  P  is  also  constant  ; 
see  eq.  (1).  As  for  the  five  unknown  reactions  (components) 
at  01  and  0»  the  bearings,  we  shall  find  that  they  too  are  con- 
stant  ;  for 

from      (IX.)  we  have  Xl  +  Xa  =  0  ;  (4) 

from       (X.)  we  have  P  +  Y,  +  Z,  =  0  ;  (5) 

from      (XL)  we  have  Za  -  G  =  0  ;  (6) 

from    (XII.)  we  have  P  .  AO+  Yl  .  ~0\0-  Y,  .  U^O  =  0  ;  (7) 
from  (XIII.)  we  have  -  X,  .  ~0fl  +  X2  .  0^0  =  0.    (8) 

Numerical  substitution  in  the  above  problem.  —  Let  the  par. 
allelopiped  be  of  wrought-iron  ;  let  P,  =  48  Ibs.;  a  =  6  in.  = 
£  ft.;  b  =  3  in.  =  £  ft.  (see  Fig.  112)  ;  ^  —  2  ft.  3  in.  =  £  ft.; 
and  h  =  4  in.  =  £  ft.  Also  iet  Ofi  —  O^O  =  18  in.  =  f  ft., 
and  A.O  =  3  in.  =  J  ft.  Selecting  the  foot-pound-second 
system  of  units,  in  which  g  =  32.2,  the  linear  dimensions  must 
be  used  in  feet,  the  heaviness,  y,  of  the  iron  must  be  used  in 
Ibs.  per  cubic  foot,  i.e.,  y  =  480  (see  §  7),  and  all  forces  in  Ibs., 
times  in  seconds. 

The  weight  of  the  iron  will  be  G  =  Vy  =  Hbjiy  =  %  .  f  .  $ 
X  480  =  90  Ibs.;  its  mass  =  90  -j-  32.2  =  2.79  ;  and  its  mo- 
ment of  inertia  about  Z  =  Iz  =  Mk*  =  M-fa(b?  +  b9)  =  2.79 
X  0.426  =  1.191.  (That  is,  the  radius  of  gyration,  kz,  = 
1/0.426  =  0.653  ft.;  or  the  moment  of  inertia,  or  any  result 
depending  solely  upon  it,  is  just  the  same  as  if  the  mass  were 
concentrated  in  a  thin  shell,  or  a  line,  or  a  point,  at  a  distance 
of  0.653  feet  from  the  axis.)  We  can  now  compute  the  an- 
gular acceleration,  6,  from  eq.  (3)  ; 


116  MECHANICS    OF   ENGINEERING. 

48  Xj  24 


"1.191  +  (48  -i-  32.2)  X  i  ~  1.191  +  0.372  " 
^-measure  units  per  "  square  second."     The  linear  acceleration 
of  Pl  is^>  =  da  =  7.68  feet  per  square  second  for  the  uniform., 
ly  accelerated  translation. 

Nothing  has  yet  been  said  of  the  velocities  and  initial  condi- 
tions of  the  motions  ;  for  what  we  have  derived  so  far  applies 
to  any  point  of  time.  Suppose,  then,  that  the  angular  velocity 
co  =  zero  when  the  time,  t  =  0  ;  and  correspondingly  the  ve- 
locity, v  =  ooa,  of  translation  of  P^  be  also  =  0  when  t  =  0. 
At  the  end  of  any  time  tf,  GO  =  6t  (§§  56  and  110)  and  v  =  pt 
=  Oat  ;  also  the  angular  space,  a  =  £6^a,  described  by  the  par- 
allelopiped  during  the  time  tf,  and  the  linear  space  s  =  %pf 
=  %6at\  through  which  the  weight  P^  has  sunk  vertically. 
For  example,  during  the  first  second  the  parallelepiped  has  ro- 
tated through  an  angle  a  =  \Qf  =  £  X  15.36x1  =  7.68  units, 
jr-measure,  i.e.,  (7.68  -f-  2?r)  =  1.22  revolutions,  while  Pl  has 
Bunk  through  s  =  %6af  =  3.84  ft.,  vertically. 

The  tension  in  the  cord,  from  (2),  is 

p  =  48(1  -  15.36  X  4  -*-  g)  =  48(1  -  0.24)  =  36.48  Ibs. 

The  pressures  at  the  bearings  will  be  as  follows,  at  any  in- 
stant :  from  (4)  and  (8),  X^  and  X3  must  individually  be  zero  ; 
from  (6)  Z,  =  0  =  Vy  =  W  Ibs.;  while  from  (5)  and  (7),  Yl 
e=  —  21.28  Ibs.,  and  Yt  =  —  15.20  Ibs.,  and  should  point  in  a 
direction  opposite  to  that  in  which  they  were  assumed  in  Fig. 
129  (see  last  lines  of  §  39). 

116.    Torsion  Balance.      A  Variably  Accel.  Rotary  Motion. 

Axis  Fixed.  —  A  homogeneous  solid  having  an  axis  of  symmetry 

is  suspended  by  an  elastic  prism, 
or  filament  (whose  mass  may  be 
neglected),  so  that  the  latter  is 
vertical  and  coincident  with  the 
axis  of  symmetry,  and  is  not  only 
supported,  but  prevented  from 
turning  at  its  upper  extremity. 
FIG.  130.  If  the  solid  is  turned  about  its 

axis  away  from  its  position  of  rest  and  set  free,  the  torsional 


DYNAMICS    OF    A    RIGID    BODY.  117 

elasticity  of  the  rod  or  filament,  which  is  fixed  in  the  solid, 
causes  an  oscillatory  rotary  motion.  Required  the  duration  of 
an  oscillation.  Fig.  130. 

Take  the  axis  Y  at  the  middle  of  the  oscillation  (the  original 
position  of  rest).  Keckon  the  time  from  the  instant  of  passing 
this  position.  Let  the  initial  angular  velocity  =  GOO.  As  the 
motion  progresses  GD  diminishes,  i.e.,  6  is  negative. 

To  consider  the  body  free,  conceive  the  rod  cut  close  to  the 
body  (in  which  it  is  firmly  inserted),  and  in  the  section  thus 
exposed  put  in  the  vertical  tension  Pr,  and  also  the  horizontal 
forces  forming  a  couple  to  which  at  any  instant  the  twisting 
action  (of  the  portion  of  rod  removed  upon  the  part  left  in  the 
free  body)  is  known  to  be  due.  Call  the  moment  of  this  couple 
Qb  (known  as  the  moment  of  torsion) ;  it  is  variable,  being 
directly  proportional  to  the  angle  a ;  hence,  if  by  experiment 
it  is  found  to  be  =  Qlbl  when  a  is  =  <*,,  for  any  value  of  .a  it 
will  be  Qb  =  (Q&  -r-  a^a  =  Ca,  in  which  C  is  the  constant 
factor. 

At  any  instant,  therefore,  the  forces  acting  are  G,  P' r  and 
those  equivalent  to  the  couple  whose  moment  =  Qb  =  Cot. 
(No  lateral  support  is  required ;  the  student  would  find  the  X^ 
YV  Xv  and  Y9  of  Fig.  129  to  be  individually  zero,  if  put  in ; 
remembering  that  here,  x  and  y  both  =  0,  as  also  fdMwz  and 
fdMyz ;  and  that  the  forces  of  the  couple  will  not  be  repre- 
sented in  any  of  the  six  summations  of  §  114,  except  in 
2  moms.z) 

From  eq.  (XIV.),  §  114,  we  have  -  Qb,  i.e.,  -  O,  =  6IZ, 
from  which 

6  —  —  (G  -f-  Iz)a,  or,  for  short,  6  =  —  Bot.    .     .     (1) 

Since  B  is  constant,  and  there  is  an  initial  (angular)  velocity 
=  GOO,  and  since  the  variables  6,  GO,  and  a,  in  angular  motion 
correspond  precisely  to  those  (p,  v,  and  s)  of  rectilinear  motion, 
it  is  evident  that  the  present  is  a  case  of  harmonic  motion, 
already  discussed  in  Problem  2  of  §  59.  Applying  the  results 
there  obtained,  since  B  of  eq.  (1)  corresponds  to  the  a  of  that 
problem,  we  find  that  the  oscillations  are  isochronal,  i.e.,  theii 


118  MECHANICS    OF   ENGINEERING. 

durations  are  the  same  whatever  the  amplitude  (provided  the 
elasticity  of  the  rod  is  not  impaired),  and  that  the  duration  of 
one  oscillation  (from  one  extreme  position  to  the  other)  is 
t'  =  TT  -i-  VlT,  or  finally, 


t'  =it  V<*JZ  -h  Q&.  ......     (2) 

117.  The  Compound  Pendulum  is  any  rigid  body  allowed  to 
oscillate  without  friction  under   the   action  of   gravity  when 
mounted  on  a  horizontal  axis.     Fig.  131  shows  the 
body  free,  in  any  position  during  the  progress  of 
the  oscillation,     C  is  the  centre  of  gravity;  let  GO 
XK\  =  s.    From  (XIV.),  §  114,  we  have  2  (mom.  about 
fixed  axis) 

=  angul.  ace,  X  mom.  of  inertia. 

.*.  —  Gs  sin  a  =  6IM 
and  6  =  —  Gs  sin  a  -j-  /0  =  —  Mgs  sin  a  -j-  Mk\, 

i.e.,  6  =  —  gs  sin  a  -f-  &0a  ......     (1) 

Hence  6  is  variable,  proportional  to  sin  a.  Let  us  see  what 
the  length  I  =  OK,  of  a  simple  circular  pendulum,  must  be,  to 
have  at  this  instant  (i.e.,  for  this  value  of  a)  the  same  angular 
acceleration  as  the  rigid  body.  The  linear  (tangential)  accelera- 
tions of  K,  the  extremity  of  the  required  simple  pendulum 
would  be  (§  77)  pt  =  —  g  sin  or,  and  hence  its  angular  accelera- 
tion* would  =  —  g  sin  a  -f-  1.  Writing  this  equal  to  0  in  eq. 
(1),  we  obtain 

l  =  k?  +  8  ........     (2) 


But  this  is  independent  of  a  ;  therefore  the  length  of  the  sim- 
ple pendulum  having  an  angular  acceleration  equal  to  that  of 
the  oscillating  body  is  the  same  in  all  positions  of  the  latter, 
and  if  the  two  begin  to  oscillate  simultaneously  from  a  position 
of  rest  at  any  given  angle  a1  with  the  vertical,  they  will  keep 
abreast  of  each  other  during  the  whole  motion,  and  hence  have 

*  Most  easily  obtained  by  considering  that  if  the  body  shrinks  into  a  mere 
point  at  K,  and  thus  becomes  a  simple  pendulum,  we  have  both  k0  and  * 
equal  to  I  ;  which  in  (1)  gives  6  =  —  g  sin  a  +  l. 


DYNAMICS    OF   A    RIGID   BODY.  119 

the  same  duration  of  oscillation  ;  which  is  /.  ,for  small  ampli- 
tudes (§  78), 


tf  =  n  VI  -r-  g  =  n  Vic?  -*-  gs,    .     .     .     .     (3) 

K  is  called  the  centre  of  oscillation  corresponding  to  the  given 
centre  of  suspension  0,  and  is  identical  with  the  centre  of  per- 
cussion (§113). 

Example.  —  Required  the  time  of  oscillation  of  a  cast-iron 
cylinder,  whose  diameter  is  2  in.  and  length  10  in.,  if  the  axis 
of  suspension  is  taken  4  in.  above  its  centre.  If  we  use  32.2 
for  <?,  all  linear  dimensions  should  be  in  feet  and  times  in 
seconds.  From  §  100,  we  have 


From  eq.  (3),  §  88, 

70  =  70+  Ms*  =  M  [^  .  \o/  +  i]  =  M  X  0.170; 


.-.  &oa  =  0.170  sq.  ft.;  .-.  tf=  7t  V0.170  -f-  (32.2xi)  =  0.395  sec. 

118.  The  Centres  of  Oscillation  and  Suspension  are  Inter- 
changeable. —  (Strictly  speaking,  these  centres  are  points  in  the 
line  through  the  centre  of  gravity  perpendicular  to  the  axis  of 
suspension.)  Refer  the  centre  of  oscillation  K  to  the  centre 
of  gravity,  thus  (Fig.  132,  at  (I.)  )  : 


~-  -- 


Now  invert  the  body  and  suspend  it  at  K\ 

required   CK^  or  *a,  to  find  the  centre  of  f~"  "/•—  01 

oscillation  corresponding  to  K  as  centre  of  I    / 

suspension.     By  analogy  from  (1)  we  have  Si]7...JK 

s2  =  ]CG  -r-  81  ;  but  from  (1),  k^  -=-«,  =  «  .*. 

sz  —  s:  in  other  words,  K^  is  identical  with          (I.)          (H.) 

').     Hence  the  proposition  is  proved.  FlGK  132> 

Advantage  may  be  taken  of  this  to  determine  the  length  Z 
y£  the  theoretical  simple  pendulum  vibrating  seconds,  and  thus 
inally  the  acceleration  of  gravity  from  formula  (3),  §  117,  viz., 


120  MECHANICS    OF   ENGINEERING. 

when  if  =  1.0  and  I  (now  =  Z)  has  been  determined  experi- 
mentally, we  have 

g  (in  ft.  per  sq.  second)  =  L  (in  ft.)  X  ^a.    .     .     (2) 

This  most  accurate  method  of  determining  g  at  any  locality 
requires  the  use  of  a  bar  of  metal,  furnished  with  a  sliding 
weight  for  shifting  the  centre  of  gravity,  and  with  two  project- 
ing blocks  provided  with  knife-edges.  These  blocks  can  also 
be  shifted  and  clamped.  By  suspending  the  bar  by  one  knife- 
-  edge  on  a  proper  support,  the  duration  of  an  oscillation  is  com- 
puted by  counting  the  total  number  in  as  long  a  period  of 
time  as  possible;  it  is  then  reversed  and  suspended  on  the 
other  with  like  observations.  By  shifting  the  blocks  between 
successive  experiments,  the  duration  of  the  oscillation  in  one 
position  is  made  the  same  as  in  the  other,  i.e.,  the  distance  be- 
twee-n  the  knife-edges  is  the  length,  I,  of  the  simple  pendulum 
vibrating  in  the  computed  time  (if  the  knife-edges  are  not  equi- 
distant from  the  centre  of  gravity),  and  is  carefully  measured. 
The  I  and  t'  of  eq.  (3),  §  117,  being  thus  known,  g  may  be  com- 
puted. Professor  Bartlett  gives  as  the  length  of  the  simple 
pendulum  vibrating  seconds  at  any  latitude  (3 

Z(in  feet)  =  3.26058  —  0.008318  cos  2/?. 

119.  Isochronal  Axes  of  Suspension. — In  any  compound 
pendulum,  for  any  axis  of  suspension,  there  are  always  three 
others,  parallel  to  it  in  the  same  gravity-plane,  for  which  the 
oscillations  are  made  in  the  same  time  as  for  the  first.  For 
any  assigned  time  of  oscillation  t',  eq.  (3),  §  117,  compute  the 
corresponding  distance  CO  =  s  of  O  from  (7; 

i.e.,  from         t'*  = 

we  have  s  =  (gt'*-z-%7r*)±  |/(^Y4^-4;7r4)  —  &cs.  .     .     (1) 

Hence  for  a  given  if,  there  are  two  positions  for  the  axis  0 
parallel  to  any  axis  through  C,  in  any  gravity-plane,  on  both 
sides;  i.e.,  four  parallel  axes  of  suspension,  in  any  gravity- 
plane,  giving  equal  times  of  vibration  ;  for  two  of  these  axes 


Mgs  Mgs 


DYNAMICS    OF   A    KIGID    BODY.  121 

we  must  reverse  the  body.  E.g.,  if  a  slender,  homogeneous, 
prismatic  rod  be  marked  off  into  thirds,  the  (small)  vibrations 
will  be  of  the  same  duration,  if  the  centre  of  suspension  is 
taken  at  either  extremity,  or  at  either  point  of  division. 

Example. — Required  the  positions  of  the  axes  of  suspension, 
parallel  to  the  base,  of  a  right  cone  of  brass,  whose  altitude  is 
six  inches,  radius  of  base,  1.20  inches,  and  weight  per  cubic  inch 
is  0.304:  Ibs.,  so  that  the  time  of  oscillation  may  be  a  half- 
second.  (N.B.  For  variety,  use  the  inch-pound-second  system 
of  units,  first  consulting  §  51.) 

120,  The  Fly- Wheel  in  Fig.  133  at  any  instant  experiences 
a  pressure  P'  against  its  crank-pin  from  the  connecting-rod 
and  a  resisting  pressure  P"  from  the  teeth  of  a  spur-wheel  with 


FIG    133. 


which  it  gears.  Its  weight  G  acts  through  C  (nearly),  and 
there  are  pressures  at  the  bearings,  but  these  latter  and  G  have 
no  moments  about  the  axis  C  (perpendicular  to  paper).  The 
figure  shows  it  free,  P"  being  assumed  constant  (in  practice 
this  depends  on  the  resistances  met  by  the  machines  which  D 
drives,  and  the  fluctuation  of  velocity  of  their  moving  parts). 
jP',  and  therefore  T  its  tangential  component,  are  variable, 
depending  on  the  effective  steam-pressure  on  the  piston  at  any 
instant,  on  the  obliquity  of  the  connecting-rod,  and  in  high- 
speed engines  on  the  masses  and  motions  of  the  piston  and  con- 
necting-rod. Let  r  =  radius  of  crank-pin  circle,  and  a  the 
perpendicular  from  C  on  P".  From  eq.  (XIV.),  §  114,  we 
have 

Tr  -  P"a  =  61^  .-.  8=(Tr-  P"a]  -*-  7C,     .     (1 


122  MECHANICS    OF   ENGINE K KING. 

as  the  angular  acceleration  at  any  instant ;  substituting  which  in 
the  general  equation  (VIII.),  §  110,  we  obtain 

Ictodao  =  Trda  -  P"ada (2) 

From  (1)  it  is  evident  that  if  at  any  position  of  the  crank-pin 
the  variable  Tr  is  equal  to  the  constant  P"a,  0  is  zero,  and 
consequently  the  angular  velocity  GO  is  either  a  maximum  or  a 
minimum.  Suppose  this  is  known  to  be  the  case  both  at  m 
and  n ;  i.e.,  suppose  T,  which  was  zero  at  the  dead-point  A, 
has  been  gradually  increasing,  till  at  ?i,  Tr  =  P"a\  and  there- 
after increases  still  further,  then  begins  to  diminish,  until  at  m 
Tr  again  —  P"a,  and  continues  to  diminish  toward  the  dead- 
point  13.  The  angular  velocity  00,  whatever  it  may  have  been 
on  passing  the  dead-point  A,  diminishes,  since  6  is  negative, 
from  A  to  n,  where  it  is  &)n,  a  minimum ;  increases  from  n  to 
m,  where  it  reaches  a  maximum  value,  &)m.  n  and  m  being- 
known  points,  and  supposing  con  known,  let  us  inquire  what 
oom  will  be.  From  eq.  (2)  we  have 

/»wm  n,m  ,*m 

Ic       vdG,=  /    Trda-P"        ada.      .     .     (3) 

t/«n  t/n  t/n 

But  rda  =  ds  =  an  element  of  the  path  of  the  crank-pin,  and 
also  the  "  virtual  velocity"  of  the  force  T,  and  ada  =  ds" ,  an 
element  of  the  path  of  a  point  in  the  pitch-circle  of  the  fly- 
wheel, the  small  space  through  which  P"  is  overcome  in  dt. 
Hence  (3)  becomes 

-/ciOm2  -  <*>n)  =f™Tds  -  P"  X  linear  arc  ~EF.    (4) 

Tds  we  might,  by  a  knowledge  of  the  vary- 
ing steam-pressure,  the  vary  ing  obliquity  of  the  connecting-rod, 
etc.,  determine  T  for  a  number  of  points  equally  spaced  along 
the  curve  nm,  and  obtain  an  approximate  value  of  this  sum  by 
Simpson's  Rule ;  but  a  simpler  method  is  possible  by  noting 
(see  eq.  (1),  §  65)  that  each  term  Tds  of  this  sum  —  the  corre- 
sponding term  Pdx  in  the  series  /  Pdx,  in  which  P  =  the 


DYNAMICS   OF   A   RIGID   BODY.  123 

effective  steam-pressure  on  the  piston  in  the  cylinder  at  any  in- 
stant, dx  the  small  distance  described  by  the  piston  while  the 
crank-pin  describes  any  ds,  and  n'  and  m'  the  positions  of  the 
piston  (or  of  cross-head,  as  in  Fig.  133)  when  the  crank-pin  is 
at  n  and  m  respectively.  (4)  may  now  be  written 

x»Wl/ 

*<£(<»*  -  G*n°)  =Jn/   Pdx  -  P"  X  linear  arc  EF,     (5) 

from  which  oom  may  be  found  as  proposed.  More  generally,  it 
is  available,  alone  (or  with  other  equations),  to  determine  any 
one  (or  more,  according  to  the  number  of  equations)  unknown 
quantity.  This  problem,  in  rotary  motion,  is  analogous  to  that 
in  §  59  (Prob.  4)  for  rectilinear  motion.  Friction  and  the  in- 
ertia of  piston  and  connecting-rod  have  been  neglected.  As 
to  the  time  of  describing  the  arc  nm,  from  equations  similar  to 
(5),  we  may  determine  values  of  GO  for  points  along  nm,  divid- 
ing it  into  an  even  number  of  equal  parts,  calling  them  c^,  G?S, 
etc.,  and  then  employ  Simpson's  Rule  for  an  approximate  value 

r~m  nim,  $a 

of  the  sum  t=Jn  ~  (from  eq.  (VI.),  §110);  e.g.,  with 
four  parts,  we  would  have 

rm         1                                      I"  1        4        2        4        1  ~| 
t  =  TTT  (angle  n  0m,  TT-meas.)     — I 1 1 1-  —    .(6) 

U      12 v  ?|^.^!^T%T:fkT-flM^' 

121.  Numerical  Example.  Fly-Wheel. — (See  Fig.  133  and 
the  equations  of  §  120.)  Suppose  the  engine  is  non-condensing 
and  non-expansive  (i.e.,  that  P  is  constant),  and  that 

P  =  55001bs.,     7- =  6  in.  =4  ft.,     a  =  2  ft., 

and  also  that  the  wheel  is  to  make  120  revolutions  per  minute, 
i.e.,  that  its  mean  angular  velocity  is  to  be 

GO'  —  J^°-  X  2nr,  i.e.,  GO'  =  ±TC  "  radians"  per  sec. 

First,  required  the  amount  of  the  resistance  P"  (constant) 
that  there  shall  be  no  permanent  change  of  speed,  i.e.,  that  the 
angular  velocity  shall  have  the  same  value  at  the  end  of  a  com- 
plete revolution  as  at  the  beginning.  Since  an  equation  of  the 
form  of  eq.  (5)  holds  good  for  any  range  of  the  motion,  let 


124  MECHANICS   OF   ENGINEERING. 

that  range  be  a  complete  revolution,  and  we  shall  have  zero  as 
the  left-hand  member  ;  fPdx  =  P  X  2  f  t.  =  5500  Ibs.  X  2  ft., 
or  11,000  foot-pounds  (as  it  may  be  called);  while  P"  is  un- 
known, and  instead  of  lin.  arc  EF  we  have  a  whole  circumfer- 
ence of  2  ft.  radius,  i.e.,  4?r  ft.; 

/.  0  =  11,000  -  P"  X  4  X  3.1416;     whence     P"  =  875  Ibs. 
Secondly,  required  the  proper  mass  to  be  given  to  the  fly- 
wheel of  2  ft.  radius  that  in  the  forward  stroke  (i.e.,  while  the 
crank-pin  is  describing  its  upper  semicircle)  the  max.  angular 
velocity  oom  shall  exceed  the  minimum  oon  by  only  -J^G/,  assum- 
ing (which  is  nearly  true)  that  %(com  +  Gon)  =  <&'.     There  be 
ing  now  three  unknowns,  we  require  three  equations,  which 
are,  including  eq.  (5)  of  §  120,  viz.: 


-  P"  X  linear  arc  EF\  (5) 

i(^m+  <»»)=  *>'=  4;r  ;  (7)     and     com  -  a>n  =  ^GO'  =  far.  (8) 

The  points  n  and  m  are  found  most  easily  and  with  sufficient 
accuracy  by  a  graphic  process.  Laying  off  the  dimensions  to 
scale,  by  trial  such  positions  of  the  crank-pin  are  found  that 
T)  the  tangential  component  of  the  thrust  P'  produced  in  the 
connecting-rod  by  the  steam-pressure  P  (which  may  be  resolved 
into  two  components,  along  the  connecting-rod  and  a  normal 
to  itself)  is  =(a  -r-  r)P",  i.e.,  is  =  3500  Ibs.  These  points  will 
be  n  and  m  (arid  two  others  on  the  lower  semicircle).  The 
positions  of  the  piston  n'  and  m',  corresponding  to  n  and  m  of 
the  crank-pin,  are  also  found  graphically  in  an  obvious  manner. 
We  thus  determine  the  angle  nCm  to  be  100°,  so  that  linear 
arc  EF=  -H^  X  2  ft.  =  i$-7t  ft.,  while 


r'  /»m/  _ 

Pdx  =  5500  Ibs.  X  /    da>  =  5500  X  *'*»'=  5500  X  0.77  ft., 
t/ra' 

n'wf  being  scaled  from  the  draft. 

Now  substitute  from  (7)  and  (8)  in  (5),  and  we  have,  with 
Tc-c  =  2  ft.  (which  assumes  that  the  mass  of  the  fly-wheel  is  con- 
centrated in  the  rim), 


DYNAMICS    OF   A    KIGID   BODY.  125 

(G  -4-  g)  X  4  X  47r  x  f  *  =  5500  X  C.77  -  875  X  ¥*> 
which  being  solved  for  (9  (with  ^  =  32.2 ;  since  we  have  used 
the  foot  and  second),  gives  G  =  600.7  Ibs. 

The  points  of  max.  and  min.  angular  velocity  on  the  back- 
stroke may  be  found  similarly,  and  their  values  for  the  fly- 
wheel as  now  determined ;  they  will  differ  but  slightly  from 
the  oom  and  oon  of  the  forward  stroke.  Professor  Cotterill  says 
that  the  rim  of  a  fly-wheel  should  never  have  a  max.  velocity 
>  80  ft.  per  sec.;  and  that  if  made  in  segments,  not  more  than 
40  to  50  feet  per  second.  In  the  present  example  we  have  for 
the  forward  stroke,  from  eqs.  (7)  and  (8),  oom=  13.2  (^r-measure 
units)  per  second;  i.e.,  the  corresponding  velocity  of  the  wheel- 
rim  is  vm  =  Goma  =  26.4  feet  per  second. 

122.  Angular  Velocity  Constant.  Fixed  Axis. — If  GO  is  con- 
stant, the  angular  acceleration,  6,  must  be  =  zero  at  all  times, 
which  requires  2  (mom.)  about  the  axis  of 
rotation  to  be  =  0  (eq.  (XIV.),  §  114).  An 
instance  of  this  occurs  when  the  only  forces 
acting  are  the  reactions  at  the  bearings  on 
the  axis,  and  the  body's  weight,  parallel  to 
or  intersecting  the  axis ;  the  values  of  these 
reactions  are  now  to  be  determined  for  dif- 
ferent forms  of  bodies,  in  various  positions  FIG.  134. 
relatively  to  the  axis.  (The  opposites  and  equals  of  these  reac- 
tions, i.e.,  the  forces  with  which  the  axis  acts  upon  the  bearings, 
are  sometimes  stated  to  be  due  to  the  "  centrifugal  forces"  or 
"  centrifugal  action."  of  the  revolving  body.) 

Take  the  axis  of  rotation  for  Z,  then,  with  0  =  0,  the  equa- 
tions of  §  114  reduce  to 

....      (IXa.) 
....       (X0.) 

=       0 ; (XI0.) 

2  moms.x  =  —  tffdMyz ;  .  .  .  (Xllff.) 
2  moms.F  =  +  offdMxz  ;  .  .  .  (Xllla.) 
2  moms.*  =  0 (XIV0.) 


126  MECHANICS   OF  ENGINEERING. 

For  greater  convenience,  let  us  suppose  the  axes  X  and  7 
(since  their  position  is  arbitrary  so  long  as  they  are  perpen- 
dicular to  each  other  and  to  Z)  to  revolve  with  the  body  in  its 
uniform  rotation. 

122a.  If  a  homogeneous   body  have  a  plane  of  symmetry 
and  rotate  uniformly  about  any  axis  Z  perpendicular  to  that 
plane  (intersecting  it  at  0],  then  the  acting  forces  are  equiva- 
lent to  a  single  force,  =  crfMp,  applied 
at  0  and  acting  in  a  gravity-line,  but 
directed    away  from    the   centre  of 
gravity.      It  is   evident   that   such  a 
force  P  =  K?Mp,  applied  as  stated 
FIG- 135-  (see  Fig.  135),  will  satisfy  all  six  con- 

ditions expressed  in  the  foregoing  equations,  taking  X  through 
the  centre  of  gravity,  so  that  x  =  p.  For,  from  (IX&.)?  P  must 
=  GifMp,  while  in  each  of  the  other  summations  the  left- 
hand  member  will  be  zero,  since  P  lies  in  the  axis  of  X',  and 
as  their  right-hand  members  will  also  be  zero  for  the  present 
body  (y  =  0  ;  and  each  of  the  snmsfdMyz  andfdMxz  is  zero, 
since  for  each  term  dMy(  -f-  z)  there  is  another  dMy(  —  z) 
to  cancel  it ;  and  similarly,  forfdMxz),  they  also  are  satisfied ; 
Q.E.D.*  Hence  a  single  point  of  support  at  0  will  suffice  to 
maintain  the  uniform  motion  of  the  body,  and  the  pressure 
against  it  will  be  equal  and  opposite  to  P. 

First  Example. — Fig.  136.  Supposing  (for  greater  safety) 
that  the  uniform  rotation  of  210  revolutions 
per  minute  of  each  segment  of  a  fly-wheel  is 
maintained  solely  by  the  tension  in  the  cor- 
responding arm,  P  ;  required  the  value  of  P 
if  the  segment  and  arm  together  weigh  -^  of 
a  ton,  and  the  distance  of  their  centre  of  FlG- 

gravity  from  the  axis  is  ~p  =  20  in.,  i.e.,  =  |  ft.  With  the  foot- 
ton-second  system  of  units,  with  g  =  32.2,  we  have 

P  =  oo\Mp  =  [^  X  27T]3  X  [-A-  •*-  32-2]  X  I  =  0.83  tons,. 
or  1660  Ibs. 

*  That  is,  neglecting  gravity. 


DYNAMICS    OF   A   RIGID   BODY. 


127 


Second  Example. — Fig.  137.  Suppose  the  uniform  rotation 
of  the  same  fly-wheel  depends  solely  on  the  tension  in  the  rim, 
required  its  amount.  The  figure  shows  the  half-  •^7 
rim  free,  with  the  two  equal  tensions,  P',  put  in  at  ^ 
the  surfaces  exposed.  Here  it  is  assumed  that  the  I  _  c  • 
arms  exert  no  tension  on  the  rim.  From  §122$  we 
have  2P'  =  afMp,  where  Jf  is  the  mass  of  the  half-  | 
rim,  and  p  its  gravity  co-ordinate,  which  may  be  ob-  FIG.  137. 
tained  approximately  by  §  26,  Problem  1,  considering  the  rim 
as  a  circular  wire,  viz.,  p  =  %r  -i-  TT. 

Let  M  =  (180  Jbs.)  -^  g,  with  r  =  2  ft.     We  have  then 

P'  =  i(22)'(180  ~  32.2X4  --  *)  =  1718.0  Ibs. 

(In  reality  neither  the  arms  nor  the  rirn  sustain  the  tensions 
just  computed  ;  in  treating  the  arms  we  have  supposed  no  duty 
done  by  the  rim,  and  vice  versa.  The  actual  stresses  are  less, 
and  depend  on  the  yielding  of  the  parts.  Then,  too,  we  have 
supposed  the  wheel  to  take  no  part  in  the  transmission  of  mo- 
tion by  belting  or  gearing,  which  would  cause  a  bending  of  the 
arms,  and  have  neglected  its  weight.) 

122b.  If  a  homogeneous  body  have  a  line  of  symmetry  and 
rotate  uniformly  about  an  axis  parallel  to  it  (0  being  the  foot 
of  the  perpendicular  from  the  centre  of  gravity  on  the  axis), 
then  the  acting  forces  are  equivalent  to  a  single  force  P^ 
=  (x?Mp,  applied  at  0  and  acting  in  a  gravity-line  away 

from  the  centre  of  gravity. 
Taking  the  axis  X  through  the 
centre  of  gravity,  Z  being  the 
axis  of  rotation,  Fig.  138,  while 
Z'  is  the  line  of  symmetry,  pass 
an  auxiliary  plane  Z '  Y'  parallel 
to  ZY.  Then  the  sum  fdMxz 
may  be  written  fdM(p  +  x')z 
which  =  pfdMz  -J~  fdMx'z. 
FlG-  138-  RutfdMz  —  Mz  —  0,  since  ~z 

=  0,  and  every  term  dM(-\-  xf)z  is  cancelled  by  a  numerically 


128 


MECHANICS    OF    P:NGINEERINa 


equal  term  dM(—  xr)z  of  opposite  sign.  Hence  fdMxz  =  0. 
K\sofdMyz  =  0,  since  each  positive  product  is  annulled  by  an 
equal  negative  one  (from  symmetry  about  Z').  Since,  also, 
y  =  0,  all  six  conditions  in  §  122  are  satisfied.  Q.  E.  D. 

•  If  the  homogeneous  body  is  any  solid  of  revolution  whose 
geometrical  axis  is  parallel  to  the  axis  of  rotation,  the  forego- 
ing is  directly  applicable. 

122c.  If  a  homogeneous  body  revolve  uniformly  about  any 
axis  lying  in  a  plane  of  symmetry,  the  acting  forces  are  equiv- 
alent to  a  single  force  P  =  crfMp,  acting  parallel  to  the  grav- 
ity-line which  is  perpendicular  to  the  axis  (Z),  and  away 
from  the  centre  of  gravity,  its  distance  from  any  origin  0  in 
the  axis  Z  being  —  [fdMxz]  -r-  Mp  (the  plane  ZX  being  a 
gravity-plane). — Fig.  139.  From  the  position  of  the  body  we 
have  p  =  x,  and  y  =  0 ;  hence  if  a 
value  a?Mp  be  given  to  P  and  it  be 
made  to  act  through  Z  and  parallel  to 
X,  and  away  from  the  centre  of  gravity, 
all  the  conditions  of  §  122  are  satisfied 
except  (XII0.)"  and  (XIII0.).  But 
symmetry  about  the  plane  XZ  makes 
fdMyz  =  0,  and  satisfies  (X1I#.),  and 
by  placing  P  at  a  distance  a  =  fdMxz  -r-  Mp  from  0  along  Z 
we  satisfy  (Xllla.)-  Q.  E.  D. 

Example. — A  slender,  homogeneous,  prismatic  rod,  of  length 
=  I,  is  to  have  a  uniform  motion,  about  a  ver-  Q. 
tical    axis    passing   through    one    extremity, 
maintained  by  a  cord-connection  with  a  fixed   p 
point  in  this  axis.     Fig.  140.     Given  GO,  (p,  I,  f 
(JcT  =  %l  cos  (p),  and  F  the  cross-section  of  the  ^ 
rod,  let  s  =  the  distance  from  0  to  any  dM  \ 
of  the  rod,  dM  being  =  Fyds  -r-  g.     The  x 
of  any  dM  =  s  cos  <p ;  its  z  =  s .  sin  cp ; 

.-.  fdMxz  =  (Fy  -r-  g)  sin  cp  cos  (pj  s*ds 

-r-  g)V  sin  cp  cos  cp  =  \M Z*  sin  <p  cos 


FIG.  139. 


Fio.  140. 


DYNAMICS   OF  A   RIGID   BODY.  129 

Hence  a,  =  fdMxz  -~  Mp,  is  =  fZ  sin  cp,  and  the  line  of  ac- 
tion of  P  (  =  <#' Mp  —  GO*  (Fyl  ~  g)  \l  cos  cp)  is  therefore 
higher  up  than  the  middle  of  the  rod.  Find  the  intersection 
I)  of  G  and  the  horizontal  drawn  through  Z  at  distance  a  from 
0.  Determine  P'  by  completing  the  parallelogram  GP',  at- 
taching the  cord  so  as  to  make  it  coincide  with  P';  for  this  will 
satisfy  the  condition  of  maintaining  the  motion,  when  once  be- 
gun, viz.,  that  the  acting  forces  G,  'and  the  cord-tension  P'9 
shall  be  equivalent  to  a  force  P  =  afMp,  applied  horizontally 
through  Z  at  a  distance  a  from  0. 

123,  Free  Axes.  Uniform  Rotation. — Referring  again  to  §  122 
and  Fig.  134,  let  us  inquire  under  what  circumstances  the 
lateral  forces,  JTn  Y^  JT0,  YQ,  with  which  the  bearings  press 
the  axis,  to  maintain  the  motion,  are  individually  zero,  i.e.,  that 
the  hearings  are  not  needed^  and  may  therefore  he  removed 
(except  a  smooth  horizontal  plane  to  sustain  the  body's  weight), 
leaving  the  motion  undisturbed  like  that  of  a  top  "asleep." 
For  this,  not  only  must  2X  and  2  Y  both  be  zero,  but  also 
(since  otherwise  Xl  and  JT0  might  form  a  couple,  or  Yl  and  Y% 
similarly)  2  (moms.)zand  2  (moms.)F  must  each  =  zero.  The 
necessary  peculiar  distribution  of  the  body's  mass  about  the 
axis  of  rotation,  then,  must  be  as  follows  (see  the  equations  of 


First,  x  and  y  each  =  0,  i.e.,  the  axis  must  he  a  gravity-axis. 

Secondly,  fdMyz  =  0,  and/$Jfe  =  0,  the  origin  being  any- 
where on  Z,  the  axis  of  rotation. 

An  axis  (Z)  (of  a  body)  fulfilling  these  conditions  is  called 
a  Free  Axis,  and  since,  if  either  one  of  the  three  Principal  Axes 
for  the  centre  of  gravity  (see  §  107)  be  made  an  axis  of  rotation 
(the  other  two  being  taken  for  X  and  Y),  the  conditions 
x  =  0,  y  =  0,  fdMxz  =  0,  and  fdMyz  =  0,  are  all  satisfied, 
it  follows  that  every  rigid  lody  has  at  least  three  free  axes, 
which  are  the  Principal  Axes  of  Inertia  of  the  centre  of 
gravity  at  right  angles  to  each  other. 

In  the  case  of  homogeneous  bodies  free  axes  can  often  be 
determined  by  inspection  :  e.g.,  any  diameter  of  a  sphere ;  any 


MECHANICS   OF   ENGINEERING. 


FIG.  141. 


transverse  diameter  of  a  right  circular  cylinder  through  its 
centre  of  gravity,  as  well  as  its  geometrical  axis ;  the  geomet- 
rical axis  of  any  solid  of  revolution  ;  etc. 

124.  Rotation  about  an  Axis  which  has  a  Motion  of  Translation. 
— Take  only  the  particular  case  where  the  moving  axis  is  a 
gravity-axis.  At  any  instant,  let  the 
velocity  and  acceleration  of  the  axis  be  v 
and  p ;  the  angular  velocity  and  accelera- 
p^»-  /,  j'tf  >  tion  about  that  axis,  GO  and  0.  Then,  since 
T  "'f\.  *  >i>j»p  the  actual  motion  of  a  dM\\\  any  dt  is 
compounded  of  its  motion  of  rotation 
about  the  gravity-axis  and  the  motion  of 
translation  in  common  with  that  axis, 
we  may,  in  forming  the  imaginary  equiva- 
lent system  in  Fig.  141,  consider  each  dM  as  subjected  to  the 
simultaneous  action  of  dP  =  dMp  parallel  to  X,  of  the  tan- 
gential dT  =  dM6p,  arid  of  the  normal  dN  =-  dM(oopf  -=-  p 
=  (*?dMp.  Take  ^Tin  the  direction  of  translation,  Z  (perpen- 
dicular to  paper  through  0)  is  the  moving  gravity-axis ;  Y 
perpendicular  to  both.  At  any  instant  we  shaft  have,  then,  the 
following  conditions  for  the  acting  forces  (remembering  that 
p  sin  cp  =  y,fdMy  —  My  =  0 ;  etc.) : 

2X  =fdP  —  fdTsin  cp  —  fdtf  cos  <p  =  Mp\   .     (1) 

2Y=fdTcos  cp  —  fdNsm  (p  =  0  ;  .     ..    (2) 

2  moms.z  =fdTp  -fdPy  =  OfdMp*  =  6IZ  =  6Mkz\    (3) 

and  three  other  equations  not  needed  in  the  folio  wing  example. 
Example. — A  homogeneous  solid  of  revolution  rolls  (with- 
out slipping)  down  a  rough  inclined 
plane.  Investigate  the  motion.  Con- 
sidering the  body  free,  the  acting  forces 
are  O  (known)  and  N  and  jP,  the  un- 
known normal  and  tangential  compo- 
nents of  the  action  of  the  plane  on  the 
roller.  If  slipping  occurs,  then  P  is  the 
gliding  friction  due  to  the  pressure  N(§  156);  here,  however,  it  is 


FlG-  14S- 


DYNAMICS   OF   A    RIGID   BODY.  131 

less  by  hypothesis  (perfect  rolling).  At  any  instant  the  four 
unknowns  are  found  by  the  equations 

2X,  i.e.,  G  sin  ft  -  P,  =  (G  -T-  g)p ;      .     (1) 
2J-,i.e.,  £cos/?-  JVJ  =  0;     ....     (2) 
^  moms.z,  i.e.,  Pa,  =  BMkj ;      .     .    (3) 

while  on  account  of  the  perfect  rolling, 

Oa=p .'  .    .    .     (4) 

Solving,  we  have,  for  the  acceleration  of  translation, 

p  =  g  sin  ft  -T-  [1  +  (kz*  ~  a8)]. 

(If  the  body  slid  without  friction, p  would  =  g  sin  ft.)  Hence 
for  a  cylinder  (§  97),  ~kz  being  =  Ja*,  we  have^?  =  %g  sin  fi- 
and  for  a  sphere  (§  103)  j?  =  ^  sin  /?. 

(If  the  plane  is  so  steep  or  so  smooth  that  both  rolling  and 
slipping  occur,  then  6a  no  longer  =  p,  but  the  ratio  of  P  to  N 
is  known  from  experiments  on  sliding  friction  ;  hence  there  are 
still  four  equations.) 

The  motion  of  translation  being  thus  found  to  be  uniformly 
accelerated,  we  may  use  the  equations  of  §  56  for  finding  dis- 
tance, time,  etc. 

Query. — How  may  we  distinguish  two  spheres  by  allowing 
them  to  roll  down  the  same  inclined  plane,  if  one  of  them  is 
silver  and  solid,  while  the  other  is  of  gold,  but  silvered  and 
hollow,  so  as  to  be  the  same  as  the  first  in  diameter,  weight, 
and  appearance? 

125.  Parallel-Rod  of  a  Locomotive. — When  the  locomotive 
moves  uniformly,  each  dJff  of  the  rod  between  the  two  (or 
three)  driving-wheels  rotates  with 
uniform  velocity  about  a  centre  of  its 
own  on  the  line  BD,  Fig.  143,  and  with 
a  velocity  v*  and  radius  r  common  ^^ 
to  all,  and  likewise  has  a  horizontal  (  j.\  ;  : 
uniform  motion  of  translation.  Hence  (ii.)" 

if  we  inquire  what  are  the  reactions  P  Fl°-  14a 

*  This  velocity  is  that  which  the  dM  has  relatively  to  the  frame  of  the 
locomotive,  in  a  circular  path.  E.g.,  if  the  locomotive  (frame;  has  a  velocity 
of  60  miles  per  hour  and  the  radius  r  is  one-third  of  the  radius  of  the  driver, 
then  v  is  20  miles  per  hour. 


132  MECHANICS    OF   ENGINEERING. 

of  its  supports,  as  induced  solely  ~by  its  weight  and  motion, 
when  in  its  lowest  position  (independently  of  any  thrust  along 
the  rod),  we  put  2Y  of  (I.)  =  2Y  of  (II.)  (II.  shows  the 
imaginary  equivalent  system),  and  obtain 

2P  -  G  =fdN  =fdM»*  -r-  r  =  (v*  H-  r}fdM  =  Mv*  +  r. 

Example. — Let  the  velocity  of  translation  =  50  miles  per 
hour,  the  radius*  of  the  pins  be  18  in.  =  f  ft.,  and  =  half  that 
of  the  driving-wheels,  while  the  weight  of  the  rod  is  200  Ibs. 
With  g  =  32.2,  we  must  use  the  foot  and  second,  and  obtain 

v  =  i[50  X  5280  -r-  3600]  ft.  per  second  =  36.6; 
while          M=  200  -f-  32.2  =  200  X  .0310  =  6.20 ; 
and  finally  P  =  4[200  +  6.2(36.6)3-7-  f]  =  2868.3  Ibs., 

or  nearly  1J  tons,  about  thirty  times  that  due  to  the  weight 
alone. 

126.  So  far  in  this  chapter  the  motion  has  been  prescribed, 
and  the  necessary  conditions  determined,  to  be  fulfilled  by  the 
acting  forces  at  any  instant.  Problems  of  a  converse  nature, 
i.e.,  where  the  initial  state  of  the  body  and  the  acting  forces 
are  given  while  the  resulting  motion  is  required,  are  of  much 
greater  complexity,  but  of  rare  occurrence  in  practice.  The 
reader  is  referred  to  Rankine's  Applied  Mechanics.  A  treat- 
ment of  the  Gyroscope  will  be  found  in  the  American  Journal 
of  Science  for  1857,  and  in  the  article  of  that  name  in  Johnson's 
Cyclopaedia.  ,  '  Worthington's  "Dynamics  of  Rotation" 
(London,  1892)  is  a  valuable  practical  book. 

*  Or,  rather,  the  radius  of  the  circular  path  of  the  pin- centre, 
velocity  in  this  path  is  25  miles  per  hour. 


WORK,   ENEKGY,   AND   POWER. 


CHAPTER  VI. 
WORK,  ENERGY,  AND  POWER. 

127.  Remark. — These  quantities  as  defined  and  developed 
in  this  chapter,  though  compounded  of  the  fundamental  ideas 
of  matter,  force,  space,  and  time,  enter  into  theorems  of  such 
wide  application  and  practical  use  as  to  more  than  justify  their 
consideration  as  separate  kinds  of  quantity. 

128.  Work  in  a  Uniform  Translation.     Definition  of  Work. — 

Let  Fig.  144  represent  a  rigid  body  having  a  motion  of  trans- 
lation parallel  to  X,  acted  on  by  a 
system  of  forces  P1?  Pa,  R^  and  R^ 
which  remain  constant.* 

(Let  s  be  any  distance  described  by 
the  body  during  its  motion  ;Jthen  2X 
must  be  zero  (§  109),  i.e.,  noting  that 
J£,  and  7?4  have  negative  X  com- 
ponents (the  supplements  of  their  Fl°-  144- 
angles  with  ^3Tare  used), 

P,  cos  a1  +  PI  cos  ^2  —  £9  cos  c*9  —  7?4  cos  at  =  0 ; 

or,  multiplying  by  s  and  transposing,  we  have  (noting  that 
s  cos  al  =  sl  the  projection  of  s  on  P,,  that  s  cos  a^  =  sa,  the 
projection  of  s  on  Pa,  and  so  on), 

PA.+  PA  =  #A  +  -ffA-  ....  (a) 
The  projections  5,,  s2,  etc.,  may  be  called  the  distances  de- 
scribed in  their  respective  directions  by  the  forces  Pn  P9,  etc.; 
P,  and  P2  having  moved  forward,  since  s,  and  s^  fall  in  front 
of  the  initial  position  of  their  points  of  application  ;  J?3  and  J?4 
backward,  since  s3  and  s4  fall  behind  the  initial  positions  in 
their  case.  (By  forward  and  backward  we  refer  to  the  direc- 

*  Constant  in  direction  as  well  as  amount. 


134  MECHANICS   OF   ENGINEERING. 

tion  of  each  force  in  turn.)  The  name  Work  is  given  to  the 
product  of  a  force  by  the  distance  described  in  the  direction 
of  the  force  ~by  the  point  of  application.  If  the  force  moves 
forward  (see  above),  it  is  called  a  working-force,  and  is  said  to 
do  the  work  (e.g.,  ^P^j)  expressed  by  this  product;  while  if 
backward,  it  is  called  a  resistance,  and  is  then  said  to  have  the 
work  (e.g.,  7?3s8),  done  upon  it,  in  overcoming  it  through  the 
distance  mentioned  (it  might  also  be  said  to  have  done  nega- 
tive work). 

Eq.  (a)  above,  then,  proves  the  theorem  that :  In  a  uniform 
translation,  the  working  forces  do  an  amount  of  work  which 
is  entirely  applied  to  overcoming  the  resistances. 

129.  Unit  of  Work. — Since  the  work  of  a  force  is  a  product 
of  force  by  distance,  it  may  logically  be  expressed  as  so  many 
foot-pounds,  inch-pounds,  kilogram-meters,  according   to   the 
system  of  units  employed.     The  ordinary  English  unit  is  the 
foot-pound,  or  ft.-lb.     It  is  of  the  same  quality  as   a  force- 
moment. 

130.  Power. — Work  as  already  defined  does  not  depend  on 
the  time  occupied,  i.e.,  the  work  Plsl  is  the  same  whether  per- 
formed in  a  long  or  short  time;  but  the  element  of  time  is  of 
so  great  importance  in  all  the  applications  of  dynamics,  as  well 
as  in  such  practical  commercial  matters  as  water-supply,  con- 
sumption of  fuel,  fatigue  of  animals,  etc.,  that  the  rate  of  work 
is  a  consideration  both  of  interest  and  necessity. 

Power  is  the  rate  at  which  work  is  done,  and  one  of  its 
units  is  one  foot-pound  per  second  in  English  practice  ;  a  larger 
one  will  be  mentioned  presently. 

The  power  exerted  by  a  working  force,  or  expended  upon  a 
resistance,  may  be  expressed  symbolically  as 

L  =  P,s,  -r-  t,     or     R^s^  -±  t, 

in  which  t  is  the  time  occupied  in  doing  the  work  P1sl  or  7?3s, 
(see  Fig.  144) ;  or  if  vl  is  the  component  in  the  direction  of 
the  force  Pl  of  the  velocity ~v  of  the  body,  we  may  also  write 

L  =  P}v, (b) 


WOKK,    ENERGY,    AND    POWER.  135 

131.  Example. — Fig.    145,  shows  as  a  free  ~body  a  sledge 
which  is  being  drawn  uniformly  up 
a   rough   inclined    plane    by  a   cord 
parallel  to  the  plane.     Required  the 
total  power  exerted  (and  expended), 
if  the  tension  in  the  cord  is  JPl  =  100  ^^\^^^\^^  (WEIGHT) 
Ibs.,  the  weight  of  sledge  R^  =  160   '  FIG.  145. 

Ibs.,  ft  =  30°,  and  the  sledge  moves  J240  ft.  each  minute.  2? 
and  Jt4  are  the  normal  and  parallel  (i.e.,  7?4  =  friction)  com- 
ponents of  the  reaction  of  the  plane  on  the  sledge.  From  eq. 
(1),  §  128,  the  work  done  while  the  sledge  advances  through 
s  =  240  ft.  may  be  obtained  either  from  the  working  forces, 
which  in  this  case  are  represented  by  PY  alone,  or  from  the 
resistances  R^  and  J?4.  Take  the  former  method  first.  Pro- 
jecting s  upon  PI  we  have  sl  =  s. 
Hence  PA  or  100  Ibs.  X  240  ft.  =  24,000  ft.-lbs. 
of  work  done  in  60  seconds.  That  is,  the  power  exerted  by  the 
working  forces  is 

L  =  P^  -T-  t  =  400  ft.-lbs.  per  second. 

As  to  the  other  method,  we  notice  that  J?3  and  J?4  are  resist- 
ances, since  the  projections  s3  =  s  sin  ft,  and  s4  =  *,  would  fall 
back  of  their  points  of  application  in  the  initial  position,  while 
N  is  neutral,  i.e.,  is  neither  a  working  force  nor  a  resistance, 
since  the  projection  of  s  upon  it  is  zero. 

From  2 X  =  0  we  have  —  7?4  —  R*  sin  ft  +  Pl  =  0, 

and  from    2  Y  =  0  (§  109)          N  —  ft*  cos  ft  =0; 

whence  R^  the  friction  =  20  Ibs.,  and  N  =  138.5  Ibs.  Also, 
since  s*  —  s  sin  ft  =  240  X  4  =  120  ft.,  and  st  =  s,  =  240  ft., 
we  have  for  the  work  done  upon  the  resistances  (i.e.,  in  over- 
coming them)  in  60  seconds 

fi^  _|_  ^A  =  160  X  120  +  20  X  240  =  24,000  ft.-lbs., 
and  the  power  expended  in  overcoming  resistances, 

L  =  24,000  -f-  60  =  400  ft.-lbs.  per  second, 
as  already  derived.     Or,  in  words  the  power  exerted  by  tho 


136  MECHANICS  OF  EJtfGINEEKING. 

tension  in  the  co^-d  is  expended  entirely  in  raising  the  weight 
a  vertical  height  of  2  feet,  and  overcoming  the  friction  through 
a  distance  of  4  feet  along  the  plane,  every  second  ;  the  motion 
l>eing  a  uniform  translation. 

132.  Horse-Power. — As  an  average,  a  horse  can  exerts  a  trac- 
tive effort  or  pull  of  100  Ibs.,  at  a  uniform  pace  of  4ft.  per  sec- 
ond, for  ten  hours  a  day  without  too  great  fatigue.     This  gives 
a  power  of  400  f  t.-lbs.  per  second ;  but  Boulton  &  Watt  in 
rating  their  engines,  and  experimenting  with   the  strong  dray- 
horses  of  London,  fixed  upon  550  ft.-lbs.  per  second,  or  33,000 
ft.-lbs.  per  minute,  as  a  convenient  large  unit  of  power.     (The 
French  horse-power,  or  cheval-vapeur,  is  slightly  less  than  the 
English,  being  75_kilogram meters-  per  second,  or  32,550  ft.-lbs. 
per  minute.)     This  value  for  the  horse-power  is  in   common 
use.     In  the  example  in  §  131,  then,  the  power  of  400  ft.-lbs, 
per  second  exerted  in  raising  the  weight  and  overcoming  fric- 
tion may  be  expressed  as  (400-^-550  =)  T8T  of  a  horse-power.    A 
man  can  work  at  a  rate  equal  to  about  -^  of  a  horse-powex , 
with  proper  intervals  for  eating  and  sleeping. 

133.  Kinetic  Energy.     Retarded  Translation. — In  a  retarded 
translation  of  a  rigid  body  whose  mass  —  M,  suppose  there 
are  no  working-forces,  and  that  the  resistances  are  constant  and 
their  resultant  is  R.     (E.g.,  Fig.   146  shows  such  a  case ;  a 

sledge,  having  an  initial  velocity  c  and  slid- 
)V  ing  on  a  rough  horizontal  plane,  is  gradu- 
ally retarded  by  the  friction  R.}     R  is  par- 
allel  to  the  direction  of  translation  (§  109) 
and  the  acceleration  is  p  =  —  R  -i-  M ; 
hence  from  vdv  =pds  we  have 

fvdv  =  -  (1  -f-  M)fRds (1) 

But  the  projection  of  each  da  of  the  motion  upon  R  is  =  ds 
itself ;  i.e.  (§  128),  Rds  is  the  work  done  upon  R,  in  overcom 
ing  it  through  the  small  distance  ds,  and  fRds  is  -the  sum  of 
all  such  amounts  of  work  throughout  any  definite  portion  of 
the  motion.     Let  the  range  of  motion  be  between  the  points 


WORK,    ENERGY,    AND    POWER.  137 

where  the  velocity  =  c,  and  where  it  =  zero  (i.e.,  the  mass 
has  come  to  rest)  With  these  limits  in  eq.  (1)  (0  and  s'  be- 
ing the  corresponding  limits  for  s),  we  have 

**'.  r*£ds. 


2 


That  is,  in  giving  up  all  its  velocity  c  the  'body  has  "been  able 
to  do  the  work/fids  (this,  if  R  remains  constant,  reduces  to 

Me* 
Us'}  or  its  equal  ~o~.     If>  then,  by  energy  we  designate  the 

ability  to  perform  work,  we  give  the  name  kinetic  energy  of 
a  moving  body  to  the  product  of  its  mass  l)y  half  the  square 

of  its  velocity  (~o~);  i-e.,  energy  due  to  motion^     (The  anti- 
quated term  vis  viva  was  once  applied  to  the  form  Mv*.) 

134.  Work  and  Kinetic  Energy  in  any  Translation.  —  Let  P 

be  the  resultant  of  the  working  forces  at  any  instant,  R  that 
of  the  resistances  ;  they  (§  109)  will  both  M 

act  in  a  gravity-line*  parallel  to  the  di-  <- 
rection  of  translation.     The  acceleration    O 

___^<U  _ 

at    any    instant    is   p  =    (^2X  -r-  M)  Fia.  147. 

=  (P  —  It)  ~  M\  hence  from  vdv  =  pds  we  have 

Mvdv  =  Pds  —  Eds.     .    .    ...    (1) 

Integrating  between  any  two  points  of  the  motion  as  O  and  0' 
where  the  velocities  are  V0  and  vf,  we  have  after  transposition 

Mv 


p 


x 


/ 
^= 


But  P  being  the  resultant  of  P,,  P»  etc.,  and  R  that  of 
R^  RV  etc.,  we  may  prove,  as  in  §  62,  that  if  du^  du^  etc.,  be 
the  respective  projections  of  any  ds  upon  P^  Pv  etc.,  while 
dw^  dWv  etc.,  are  those  upon  R^  R»  etc.,  then 

....    and 


and  (d)  may  be  rewritten 


*  That  is,  a  line  passing  through  the  centre  of  gravity. 


138  MECHANICS   OF   ENGINEERING. 


or,  in  words  :  In  any  translation,  a  portion  of  the  work  done 
by  the  working  forces  is  applied  in  overcoming  the  resistances 
while  the  remainder  equals  the  change  in  the  kinetic  energy  of 
the  body. 

It  will  be  noted  that  the  bracket  in  (e)  depends  only  on  the 
initial  and  final  velocities,  and  not  upon  any  intermediate 
values  ;  hence,  if  the  initial  state  is  one  of  rest,  and  also  the 
final,  the  total  change  in  kinetic  energy  is  zero,  and  the  work 
of  the  working  forces  has  been  entirely  expended  in  the  work 
of  overcoming  the  resistances  ;  but  at  intermediate  stages  the 
former  exceeds  the  work  so  far  needed  to  overcome  resistances, 
and  this  excess  is  said  to  be  stored  in  the  moving  mass  ;  and  as 
the  velocity  gradually  becomes  zero,  this  stored  energy  becomes 
available  for  aiding  the  working  forces  (which  of  themselves 
are  then  insufficient)  in  overcoming  the  resistances,  and  is  then 
said  to  be  restored.  (The  function  of  a  fly-wheel  might  be 
stated  in  similar  terms,  but  as  that  involves  rotary  motion  it 
will  be  deferred.) 

Work  applied  in  increasing  the  kinetic  energy  of  a  body  is 
sometimes  called  "  work  of  inertia,"  as  also  the  work  done  by 
a  moving  body  in  overcoming  resistances,  and  thereby  losing 
speed. 

135.  Example  of  Steam-Hammer.  —  Let  us  apply  eq.  (e)  to 
determine  the  velocity  v'  attained  by  a  steam-hammer  at  the 
lower  end  of  its  stroke  (the  initial  velocity  being  =  0),  just 
before  delivering  its  blow  upon  a  forging,  supposing  that 
the  steam-  pressure  JP9  at  all  stages  of  the  downward  stroke  is 
given  by  an  indicator.  Fig.  148.  Weight  of  moving  mass 
is  322  Ibs.;  /.  Jlf  =  10  (foot-pound-second  system),  I  =  1  foot. 
The  working  forces  at  any  instant  are  Pl  =  G  =  322  Ibs.;  P2, 
which  is  variable,  but  whose  values  at  the  seven  equally  spaced 


WORK,    ENERGY,    AND    POWER. 


139 


points  a,  I,  c,  d,  e,  /,  g,  are  800,  900,  900,  800,  600,  500,  450 

Ibs.,  respectively.      Rl  the  exhaust-pressure  (16 

Ibs.  per  sq.  inch  X  20  sq.  inches  piston-area)  = 

320  Ibs.,  is  the  only  resistance,  and  is  constant. 

Hence  from   eq.  (e),  since   here  the  projections 

diii,  etc.,  of  any  ds  upon  the    respective  forces 

are  equal  to  each  other  and  =  ds, 


(i)    F 


The  term  fP^ds  can  be  obtained  approximately 
by  Simpson's  Rule,  using  the  above  values  for 
six  equal  divisions,  which  gives 

TV[800  +  4(900  +  800  +  500) 
+  2(900 +  600) +450] 
=  725  ft.-lbs.  of  work.     Hence,  making  all  the  substitutions, 


FIG.  148. 


we  have,  since  /  ds  =  1  ft., 

I/O 


322  X  1  +  725  =  320  X  1  +  %Mv"\  .-.  $Mvn  =  727  ft.-lbs. 
of  energy  to  be  expended  in  the  forging.  (Energy  is  evi- 
dently expressed  in  the  same  kind  of  unit  as  work.)  We  may 
then  say  that  the  forging  receives  a  blow  of  727  ft.-lbs. 
energy.  The  pressure  actually  felt  at  the  surface  of  the  ham- 
mer varies  from  instant  to  instant  during  the  compression  of 
the  forging  and  the  gradual  stopping  of  the  hammer,  and 
depends  on  the  readiness  with  which  the  hot  metal  yields. 

If  the  mean  resistance  encountered  is  Rm,  and  the  depth  of 
compression  s",  we  would  have  (neglecting  the  force  of  gravity, 
and  noting  that  now  the  initial  velocity  is  v',  and  the  final 
zero),  from  eq.  (<?), 

i  W  =  Rms" ;  i.e.,  Rm  =  [727  -;-  s"  (ft.)]  Ibs. 

E.g.,  if  «"  =  |  of  an  inch  =  ^  of  a  foot,  Rm  =  43620  Ibs., 
and  the  maximum  value  of  R  would  probably  be  about  double 
this  near  the  end  of  the  impact.  If  the  anvil  also  sinks  during 
the  impact  a  distance  s'",  we  must  substitute  s'"  +  s"  instead 
of  s" ;  this  will  give  a  smaller  value  for  Rm. 


140 


MECHANICS   OF   ENGINEERING. 


By  mean  value  for  R  is  meant  [eq.  (c)]  that  value,  Rm,  which 
satisfies  the  relation 


Rmsf  =  /    Rds. 
«A 

This  may  be  called  more  explicitly  a  space-average,  to  dis- 
tinguish it  from  a  time- average,  which  might  appear  in  some 
problems,  viz,,  a  value  Rtm,  to  satisfy  the  relation  (if  being  the 
duration  of  the  impact) 

RtJ  =/* '  Bdt, 

and  is  different  from  Rm. 

From  %Mv'*  =  727  ft.-lbs.,  we  have  v'  =  12.06  ft.  per  sec., 
whereas  for  a  free  fall  it  would  have  been 


.2x1  =  8.03. 

(This  example  is  virtually  of  the  same  kind  as  Prob.  4,  §  59, 
differing  chiefly  in  phraseology.) 

136.  Pile-Driving.*— The  safe  load  to  be  placed  upon  a  pile 
after  the  driving  is  finished  is  generally  taken  as  a  fraction  (from 
J  to  -J-)  of  the  resistance  of  the  earth  to  the  passage  of  the  pile  as 
indicated  by  the  effect  of  the  last  few  blows  of  the  ram,  in  ac- 
cordance with  the  following  approximate  theory  :  Toward  the 
end  of  the  driving  the  resistance  R  encountered  by 
the  pile  is  nearly  constant,  and  is  assumed  to  be  that 
met  by  the  ram  at  the  head  of  the  pile;  the  distance 
s'  through  which  the  head  of  the  pile  sinks  as  an 
effect  of  the  last  blow  is  observed.  If  G,  then,  is 
the  weight  of  the  ram,  =  Mg,  and  h  the  height  of 
free  fall,  the  velocity  due  to  A,  on  striking  the  pile, 
is  c  =  V%gh  (§  52),  and  we  have,  from  eq.  (c), 


?a,  i.e.,  Gh,  = 


=  Rs' 


(1) 


(R  being  considered  constant)  ;  hence  R  =  Gh  -=- 
and  the  safe  load  (for  ordinary  wooden  piles), 


P  =  from      to  i  of  Gh  •+•  s' 


(2) 


Maj.  Sanders  recommends  •$•  from  experiments  made  at  Fort 

*  See  also  p.  87  of  the  author's  Notes  and  Examples  in  Mechanics. 


WORK,    ENEKGY,    AND    POWER.  141 

Delaware  in  1851;  Molesworth,  £;  General  Barnard,  -J-,  from 
extensive  experiments  made  in  Holland. 

Of  course  from  eq.  (2),  given  JP9  we  can  compute  s'. 

(Owing  to  the  uncertainty  as  to  how  much  of  the  resistance 
R  is  due  to  friction  of  the  soil  on  the  sides  of  the  pile,  and 
how  much  to  the  inertia  of  the  soil  around  the  shoe,  the  more 
elaborate  theories  of  Weisbach  and  Rankine  seem  of  little 
practical  account.) 

137.  Example.  —  In  preparing  the  foundation  of  a  bridge-pier 
it  is  found  that  each  pile  (placing  them  4  ft.  apart)  must  bear 
safely  a  load  of  72  tons.     If  the  ram  weighs  one  ton,  and  falls 
12  ft.,  what  should  be  the  effect  of  the  last  blow  on  each  pile? 
Using  the  foot-ton-second  system  of  units,  and  Molesworth 
factor  -J-,  eq.  (2)  gives 

s'  =  -J(l  X  12  -r-  72)  =  Jy  of  a  foot  =  J  of  an  inch. 

That  is,  the  pile  should  be  driven  until  it  sinks  only  J  inch 
Binder  each  of  the  last  few  blows. 

138.  Kinetic  Energy  Lost  in  Inelastic  Direct  Central  Impact.  — 

Referring  to  §  60,  and  using  the  same  notation  as  there  given, 
we  find  that  if  the  united  kinetic  energy  possessed  by  two  in- 
elastic bodies  after  their  impact,  viz.,  %M^C*  +  •JJ/261'a,  Shav- 
ing the  value  (Mlcl  +  J/X)  -=-  (-3/J  +  -^4),  be  deducted  from 
the  amount  before  impact,  viz.,  ^M^c*  +  \M$*,  the  loss  of 
kinetic  energy  during  impact  of  two  inelastic  bodies  is  * 


An  equal  amount  of  energy  is  also  lost  by  partially  elastic 
bodies  during  the  first  period  of  the  impact,  but  is  partly  re- 
gained in  the  second.  If  the  bodies  were  perfectly  elastic,  we 
would  find  it  wholly  regained  and  the  resultant  loss  zero,  from 
the  equations  of  §  60  ;  but  this  is  not  quite  the  reality,  on 
account  of  internal  vibrations. 

The  Tcvnetic  energy  still  remaining  in  two  inelastic  bodies 
after  impact  (they  move  together  as  one  mass)  is 
*  See  Eng,  News,  July,  1888,  pp.  33  and  34. 


142  MECHANICS   OF   ENGINEERING. 

£( Ml  +  MI)  C\  or,  after  inserting  the  value  of 
O  =  (M.c,  +  Jfaca)  -T-  (Ml  +  JQ,  we  have 


-2  • 


M 


Example  1. — The  weight  Gl  =  M,g  falls  freely 
through  a  height  A,  impinging  upon  a  weight  Gt 
=  M.$,  which  was  initially  at  rest.     After  their  (in- 
elastic) impact  they  move  on  together  with  the  com- 
bined kinetic  energy  just  given  in  (2),  which,  since 
c^  and  <?3,  the  velocities  before  impact,  are  respectively 
V%gh  and  0,  may  be  reduced  to  a  simpler  form. 
This  energy  is  soon  absorbed  in  overcoming  the 
flange-pressure  R,  which  is  proportional  (so  long  as 
i^j8        the  elasticity  of  the  rod  is  not  impaired)  to   the 
elongation  s,  as  with  an  ordinary  spring.     If  from 
FIG.  150.      previous  experiment  it  is  known   that  a  force  7?0 
produces  an  elongation  s0,  then  the  variable  R  =  (J?0  -^-  s0)s. 
Neglecting  the  weight  of  the  two  bodies  as  a  working  force, 
we  now  have,  from  eq.  (d), 

*°'  •*+<>- aS? 


*. 

When  s  =  /,  i.e.,  when  the  masses  are  (momentarily)  at  rest 
in  the  lowest  position,  the  flange-pressure  or  tensile  stress  in  the 
rod  is  a  maximum,  E'  =  (E^  ~  s0)s'  ',  whence  s'  =  It's,  -~  R  \  ; 
and  (3)  may  be  written 


2 


R>\_      M?gh  ' 

~        ~  ' 


Eq.  (3)  gives  the  final  elongation  of  the  rod,  and  (5)  the  greatest 
tensile  force  upon  it,  provided  the  elasticity  of  the  rod  is  not 


WORK,   ENERGY,   AND   POWER.  143 

impaired.    The  form  £7?Y  in  (4)  may  be  looked  upon  as  a  direct 
integration  of    /    Rds,  viz.,  the  mean  resistance  (i-/?')  multi- 

plied by  the  whole  distance  (s1)  gives  the  work  done  in  over- 
coming the  variable  R  through  the  successive  <&'s. 

If  the  elongation  is  considerable,  the  working-forces  Gl  and 
GI  cannot  be  neglected,  and  would  appear  in  the  term  -f-  (Gl 
+  G^s'  in  the  right-hand  members  of  (3s),  (4),  and  (5).  The 
upper  end  of  the  rod  is  firmly  fixed,  and  the  rod  itself  is  of 
small  mass  compared  with  Ml  and  MY 

Example  2.  —  Two  cars,  Fig.  151,  are  connected  by  an  elastic 
chain  on  a  horizontal  track.  Velocities  before  impact  (i.e., 
before  the  stretching  of  the  chain  be- 
gins,  by  means  of  which  they  are 
brought  to  a  common  velocity  at  the  " 
instant  of  greatest  tension  12',  and  FIG.  151. 

elongation  s'  of  the  chain)  are  cl  =  <?t,  and  £a  =  0. 

During  the  stretching,  i.e.,  the  first  period  of  the  impact,  the 
kinetic  energy  lost  by  the  masses  has  been  expended  in  stretch- 
ing the  chain,  i.e.,  in  doing  the  work  j72Y  ;  hence  we  may 
write  (the  elasticity  of  the  chain  not  being  impaired)  (see  eq.  (1)  ) 

•  • 


in  which  the  different  symbols  have  the  same  meaning  as  in 
Example  1,  in  which  the  rod  corresponds  to  the  chain  of  this 
example. 

In  this  case  the  mutual  accommodation  of  velocities  is  due 
to  the  presence  of  the  chain,  whose  stretching  corresponds  to 
the  compression  (of  the  parts  in  contact)  in  an  ordinary  impact. 

In  numerical  substitution.  32.2  for  g  requires  the  use  of  the 
units  foot  and  second  for  space  and  time,  while  the  unit  of 
force  may  be  anything  convenient. 

139.   Work  and  Energy  in   Rotary  Motion.     Axis  Fixed.— 

The  rigid  body  being  considered  free,  let  an  axis  through  0 
perpendicular  to  the  paper  be  the  axis  of  rotation,  and  resolve 
all  forces  not  intersecting  the  axis  into  components  parallel 


144 


MECHANICS    OF   ENGINEERING. 


and  perpendicular  to  the  axis,  and  the  latter  again  into  com 
ponents  tangent  and  normal  to  the  circular  path  of  the  point 

of  application.  These  tangential  com. 
ponents  are  evidently  the  only  ones 
of  the  three  sets  mentioned  which 
have  moments  about  the  axis,  those 
having  moments  of  the  same  sign  as 
GO  (the  angular  velocity  at  any  instant) 
being  called  working  forces,  T^,  T^ 
etc. ;  those  of  opposite  sign,  resist- 
ances, T^,  T^,  etc.;  for  when  in  time 
dt  the  point  of  application  B^,  of  T^,  describes  the  small  arc 
dsl  =  a^da,  whose  projection  on  T^  is  —  ds19  this  projection 
falls  ahead  (i.e.,  in  direction  of  force)  of  the  position  of  the 
point  at  the  beginning  of  dt,  while  the  reverse  is  true  for  Tf+ 
From  eq.  (XIY.),  §  114,  we  have  for  8  (angul.  accel.) 


(9  = 


FIG.  152. 


which  substituted  in  coda)  =  Oda  (from  §  110)  gives  (remem- 
bering that  atda  =  d8v  etc.),  after  integration  and  transposition, 


f 


etc. 


etc. 


,  (2) 


where  0  and  n  refer  to  any  two  (initial  and  final)  positions  of 
the  rotating  body.     Eq.  (4),  §  120,  is  an  example  of  this. 

Now  \&nl  —  -kGOnfdMp*  =  f^dM(c>}npf  ,  which,  since  conp 
is  the  actual  velocity  of  any  dM  \it  this  (final)  instant,  is  nothing 
more  than  the  sum  of  the  amounts  of  kinetic  energy  possessed 
at  this  instant  by  all  the  particles  of  the  body;  a  similar  state- 
ment may  be  made  for  -J-^2/. 

Eq.  (2)  therefore  may  be  put  into  words  as  follows: 

Between  any  two  positions  of  a  rigid  body  rotating  about  a 

fixed  axis,  the  work  done  by  the  working  forces  is  partly  used 

in  overcoming  the  resistances,  and  the  remainder  in  changing 

the  kinetic  energy  of  the  individual  particles.     If  in  any  case 


WORK,   ENERGY,   AND   POWER.  145 

tins  remainder  is  negative,  the  final  kinetic  energy  is  less  than 
the  initial,  i.e.,  the  work  done  by  the  working  forces  is  less  than 
that  necessary  to  overcome  the  resistances  through  their  respec- 
tive spaces,  and  the  deficiency  is  made  up  by  the  restoring  of 
some  of  the  initial  kinetic  energy  of  the  rotating  body.  A 
moving  fly-wheel,  then,  is  a  reservoir  of  kinetic  energy. 

Eq.  (2)  has  already  been  illustrated  numerically  in  §  121, 
where  the  additional  relation  was  utilized  (for  a  connecting-rod 
and  piston  of  small  mass),  that  the  work  done  in  the  steam- 
cylinder  is  the  same  as  fliat  done  directly  at  the  crank-pin  by 
the  working-force  there. 

140.  Work  of  Equivalent  Systems  the  Same. — If  two  plane 
systems  of  forces  acting  on  a  rigid  body  are  equivalent  (§  1 5#), 
the  aggregate  work  done  ~by  either  of  them  during  a  given  slight 
displacement  or  motion  of  the  body  parallel  to  their  plane  is 
the  same.  By  aggregate  work  is  meant  what  has  already  been 
defined  as  the  sum  of  the  "  virtual  moments"  (§§  61  to  64),  in 
any  small  displacement  of  the  body,  viz.,  the  algebraic  sum  of 
the  products,  2  (Pdu\  obtained  by  multiplying  each  force  by 
the  projection  (du)  of  the  displacement  of  (or  small  space 
described  by)  its  point  of  application  upon  the  force.  (We 
here  class  resistances  as  negative  working  forces.) 

Call  the  systems  A  and  B ;  then,  if  all  the  forces  of  B  were 
reversed  in  direction  and  applied  to  the  body  along  with  those 
of  A9  the  com  pound  system  would  be  a  balanced  system,  and 
hence  we  should  have  (§  64),  for  a  small  motion  parallel  to  the 
plane  of  the  forces, 

2(Pdu)  =  0,  i.e.,  2(Pdu)  for  A  -  2(Pdu)  for  B  =  0, 
or  +  2(Pdu)  for  A  =  +  2(Pdu)  for  B. 

But  -|-  2  (Pdu)  for  A  is  the  aggregate  work  done  by  the  forces 
of  A  during  the  given  motion,  and  -f-  2(Pdu)  for  B  is  a 
similar  quantity  for  the  forces  of  B  (not  reversed)  during  the 
same  small  motion  if  B  acted  alone.  Hence  the  theorem  is 
proved,  and  could  easily  be  extended  to  space  of  three  dimen- 
sions. 

10 


146 


MECHANICS   OF   ENGINEERING. 


FlG- 


141.  Relation  of  Work  and  Kinetic  Energy  for  any  Extended 
Motion  of  a  Rigid  Body  Parallel  to  a  Plane. — (If  at  any  instant 

any  of  the  forces  acting  are  not 
parallel  to  the  plane  mentioned, 
their  components  lying  in  or 
parallel  to  that  plane,  will  be  used 
instead,  since  the  other  compo- 
nents obviously  would  be  neither 
working  forces  nor  resistances.) 
Fig.  153  shows  an  initial  position, 
o,  of  the  body ;  a  final,  n ;  and  any  intermediate,  as  q.  The 
forces  of  the  system  acting  may  vary  in  any  manner  during 
the  motion. 

In  this  motion  each  dM  describes  a  curve  of  its  own  with 
varying  velocity  v,  tangential  acceleration  pt,  and  radius  of 
curvature  r ;  hence  in  any  position  §-,  an  imaginary  system  B 
(see  Fig.  154),  equivalent  to  the  actual  system  A  (at  q  in  Fig. 
153),  would  be  formed  by  applying  to  each  dM  a 
tangential  force  dT  =  dMpt,  and  a  normal  force 
dN  •=.  dMv*  -r-  r.  By  an  infinite  number  of  con- 
secutive small  displacements,  the  body  passes  from 
o  to  n.  In  the  small  displacement  of  which  q  is  the 
initial  position,  each  dM  describes  a  space  ds,  and  FlGK  154. 
dT  does  the  work  dTds  —  dMvdv,  while  dN  does  the  work- 
dN  X  0  =  0.  Hence  the  total  work  done  by  B  in  the  small 
displacement  at  q  would  be 


'  +  dM"v"dv"  +  etc., 


(1) 


including  all  the  dM  's  of  the  body  and  their  respective  veloci- 
ties at  this  instant. 

But  the  work  at  q  in  Fig.  153  by  the  actual  forces  (i.e.,  of 
system  A)  during  the  same  small  displacement  must  (by  §  140) 
be  equal  to  that  done  by  B.  hence 


,  +  P,du,  +  etc.  =  dM'v'dv'  +  dM"v"dv"  +  etc.  (q) 
Now  conceive  an  equation  like  (q)  written  out  for  each  of 


WORK,    ENERGY,   AND   POWER.  147 

the  small  consecutive  displacements  between  positions  o  and 
n  and  corresponding  terms  to  be  added  ;  this  will  give 

etc. 


v'dv'  +  dM"fnv"dv"  +  etc. 


n"a  -  <")  +  etc. 
The  second  member  may  be  rewritten  so  as  to  give,  finally, 

"  P.du.+f"  P,dut+etc.=2$dMvn')-2$dMv:),  (XY.) 

or,  in  words,  the  work  done  by  the  acting  forces  (treating  a  re- 
sistance as  a  negative  working  force)  between  any  two  posi- 
tions is  equal  to  the  gain  (or  loss)  in  the  aggregate  kinetic 
energy  of  the  particles  of  the  body  between  the  two  positions. 
To  avoid  confusion,  2  has  been  used  instead  of  the  signy  in 
one  member  of  (XV.),  in  which  vn  is  the  final  velocity  of  any 
dM  (not  the  same  for  all  necessarily)  and  v0  the  initial. 

(The  same  method  of  proof  can  be  extended  to  three  dimen- 
sions.) 

Since  kinetic  energy  is  always  essentially  positive,  if  an  ex- 
pression for  it  comes  out  negative  as  the  solution  of  a  problem, 
some  impossible  conditions  have  been  imposed. 

142.    Work  and   Kinetic   Energy  in   a  Moving  Machine.  — 

Defining  a  mechanism  or  machine  as  a  series  of  rigid  bodies 
jointed  or  connected  together,  so  that  working-forces  applied 
to  one  or  more  may  be  the  means  of  overcoming  resistances 
occurring  anywhere  in  the  system,  and  also  of  changing  the 
amount  of  kinetic  energy  of  the  moving  masses,  let  us  for 
simplicity  consider  a  machine  the  motions  of  whose  parts  are 
all  parallel  to  a  plane,  and  let  all  the  forces  acting  on  any  one 
piece,  considered  free,  at  any  instant  be  parallel  to  the  same 
plane. 

Now  consider  each  piece  of  the  machine,  or  of  any  series  of 
its  pieces,  as  a  free  body,  and  write  out  eq.  (XV.)  for  it  be- 
tween any  two  positions  (whatever  initial  and  final  positions  are 


348 


MECHANICS    OF   ENGINEERING. 


selected  for  the  first  piece,  those  of  the  others  must  be  corre- 
sponding initial  and  corresponding  final  positions),  and  it  will 
be  found,  on  adding  up  corresponding  members  of  these  equa- 
tions, that  the  terms  involving  those  components  of  the  mutual 
pressures  (between  the  pieces  considered)  which  are  normal 
to  the  rubbing  surfaces  at  any  instant  will  cancel  out,  while 
their  components  tangential  to  the  rubbing  surfaces  (i.e.,/Ho 
tion,  since  if  the  surfaces  are  perfectly  smooth  there  can  be 
no  tangential  action)  will  appear  in  the  algebraic  addition  as 
resistances  multiplied  by  the  distances  rubbed  through,  meas- 
ured on  the  rubbing  surfaces.  For  example,  Fig.  155,  where 
one  rotating  piece  both  presses  an$  rubs  on  another.  Let  the 
normal  pressure  between  them  at  A  be  7?2  =  Pz ;  it  is  a  work- 
ing force  for  the  body  of  mass  Mf> ',  but  a  resistance  for  M' , 
hence  the  separate  symbols  for  the  numerically  equal  forces 
(action  and  reaction). 

Similarly,  the  friction  at  A  is  2?3  =  P9 ;  a  resistance  for  M', 
a  working-force  for  M".  (In  some  cases,  of  course,  friction 
may  be  a  resistance  for  both  bodies.)  For  a  small  motion,  A 
describes  the  small  arc  A  A'  about  0'  in  dealing  with  M'<  but 
for  M"  it  describes  the  arc  AA"  about  0" ,  A' A'  being 
parallel  to  the  surface  of  contact  AD,  while  AB  is  perpen- 


Fia.  156. 


FIG.  157. 


FIG.  155. 

dicnlar  to  A'  A".  In  Figs.  156  and  157  we  see  M'  and  M" 
free,  and  their  corresponding  small  rotations  indicated.  During 
these  motions  the  kinetic  energy  (K.  E.)  of  each  mass  has 
changed  by  amounts  6?(K.  E.)jf/  and  d(K.  E.)JH»/  respectively,  and 
hence  eq.  (XV.)  gives,  for  each  free  body  in  turn, 

P~aaf  - 


=  d(K.  E.)*' 
=  d(K.  E.  V 


(1) 
(2) 


WO  UK,    ENERGY,    AND    POWER.  149 

Now  add  (1)  and  (2),  member  to  member,  remembering  that 
jPa  =  j?3  and  P3  =  7?3  =  Ft  =  friction,  and  we  have 


P,aar-  F9A7A/  -  BT7  =  d(K.  E.)M,  +  d(K.  E.)ji",   (3) 

in  which  the  mutual  actions  of  M1  and  M"  do  not  appear, 
except  the  friction,  the  work  done  in  overcoming  which,  when 
the  two  bodies  are  thus  considered  collectively,  is  the  product 
of  the  friction  ~by  the  distance  A'  A"  of  actual  rubbing  meas- 
ured on  the  rubbing  surface.  For  any  number  of  pieces,  then, 
considered  free  collectively,  the  assertion  made  at  the  beginning 
of  this  article  is  true,  since  any  finite  motion  consists  of  an 
infinite  number  of  small  motions  to  each  one  of  which  an  equa- 
tion like  (3)  is  applicable. 

Summing  the  corresponding  terms  of  all  such  equations,  we 
have 


This  is  of  the  same  form  as  (XV.),  but  instead  of  applying  to  a 
single  rigid  body,  deals  with  any  assemblage  of  rigid  parts 
forming  a  machine,  or  any  part  of  a  machine  (a  similar  proof 
will  apply  to  three  dimensions  of  space);  but  it  must  be  remem- 
bered that  it  excludes  all  the  mutual  actions*  of  the  pieces  con- 
sidered except  friction,  which  is  to  be  introduced  in  the  manner 
just  illustrated.  A  flexible  inextensible  cord  may  be  considered 
as  made  up  of  a  great  number  of  short  rigid  bodies  jointed 
without  friction,  and  hence  may  form  part  of  a  machine  with- 
out vitiating  the  truth  of  (XVI.). 

2(K.  E.)n  signifies  the  sum  obtained  by  adding  the  amounts 
of  kinetic  energy  (^dMv^  for  each  elementary  mass)  possessed 
by  all  the  particles  of  all  the  rigid  bodies  at  their  final  posi- 
tions ;  2(K.  E.)0,  a  similar  sum  at  their  initial  positions.  For 
example,  the  K.  E.  of  a  rigid  body  having  a  motion  of  transla- 
tion of  velocity  v,  =  ^v]fdM  =  %Mv*  ;  that  of  a  rigid  body 
having  an  angular  velocity  OD  about  a  fixed  axis  Z,  =  ^<x?Iz 
(§  139)  ;  while,  if  it,  has  an  ansrnlar  velocity  &  about  a  gravity- 

*  These  mutual  actions  consist  only  of  actions  by  contact  (pressure,  rub, 
etc.).   No  magnetic  or  electrical  attractions  or  repulsions  are  here  considered. 


150  MECHANICS   OF   ENGINEEKING. 

axis  Z,  which  has  a  velocity  vz  of  translation  at  right  angles  to 
itself,  the  (K.  E.)  at  this  instant  may  be  proved  to  be 


i.e.,  is  the  sum  of  the  amounts  due  to  the  two  motions  sepa- 
rately. 

143.  K.  E.  of  Combined  Rotation  and  Translation.  —  The  last 

statement  may  be  thus  proved.  Fig.  158. 
At  a  given  instant  the  velocity  of  any  cO^is 
v,  the  diagonal  formed  on  the  velocity  vz  of 
translation,  and  the  rotary  velocity  cop  rela- 
tively to  the  moving  gravity-axis  Z  (per- 
pendicular  to  paper)  (see  §  71), 

i.e.,  v*  =  vz*  +  (<*P)*  —  Z(™p)vz  cos  cp  ; 
hence  we  have  K.  E.,  at  this  instant, 
=  f%dMv*  =  &zfdM  +  \tffdMtf  -  covzfdMp  cos  <?, 

but  p  cos  <p  =  y,  and  fdMy  =  My  =  0,  since  Z  is  a  gravity- 
axis, 

/.  K.  E.  =  $Mvz*  +  ^Iz.     Q.  E.  D. 

It  is  interesting  to  notice  that  the  K.  E.  due  to  rotation,  viz., 
^GO^IZ  =  -|Jf(rf)3,  is  the  same  as  if  the  whole  mass  were  con- 
centrated in  a  point,  line,  or  thin  shell,  at  a  distance  &,  the 
radius  of  gyration,  from  the  axis. 

144.  Example  of  a  Machine  in  Operation.  —  Fig.  159.     Con- 
sider the  four  consecutive  moving  masses,  M',  M"  ,  Mff>  ',  and 
Mlv  (being  the  piston  ;  connecting-rod  ;  fly-wheel,  crank,  drum, 
and  chain  ;  and  weight  on  inclined  plane)  as  free,  collectively. 
Let  us  apply  eq.  (XVL),  the  initial  and  final  positions  being 
taken  when  the  crank-pin  is  at  its  dead-points  o  and  n  ;  i.e.,  we 
deal  with  the  progress  of  the  pieces  made  while  the  crank-pin 
describes  its  upper  semicircle.      Remembering  that  the  mutual 
actions  between  any  two  of  these  four  masses  can  be  left  out 
of  account  (except  friction),  the  only  forces  to  be  put  in  are 
the  actions  of  other  bodies  on  each  one  of  these  four,  and  are 


WORK,    ENERGY,    AND    POWER. 


151 


shown  in  the  figure.  The  only  mutual  friction  considered  will 
be  at  the  crank-pin,  and  if  this  as  an  average  —  F" ,  the  work 
done  on  it  between  o  and  n  =  F"rtr" ,  where  r"  =  radius  of 
crank-pin.  The  work  done  by  Pl  the  effective  steam- pressure 
(let  it  be  constant)  during  this  period  is  =  PJ/ ;  that  done  in 
overcoming^,  the  friction  between  piston  and  cylinder,  =  FJf ; 
that  done  upon  the  weight  6r''of  connecting-rod  is  cancelled  by 
the  work  done  by  it  in  the  descent  following ;  the  work  done 


FIG.  159. 

upon  6riv,  =  G^na  sin  /?,  where  a  =  radius  of  drum ; 
upon  the  friction  F^  =  Fjta.  The  pressures  N,  Nr,  N'1*,  and 
N'",  and  weights  G  and  #'",  are  neutral,  i.e.,  do  no  work  either 
positive  or  negative.  Hence  the  left-hand  member  of  (XYI.) 
becomes,  between  o  and  n, 

PJ -FJ,' -F"7tr"-G™7tasmp -Fjta,  .     .    (1) 

provided  the  respective  distances  are  actually  described  by 
these  forces,  i.e.,  if  the  masses  have  sufficient  initial  kinetic 
energy  to  carry  the  crank-pin  beyond  the  point  of  minimum 
velocity,  with  the  aid  of  the  working  force  jP1?  whose  effect  is 
small  up  to  that  instant. 

As  for  the  total  initial  kinetic  energy,  i.e.,  ^(K.  E.)0,  let  us 
express  it  in  terms  of  the  velocity  of  crank-pin  at  o,  viz.,  V9. 
The  (K.E.).  of  M'  is  nothing  ;  that  of  M" ,  which  at  this  in- 
stant is  rotating  about  its  right  extremity  (fixed  for  the  instant) 
with  angular  velocity  GO"  =  V0  -4- 1",  is  &«"*£";  that  of  W" 
—  -i&?"/a/c"',  in  which  a/"  =  V0  +  r ;  that  of  Jf iv  (translation) 
"iv<y0iva>  in  which  v0iv  =  (a  -5-  r)  V0.  SQL.  E.)n  is  expressed 


152 


MECHANICS    OF   ENGINEERING. 


in  a  corresponding  manner  with  Vn  (final  velocity  of  crank-pin) 
instead  of  VQ.  Hence  the  right-hand  member  of  (XYI.)  will 
give  (putting  the  radius  of  gyration  of  M"  about  0"  =  &", 
and  that  of  M"  about  C  =  k) 

.    .    (2) 

By  writing  (1)=(2),  we  have  an  equation  of  condition,  capa- 
ble of  solution  for  any  one  unknown  quantity,  to  be  satisfied 
for  the  extent  of  motion  considered.  It  is  understood  that  the 
chain  is  always  taut,  and  that  its  weight  and  mass  are  neg- 
lected. 

• 

145.  Numerical  Case  of  the  Foregoing. — (Foot-pound-second 
system  of  units  for  space,  force,  and  time;  this  requires  g 
=  32.2.) 

Suppose  the  following  data  : 


FEET. 

LBS. 

LBS. 

MASS  UNITS. 

V  =  2.0 
I"  =  4.0 
a  =  1.5 
r  =  1.0 

z-  ••  ..  1  s 

Pi   =  6000 

Fi  =    200 
PI  (av'ge)  =    400 
Ft  =   300 

0'  =      60 
G"  =      50 
G'"  =    400 
G"  =  3220 

(and  .-.) 
M  '  =      1.86 
M"  =      1.55 
M'"  =    12.4 
M"  =  100.0 

k"  =  2.3 
r"  =  0.1 

Also  let  Fo  =  4ft.  per  sec.  ;  /?=30° 

Denote  (1)  by  W  and  the  large  bracket  in  (2)  by  M  (this  by 
some  is  called  the  total  mass  "  reduced"  to  the  crank-pin). 
Putting  (1)  =  (2)  we  have,  solving  for  the  unknown  Vw 


(3) 


For  above  values, 

W  =  12,000  -  400  -  125.7  -  7590.0  -  1417.3 

=  2467  foot-pounds ; 

while  M  =  0.5  +  40.3  +  225.0  =  265.8  mass-units ; 
whence 


Vn  =  1/13.56  +  16  =  1/34.56  =  5.88  ft.  per  second. 


WORK,    ENERGY,    AND    POWER.  153 

As  to  whether  the  crank-pin  actually  reaches  the  dead-point 
;i,  requires  separate  investigations  to  see  whether  V  becomes 
zero  or  negative  between  o  and  n  (a  negative  value  is  inad- 
missible, since  a  reversal  of  direction  implies  a  different 
value  for  TF"),  i.e.,  whether  the  proposed  extent  of  motion  is 
realized ;  and  these  are  made  by  assigning  some  other  inter- 
mediate position  m,  as  a  final  one,  and  computing  Vm,  remem- 
bering that  when  m  is  not  a  dead-point  the  (K.  E.)w  of  M'  is  not 
zero,  and  must  be  expressed  in  terms  of  Vm,  and  that  the 
(K.  E.)OT  of  the  connecting-rod  M' 'must  be  obtained  from  §  143. 

146.  Regulation   of  Machines. — As    already   illustrated   in 
several  examples  (§  121),  a  fly-wheel  of  sufficient  weight  and 
radius  may  prevent  too  great  fluctuation  of  speed  in  a  single 
stroke  of  an  engine  ;  but  to  prevent  a  permanent  change,  which 
must  occur  if  the  work  of  the  working  force  or  forces  (such  as 
the  steam-pressure  on  a  piston,  or  water-impulse  in  a  turbine) 
exceeds  for  several  successive  strokes  or  revolutions  the  work 
required  to  overcome  resistances  (such  as  friction,  gravity,  re- 
sistance at  the  teeth  of  saws,  etc.,  etc.)  through  their  respective 
spaces,  automatic  governors  are   employed    to   diminish   the 
working  force,  or  the  distance  through  which  it  acts  per  stroke, 
until  the  normal  speed  is  restored  ;  or  vice  versa,  if  the  speed 
slackens,  as  when  new  resistances  are  temporarily  brought  into 
play.     Hence  when  several  successive  periods,  strokes  (or  other 
cycle),  are  considered,  the  kinetic  energy  of  the  moving  parts 
will  disappear  from  eq.  (XVI.),  leaving  it  in  this  form  : 

work  of  working-forces  =  work  done  upon  resistances. 

147.  Power  of  Motors. — In  a  mill  where  the  same  number  of 
machines  are  run  continuously  at  a  constant  speed  proper  for 
their  work,  turning  out  per  hour  the  same  number  of  barrels 
of  flour,  feet  of  lumber,  or  other  commodity,  the  motor  (e.g., 
a  steam-engine,  or  turbine)  works  at  a  constant  rate,  i.e.,  de- 
velops a  definite  horse-power  (H.P.),  which  is  thus  found  in 
the  case  of  steam-engines  (double-acting) : 


154  MECHANICS    OF    ENGINEERING. 

H.P.  =  total  mean  effective  )        (  distance  in  feet  ) 

steam-pressure  on  v  X  -j  travelled  by  pis-  V  -r-  550, 
piston  in  Ibs.          )        (  ton  per  second.   ) 

i.e.,  the  work  (in  ft.-lbs)  done  per  second  by  the  working  force 
divided  by  550  (see  §  132).  The  total  effective  pressure  at  any 
instant  is  the  excess  of  the  forward  over  the  back-pressure, 
and  by  its  mean  value  (since  steam  is  usually  used  expansively) 
is  meant  such  a  value  P'  as,  multiplied  by  the  length  of  stroke 
I,  shall  give 

P'l  = 

where  P  is  the  variable  effective  pressure  and  dx  an  element 
of  its  path.  If  u  is  the  number  of  strokes  per  second,  we  may 
also  write  (foot-pound-  second  system) 


H.P.  =  P'lu  -f-  550  =  ^pdxu  -T-  550.  (XVII.) 

Yery  often  the  number  of  revolutions  jp^r  minute,  m,  of  the 
crank  is  given,  and  then 

H.P.  =  P!  (Ibs.)  X  2Z  (feet)  X  m  -5-  33,000. 

If  F  =  area  of  piston  we  may  also  write  Pr  =  Fp',  where pf 
is  the  mean  effective  steam-pressure  per  unit  of  area.  Evi- 
dently, to  obtain  P'  in  Ibs.,  we  multiply  Fin  sq.  in.  byj/  in 
Ibs.  per  sq.  in.,  or  F  in  sq.  ft.  by  p'  in  Ibs.  per  sq.  foot ;  the 
former  is  customary,  p'  in  practice  is  obtained  by  measure- 
ments and  computations  from  "indicator-cards"  (see  §135,  in 
which  (jPt  —  12 ^  corresponds  to  P  of  this  section)  ;  or  P'l,  i.e., 

/Pdx,  may  be  computed  theoretically  as  in  §  59,  Problem  4. 
-  „ 

The  power  as  thus  found  is  expended  in  overcoming  the 
friction  of  all  moving  parts  (which  is  sometimes  a  large  item), 
and  the  resistances  peculiar  to  the  kind  of  work  done  by  the  ma- 
chines. The  work  periodically  stored  in  'the  increased  kinetic 
energy  of  the  moving  masses  is  restored  as  they  periodically 
resume  their  minimum  velocities. 


WORK,   ENERGY,   AND   POWER. 

148.  Potential  Energy. — There  are  other  ways  in  which  work 
or  energy  is  stored  and  then  restored,  as  follows : 

First.  In  raising  a  weight  G  through  a  height  A,  an  amount 
of  work  =  Oh  is  done  upon  G,  as  a  resistance,  and  if  at  any 
subsequent  time  the  weight  is  allowed  to  descend  through  the 
same  vertical  distance  A  (the  form  of  path  is  of  no  account),  G, 
now  a  working  force,  does  the  work  Gh,  and  thus  in  aiding  the 
motor  repays,  or  restores,  the  Gh  expended  by  the  motor  in 
raising  it.  If  A  is  the  vertical  height  through  which  the  centre 
of  gravity  rises  'and  sinks  periodically  in  the  motion  of  the 
machine,  the  force  G  may  be  left  out  of  account  in  reckoning 
the  expenditure  of  the  motor's  work,  and  the  body  when  at  its 
highest  point  is  said  to  possess  an  amount  Gh  of  potential 
energy,  i.e.,  energy  of  position,  since  it  is  capable  of  doing  the 
work  Gh  in  sinking  through  its  vertical  range  of  motion. 

Second.  So  far,  all  bodies  considered  have  been  by  express 
stipulation  rigid,  i.e.,  incapable  of  changing  shape,  To  see 
the  effect  of  a  lack  of  rigidity  as  affecting  the  principle  of 
work  and  energy  in  machines, 
take  the  simple  case  in  Fig.  160. 


A  helical  spring  at  a  given  in- 
stant is  acted  on  at  each  end  by 
a  force  P  in  an  axial  direction 
(they  are  equal,  supposing  the  FIG.  IGO. 

mass  of  the  spring  small).  As  the  machine  operates  of  which 
it  is  a  member,  it  moves  to  a  new  consecutive  position  B, 
suffering  a  further  elongation  dX  in  its  length  (if  P  is  increas- 
ing). P  on  the  right,  a  working  force,  does  the  work  Pdx'; 
how  is  this  expended  ?  P  on  the  left  has  the  work  Pdx  done 
upon  it,  and  the  mass  is  too  small  to  absorb  kinetic  energy  or 
to  bring  its  weight  into  consideration.  The  remainder,  Pdx' 
—  Pdx  —  PdX,  is  expended  in  stretching  the  spring  an  addi- 
tional amount  dX,  and  is  capable  of  restoration  if  the  spring 
retains  its  elasticity.  Hence  the  work  done  in  changing  the 
form  of  bodies  if  they  are  elastic  is  said  to  be  stored  in  the 
form  of  potential  energy.  That  is,  in  the  operation  of  ma- 
chines, the  name  potential  energy  is  also  given  to  the  energy 


156  MECHANICS    OF    ENGINEERING. 

stored  and  restored  periodically  in  the  changing  and  regaining 
of  form  of  elastic  bodies. 

149.  Other  Forms  of  Energy. — Numerous  experiments  with 
various  kinds  of  apparatus   have  proved   that  for  every  772 
(about)  ft.-lbs.  of  work  spent  in  overcoming  friction,  one  British 
unit  of  heat  is  produced  (viz.,  the  quantity  of  heat  necessary  to 
raise  the  temperature  of  one  pound  of  water  from  32°  to  33° 
Fahrenheit);  while  from  converse  experiments,  in  which  the 
amount  of  heat  used  in  operating  a  steam-engine  was  all  carefully 
estimated,  the  disappearance  of  a  certain  portion  of  it  could  only 
be  accounted  for  by  assuming  that  it  had  been  converted  into 
work  at  the  same  rate  of  (about)  772  ft.-lbs.  of  work  to  each 
unit  of  heat  (or  425  kilogrammetres  to  each  French   unit  of 
heat).     This  number  77.2,  or  425,  according  to  the  system  of 
units  employed,  is  called  the  Mechanical  Equivalent  of  Heat, 
first  discovered  by  Joule  and  confirmed  by  Him.* 

Heat  then  is  energy,  and  is  supposed  to  be  of  the  kinetic 
form  due  to  the  rapid  motion  or  vibration  of  the  molecules  of 
a  substance.  A  similar  agitation  among  the  molecules  of  the 
(hypothetical)  ether  diffused  through  space  is  supposed  to  pro- 
duce the  phenomena  of  light,  electricity,  and  magnetism. 
Chemical  action  being  also  considered  a  method  of  transform- 
ing energy  (its  possible  future  occurrence  as  in  the  case  of  coal 
and  oxygen  being  called  potential  energy),  the  well-known 
doctrine  of  the  Conservation  of  Energy,  in  accordance  with 
which  energy  is  indestructible,  and  the  doing  of  work  is  simply 
the  conversion  of  one  or  more  kinds  of  energy  into  equivalent 
amounts  of  others,  is  now  one  of  the  accepted  hypotheses  of 
physics. 

Work  consumed  in  friction,  though  practically  lost,  still  re- 
mains in  the  universe  as  heat,  electricity,  or  some  other  subtile 
form  of  energy. 

150.  Power  Required  for  Individual  Machines.     Dynamome- 
ters of  Transmission. — If  a  machine  is  driven  by  an  endless 
belt  from  the  main-shaft,  A,  Fig.  161,  being  the  driving-pulley 

*  Prof.  Rowland's  recent  experiments  result  in  the  value  429  8  kilogram- 
metres  at  a  temperature  of  5°  Cent. 


WOHK,    ENEKGY,    AND    POWER. 


157 


FIG.  161. 


on  the  machine,  the  working  force  which  drives  the  machine, 
in  other  words  the  "  grip"  with  which  the 
belt  takes  hold  of  the  pulley  tangential  ly,  p" 
•—P  —  P',  P  and  P'  being  the  tensions 
in  the  "  driving"  and  "  following"  sides  of 
the  belt  respectively.    The  belt  is  supposed      , 
not  to  slip  on  the  pulley.     If  v  is  the  ve- 
locity  of   the    pulley -circumference,    the 
work  expended  on  the  machine  per  second,  i.e.,  the  power,  is 

L  =  (P-Pf)v (1) 

To  measure  the  force  (P  —  Pf\  an  apparatus  called  a  Dy- 
namometer of  Transmission  may  be  placed  between  the  main 
shaft  and  the  machine,  and  the  belt  made  to  pass  through  it  in 
such  a  way  as  to  measure  the  tensions  P  and  P ',  or  princi- 
pally their  difference,  without  meeting  any  resistance  in  so  do- 
ing ;  that  is,  the  power  is  transmitted,  not  absorbed,  by  the 
apparatus.  One  invention  for  this  purpose  (mentioned  in  the 
Journal  of  the  Franklin  Institute  some  years  ago)  is  shown 

(in  principle]  in  Fig.  162.     A  ver- 
tical plate  carrying  four  pulleys  arid 
a  scale-pan  is  first  balanced  on  the 
pivot  C.     The  belt  being  then  ad- 
justed,   as   shown,   and  the   power 
.turned  on,  a  sufficient  weight  G  is 
placed  in  the  scale-pan   to  balance 
FIG.  162.  the  plate  again,  for  whose  equilib- 

rium we  must  have  Go  =  Pa  —  P'a,  since  the  P  and  Pf  on 
the  right  are  purposely  given  no  leverage  about  C.  The  ve- 
locity of  belt,  'y,  is  obtained  by  a  simple  counting  device. 
Hence  (P  —  P')  and  v  become  known,  and  .*.  L  from  (1). 

Many  other  forms  of  transmission-dynamometers  are  in  use, 
some  applicable  whether  the  machine  is  driven  by  belting  or 
gearing  from  the  main  shaft.  Emerson's  Hydrodynamics  de- 
scribes his  own  invention  on  p.  283,  and  gives  results  of  meas- 
urements with  it ;  e.g.,  at  Lowell,  Mass.,  the  power  required 
to  drive  112  looms,  weaving  36-inch  sheetings,  No.  20  yarn, 


158  MECHANICS    OF   ENGINEERING. 

60  threads  to  the  inch,  speed  130  picks  to  the  minute,  was 
found  to  be  16  H.P.,  i.e.,  |  H.P.  to  each  loom  (p.  335). 

151.  Dynamometers  of  Absorption. — These  are  so  named 
since  they  furnish  in  themselves  the  resistance  (friction  or  a 
weight)  in  the  overcoming  (or  raising)  of  which  the  power  is 
expended  or  absorbed.  Of  these  the  Prony  Friction  Brake 
is  the  most  common,  and  is  used  for  measuring  the  power 
developed  by  a  given  motor  (e.g.,  a  steam-engine  or  turbine) 
not  absorbed  in  the  friction  of  the  motor  itself.  Fig.  163 


shows  one  fitted  to  a  vertical  pulley  driven  by  the  motor.  By 
tightening  the  bolt  J?,  the  velocity  v  of  pulley-rim  may  be 
made  constant  at  any  desired  value  (within  certain  limits)  by 
the  consequent  friction,  v  is  measured  by  a  counting  appara- 
tus, while  the  friction  (or  tangential  components  of  action  be- 
tween pulley  and  brake),  =  F,  becomes  known  by  noting  the 
weight  G  which  must  be  placed  in  the  scale-pan  to  balance  the 
arm  between  the  checks  ;  then 

Fa=Gb, (1) 

for  the  equilibrium  of  the  brake  (supposing  the  weight  of 
brake  and  scale-pan  previously  balanced  on  C)  and  the  work 
done  per  unit  of  time,  or  power,  is 

L-Fo (2) 

A  "  dash-pot  "  is  frequently  connected  with  the  arm  to  prevent 
sudden  oscillations.  In  case  the  pulley  is  horizontal,  a  bell- 
crank  lever  is  added  between  the  arm  and  the  scale-pan,  and 
then  eq.  (1)  will  contain  two  additional  lever-arms. 


WORK,    ENERGY,    AND    POWER. 


159 


152.  The  Indicator,  used  with  steam  and  other  fluid  engines, 
is  a  special  kind  of  dynamometer  in  which  the  automatic  mo- 
tion of  a  pencil  describes  a  curve 
on  paper  whose  ordinates  are 
proportional  to  the  fluid  pres- 
sures exerted  in  the  cylinder  at 
successive  points  of  the  stroke. 
Thus, .Fig.  164,  the  back-pres- 
sure being  constant  and  =  P6,  FIG.  164. 
the  ordinates  P0,  P1?  etc.,  represent  the  effective  pressures  at 
equally  spaced  points  of  division.  The  mean  effective  pressure 
P'  (see  §  147)  is,  for  this  figure,  by  Simpson's  /Rule  (six  equal 
spaces), 


ZERO  LINE 


+  4(P,  +  P3  +  P.)  +  2(P,  +  P4)  +  PJ. 

This  gives  a  near  approximation.  The  power  is  now  found  by 
§147. 

153.  The  theory  of  Atwood's  Machine  is  most  directly  ex- 
pressed by  the  principle  of  work  and  energy ;  i.e.,  by  eq. 
(XVI.),  §142.  Fig.  165.  The  parts 
considered  free,  collectively,  are  the 
rigid  bodies  P,  Q,  G,  and  four  friction- 
wheels  like  GV  and  the  flexible  cord, 
which  does  not  slip  on  the  upper  pul- 
ley. There  is  no  slipping  at  Z>,  hence 
no  sliding  friction  there.  The  actions 
of  external  bodies  on  these  eight  consist 
of  the  working  force  P,  the  resistances 
Q  and  the  four  JP's  (at  bearings  of  f ric- 
IBS.  tion-wheel  axles);  all  others  (6r,  46r,, 

and  the  four  J2's)  are  neutral.  Since  there  is  no  rubbing  be- 
tween any  two  of  the  eight  bodies,  no  mutual  actions  whatever 
will  enter  the  equation.  Let  P  >  Q,  and /and  1^  be  the  mo- 
ments of  inertia  of  G  and  #n  respectively,  about  their  respec- 
tive axes  of  figure.  Let  the  apparatus  start  from  rest,  then 
when  P  has  descended  through  any  vertical  distance  s,  and  ac- 


1(50  MECHANICS    OF    ENGINEERING. 

quired  the  velocity  v,  Q  has  been  drawn  up  an  equal  distance 
and  acquired  the  same  velocity,  while  the  pulley  G  has  ac- 
quired an  angular  velocity  GO  =  v  -=-  #,  each  friction-pulley  an 
angular  velocity  ool  =.  (r  :  a)v  -7-  ar  As  to  the  forces,  P  has 
done  the  work  Ps,  Q  has  had  the  work  Qs  done  upon  it,  while 
each  j^has  been  overcome  through  the  space  (rl  :•«,)(?•  :  a)8\ 
all  the  other  forces  are  neutral.  Hence,  from  eq.  (XVL),  §  142 
(see  also  §  139),  we  have 


Evidently  v  =  Vs  X  constant,  i.e.,  the  motion  of  P  and  Q  is 
uniformly  accelerated.  If,  after  the  observed  spaces  has  been 
described,  P  is  suddenly  diminished  to  such  a  value  P'  that 
the  motion  continues  with  a  constant  velocity  =  v,  we  shall 
have,  for  any  further  space  s', 

P's'  -Qsf  -±F-1-  .-sf  =  Q, 
a,    a 

from  which  F  can  be  obtained  (nearly)  ;  while  if  tf  be  the  ob- 
served time  of  describing  «',  v  =  s'  -r-  t'  becomes  known. 
Also  we  may  write  1=  (G  -r-  g)%?  and  1^  =  (Gl  -f-  g)k*9  and 
thus  finally  compute  the  acceleration  of  gravity,  ^,  from  our 
first  equation  above. 

154.  Boat-Rowing.—  x^ig.  166.  During  the  stroke  proper, 
let  P  =  mean  pressure  on  one  oar-handle  ;  hence  the  pressures 
on  the  foot-rest  are  2P,  resistances.  Let  M  =  mass  of  boat 
and  load,  v0  and  vn  its  velocities  at  beginning  and  end  of  stroke. 
Pl  =.  pressures  between  oar-  blade  and  water.  12  =  mean  re- 
sistance of  water  to  the  boat's  passage  at  this  (mean)  speed. 
These  are  the  only  (horizontal)  forces  to  be  considered  as  act- 
ing on  the  boat  and  two  oars,  considered  free  collectively. 
During  the  stroke  the  boat  describes  the  space  s3  =  CD,  the 
oar-handle  the  space  #a  =  AJB,  while  the  oar-blade  slips  back- 


WORK,    ENERGY,    AND    POWER.  161 

ward   through   the  small   space  (the  "  slip")   —  s,  (average). 
Hence  by  eq.  (XVI.),  §  142, 


.e., 

or,  in  words,  the  product  of  the  oar-handle  pressures  into  the 
distance  described  by  them  measured  on  the  boat,  i.e.,  the  work 
done  by  these  pressures  relatively  to  the  boat,  is  entirely  ac- 
counted for  in  the  work  of  slip  and  of  liquid-resistance,  and  in- 


FIG.  166. 

creasing  the  kinetic  energy  of  the  mass.  (The  useless  work 
due  to  slip  is  inevitable  in  all  paddle  or  screw  propulsion,  as 
well  as  a  certain  amount  lost  in  machine-friction,  not  considered 
in  the  present  problem.)  During  the  "  recover"  the  velocity 
decreases  again  -to  v0. 

155.  Examples. — 1.  What  work  is  done*  on  a  level  track,  in 
bringing  up  the  velocity  of  a  train  weighing  200  tons,  from 
zero  to  30  miles  per  hour,  if  the  total  frictional  resistance  (at 
any  velocity,  say)  is  10  Ibs.  per  ton,  and  if  the  change  of  speed 
is  accomplished  in  a  length  of  3000  feet  ? 

(Foot-ton-second  system.)  30  miles  per  hour  =  44  ft.  per 
sec.  The  mass 

=  200  -v-  32.2  =  6.2  ; 
.'.  the  change  in  kinetic  energy, 

(=  -p/^>  _  ij/  x    O8), 
=  i(6.2)  X  442  =  6001.6  ft.-tons. 

*  That  is,  what  work  is  done  by  the  pull,  or  tension,  Pt  in  the  draw-bar 
between  the  locomotive  and  the  "tender." 


162  MECHANICS    OF    ENGINEERING. 

The  work  done  in  overcoming  friction  =  7's,  i.e., 

=  200  X  10  X  3000  =  6,000,000  f t.-lbs.  =  3000  ft.-tons ; 
/.  total  work  =  6001.6  +  3000     ±=  9001.6  ft.-tons.    • 

(If  the  track  were  an  up-grade,  1  in  10(5  say,  the  item  of 
200  X  30  =  6000  ft.-tons  would  be  added.) 

Example  2. — Required  the  rate  of  work,  or  power,  in  Ex- 
ample 1.  The  power  is  variable,  depending  on  the  velocity  of 
the  train  at  any  instant.  Assume  the  motion  to  be  uniformly 
accelerated,  then  the  working  force  is  constant ;  call  it  P. 
The  acceleration  (§  56)  will  be p=v*-i- 2s= 1936-5-6000=0.322 
ft.  per  sq.  sec.;  and  since  P  —  F  =  Mp,  we  have 

P  =  1  ton  +  (200  -r-  32.2)  X  0.322  =  3  tons, 

which  is  6000  -r-  200  =  30  Ibs.  per  ton  of  train,  of  which  20  is 
due  to  its  inertia,  since  when  the  speed  becomes  uniform  the 
work  of  the  engine  is  expended  on  friction  alone. 

Hence  when  the  velocity  is  44  ft.  per  sec.,  the  engine  is 
working  at  the  rate  of  Pv  =  264,000  f  t.-lbs.  per  sec.,  i.e.,  at  the 
rate  of  480  H.  P.; 

At  i  of  3000  ft.  from  the  start,  at  the  rate  of  240  H.  P.,  half 
as  much  ; 

At  a  uniform  speed  of  30  miles  an  hour  the  power  would  be 
simply  1  X  44  =  44  ft. -tons  per  sec.  =  160  H.  P. 

Example  3. — The  resistance  offered  by  still  water  to  the 
passage  of  a  certain  steamer  at  10  knots  an  hour  is  15,000  Ibs. 
What  power  must  be  developed  by  its  engines,  at  this  uniform 
speed,  considering  no  loss  in  "  slip"  nor  in  friction  of  ma- 
chinery ?  Ans.  461  H.  P. 

Example  4. — Same  as  3,  except  that  the  speed  is  to  be  15 
knots  (i.e.,  nautical  miles ;  each  —  6086  feet)  an  hour,  assum- 
ing that  the  resistances  are  as  the  square  of  the  speed  (approxi- 
mately true).  Ans.  1556  H,  P. 

Example  5. — Same  as  3,  except  that  12$  of  the  power  is  ab- 
sorbed in  the  "  slip"  (i.e.,  in  pushing  aside  and  backwards  the 
water  acted  on  by  the  screw  or  paddle),  and  8$  in  friction  of 
machinery.  Ans.  576  H.  P. 

Example  6. — In  Example  3,  if  the   crank-shaft   makes  60 


WORK,    ENERGY,    AND   POWER.  163 

revolutions  per  minute,  the  crank-pin  describing  a  circle  of  18 
inches  radius,  required  the  average*  value  of  the  tangential 
component  of  the  thrust  (or  pull)  of  the  connecting-rod  against 
the  crank-pin.  Ans.  26890  Ibs. 

Example  7. — A  solid  sphere  of  cast-iron  is  rolling  up  an  in- 
cline of  30°,  and  at  a  certain  instant  its  centre  has  a  velocity  of 
36  inches  per  second.  Neglecting  friction  of  all  kinds,  how 
much  further  will  the  ball  mount  the  incline  (see  §  143)  ? 

Ans.  0.390  ft. 

Example  8. — In  Fig.  163,  with  5  =  4  f  t.  and  a  =  16  inches, 
it  is  found  in  one  experiment  that  the  friction  which  keeps  the 
speed  of  the  pulley  at  120  revolutions  per  minute  is  balanced 
by  a  weight  G  =  160  Ibs.  Kequired  the  power  thus  measured. 

Ans.  14.6  H.  P. 

Although  in  Examples  1  to  6  the  steam  cylinder  is  itself  in 
motion,  the  work  per  stroke  is  still  =  mean  effective  steam- 
pressure  on  piston  X  length  of  stroke,  for  this  is  the  final  form 
to  which  the  separate  amounts  of  work  done  by,  or  upon,  the 
two  cylinder  heads  and  the  two  sides  of  the  piston  will  re- 
duce, when  added  algebraically.  See  §  154. 

*  By  "  average  value"  is  meant  such  a  value,  Tm,  as  multiplied  into  the 
length  of  path  described  by  the  crank-pin  per  unit  of  time  shall  give  the 
power  exerted. 


164  MECHANICS   OF  ENGINEEKINGh. 


CHAPTER  YII. 

FRICTION. 

156.  Sliding  Friction. — When  the  surfaces  of  contact  of  two 
bodies  are  perfectly  smooth,  the  direction  of  the  pressure  or  pair 
of  forces  between  them  is  normal  to  these  surfaces,  i.e.,  to  their 
tangent-plane ;  but  when  they  are  rough,  and 
moving  one  on  the  other,  the  forces  or  ac- 
tions between  them  incline  away  from  the 
normal,  each  on  the  side  opposite  to  the  di- 
rection of  the  (relative)  motion  of  the  body 
on  which  it  acts.     Thus,  Fig.  167,  a  block 
FIG.  167.  whose  weight  is  G,  is  drawn  on  a  rough 

horizontal  table  by  a  horizontal  cord,  the  tension  in  which  is 
P.  On  account  of  the  roughness  of  one  or  both  bodies  the  ac- 
tion of  the  table  upon  the  block  is  a  force  jP,,  inclined  to  the 
normal  (which  is  vertical  in  this  case)  at  an  angle  =  q>  away 
from  the  direction  of  the  relative  velocity  v.  This  angle  q>  is 
called  the  angle  of  friction,  while  the  tangential  component  of 
jP,  is  called  the  friction  =  F.  The  normal  component  N, 
which  in  this  case  is  equal  and  opposite  to  G  the  weight  of  the 
body,  is  called  the  normal  pressure. 

Obviously  F  •=•  jVtan  <p,  and  denoting  tan  q>  by/*,  we  have 

F=fN. (1) 

/"is  called  the  coefficient  of  friction,  and  may  also  be  defined 
as  the  ratio  of  the  friction  F  to  the  normal  pressure  N  which 
produces  it. 


FRICTION.  165 

In  Fig.  167,  if  the  motion  is  accelerated  (ace.  =  p),  we  have 
(eq.  (IV.),  § 55)  P  —  F '=  Mp ;  if  uniform,  P  —  F=0',  from 
which  equations  (see  also  (I))/1  may  be  computed.  In  the 
latter  case/"  may  be  found  to  be  different  with  different  veloci- 
ties (the  surfaces  retaining  the  same  character  of  course),  and 
then  a  uniformly  accelerated  motion  is  impossible  unless  P 
—  F  were  constant. 

As  for  the  lower  block  or  table,  forces  the  equals  and  op- 
posites  of  N  and  F  (or  a  single  force  equal  and  opposite  to  P^ 
are  comprised  in  the  system  of.  forces  acting  upon  it. 

As  to  whether  F  is  a  working  force  or  a  resistance,  when 
either  of  the  two  bodies  is  considered  free,  depends  on  the  cir- 
cumstances of  its  motion.  For  example,  in  friction-gearing 
the  "tangential  action  between  the  two  pulleys  is  a  resistance 
for  one,  a  working  force  for  the  other. 

If  the  force  P,  Fig.  167,  is  just  sufficient  to  start  the  body, 
or  is  just  on  the  point  of  starting  it  (this  will  be  called  impending 
motion),  F\&  called  the  friction  of  rest.  If  the  body  is  at  rest 
and  P  is  not  sufficient  to  start  it,  the  tangential  component  will 
then  be  <  the  friction  of  rest,  viz.,  just  =  P.  As  P  increases, 
this  component  continually  equals  it  in  value,  and  P1  acquires 
a  direction  more  and  more  inclined  from  the  normal,  until  the 
instant  of  impending  motion,  when  the  tangential  component 
•=fJV=  the  friction  of  rest.  When  motion  is  once  in  prog- 
ress, the  friction,  called  then  the  friction  of  motion,  =fJW, 
in  which/*  is  not  necessarily  the  same  as  in  the  friction  of  rest. 

157.  Laws  of  Sliding  Friction.* — Experiment  has  demon- 
strated the  following  relations  approximately,  for  two  given 
rubbing  surfaces  :  (bee  §  175.) 

(1)  The  coefficient,/1,  is  independent  of  the  normal  pressure 
N. 

(2)  The  coefficient,/,  for  friction  of  motion,  is  the  same  at 
all  velocities. 

(3)  The  coefficient, /,  for 'friction   of  rest  (i.e.,  impending 
motion)   is  usually  greater  than   that  for   friction   of  motion 
( probably  on  account  of  adhesion). 

*  These  "laws"  apply,  with  only  a  rude  approximation,  to  dry  surfaces 
(except  5)  under  moderate  pressures  and  at  low  velocities.  See  pp.  167, 
168,  uiid  192.  Also  see  Eng.  News,  May  1895,  p.  322,  for  experiments  on 
wood. 


166 


MECHANICS    OF   ENGINEERING. 


(4)  The  coefficient, jf,  is  independent  of  the  extent  of  rub- 
bing surface. 

(5)  The  interposition  of  an  unguent  (such  as  oil,  lard,  tallow, 
etc.)  diminishes  the  friction  very  considerably. 

158.  Experiments  on  Sliding  Friction. — These  may  be  made 
with  simple  apparatus.  If  a  block  of  weight  —  G,  Fig.  168, 
be  placed  on  an  inclined  plane  of  uniformly  rough  surface, 
and  the  latter  be  gradually  more  and  more  inclined  from  the 
horizontal  until  the  block  begins  to  move,  the  value  of  ft  at 


FIG.  168. 


FIG.  169. 


this  instant  =  ^>,  and  tan  (p=f=  coefficient  of  friction  of 
rest.  For  from  ^2X  =  0  we  have  F,  i.e.,  fN,  =  G  sin  ft\ 
from  2Y=  0,  N=  G  cos  ft ;  whence  tan  ft  =/,  .-.  ft  must 

=  9- 

Suppose  ft  so  great  that  the  motion  is  accelerated,  the  body 

starting  from  rest  at  0,  Fig.  169.  It  will  be  found  that  the 
distance  x  varies  as  the  square  of  the  time,  hence  (§  56)  the 
motion  is  uniformly  accelerated  (along  the  axis  X).  (Notice 
fin  the  figure  that  G  is  no  longer  equal  and  opposite  to  jP,,  the 
resultant  of  ^and  F^  as  in  Fig.  168.) 

2  Y  —  0,      which  gives  N  —  G  cos  ft  =  0  ; 
2X  =  Mp,  which  gives  G  sin  /3—fJ^=  (G  +  g)p\ 
while  (from  §  56)  p=2x+  f. 

Hence,  by  elimination,  x  and  the  corresponding  time  t  having 
been  observed,  we  have  for  the  coefficient  of  friction  of  motion 


/  =  tan  ft  — 


gt*  cos  ft' 


FRICTION.  167 

In  view  of  (3),  §  157,  it  is  evident  that  if  a  value  ftm  has  been 
found  experimentally  for  ft  such  that  the  block,  once  started  ly 
hand,  preserves  a  uniform  motion  down  the  plane,  then,  since 
tan  ftm  =f  for  friction  of  motion,  ftm  may  be  less  than  the  ft 
in  Fig.  168,  for  friction  of  rest. 

159.  Another  apparatus  consists  of  a  horizontal  plane,  a  pul- 
ley, cord,  and  two  weights,  as  shown  in  Fig.  59.  The  masses 
of  the  cord  and  pulley  being  small  and  hence  neglected,  the 
analysis  of  the  problem  when  G  is  so  large  as  to  cause  an  ac- 
celerated motion  is  the  same  as  in  that  example  [(2)  in  §  57], 
except  in  Fig.  60,  where  the  frictional  resistance  /LZV  should  be 
put  in  pointing  toward  the  left.  N  still  =  Ga  and  /. 

8  -/&>=(&,  +  fa;    .    ,,••'.'  .    .    (1) 
while  for  the  other  free  body  in  Fig.  61  we  have,  as  before, 

G-S=(G  +  g}p..    .  V  .    ,     .     (2) 

From  (1)  and  (2),  8  the  cord-tension  can  be  eliminated,  and 
solving  for  /?,  writing  it  equal  to  2s  ~  t\  s  and  t  being  the  ob- 
served distance  described  '(from  rest)  and  corresponding  time, 
we  have  finally  for  friction  of  motion 


If  G,  Fig.  59,  is  made  just  sufficient  to  start  the  block,  or 
sledge,  GU  we  have  for  the  friction  of  rest 

I  /-£.  .:.....<„ 


160.  Results  of  Experiments  on  Sliding  Friction.  —  Professor 
Thurston  in  his  article  on  Friction  (which  the  student  will  do 
well  to  read)  in  Johnson's  Cyclopaedia  gives  the  following 
epitome  of  results  from  General  Morin's  experiments  (made 
for  the  French  Government  in  1833)  (low  velocities  and 
pressures)  : 


168 


MECHANICS    OF   ENGINEERING. 
TABLE  FOR  FRICTION  OF  MOTION. 


No. 

Surfaces. 

Unguent. 

Angle  <f>. 

/=tan«{,. 

1 

Wood  on  wood. 

None. 

14°    to  26£° 

0.25  to  0.50 

2 

Wood  on  wood. 

Soap. 

2°    to  1H° 

0.04  to  0.20 

3 

Metal  on  wood. 

Nane. 

26|°  to  3H° 

0.50  to  0.60 

4 

Metal  on  wood. 

Water. 

15°   to  20° 

0.25  to  0.35 

5 

Metal  on  wood. 

Soap. 

nr 

0.20 

6 

Leather  on  metal. 

None. 

29r 

0.56 

7 

Leather  on  metal. 

Greased. 

13° 

0.23 

8 

Leather  on  metal. 

Water. 

20° 

0.36 

9 

Leather  on  metal. 

Oil. 

8|° 

0.15 

10 

Smoothest  and  best 

lubricated  surfaces. 

lf°  to    2° 

0.03  to  0.036 

For  friction  of  rest,  about  40$  may  be  added  to  the  coeffi- 
cients in  the  above  table. 

In  dealing  with  the  stone  blocks  of  an  arch-ring,  cp  is  com- 
monly taken  =  30°,  i.e.,  /  =  tan  30°  =  0.58  as  a  low  safe 
value ;  it  is  considered  that  if  the  direction  of  pressure  between 
two  stones  makes  an  angle  >  30°  with  the  normal  to  the  joint 
(see  §  161)  slipping  may  take  place  (the  adhesion  of  cement 
being  neglected). 

General  Morin  states  that  for  a  sledge  on  dry  ground  f  = 
about  0.66. 

Weisbach  gives  for  metal  on  metal,  dry  (R.  R.  brakes*  for 
example), y  =  from  0.15  to  0.24".  Trautwine's  Pocket-Book 
gives  values  off  for  numerous  cases  of  friction. 


-U, 


161.  Cone  of  Friction.— Fig.  170.     Let  A  and  B  be  two 

rough  blocks,  of  which  B  is  immovable,  and  P  the  resultant 
of  all  the  forces  acting  on  A,  except  the  pres- 
sure from  B.  B  can  furnish  any  required 
normal  pressure  N  to  balance  P  cos  /?,  but 
the  limit  of  its  tangential  resistance  is  fN. 
So  long  then  as  /3  is  <  cp  the  angle  of  fric- 
tion, or  in  other  words,  so  long  as  the  line  of 
action  of  P  is  within  the  "  cone  of  friction" 

*  Capt.  Galton's  experiments  in  1878,  with  cast-iron  bnike-shoes  and 
steel-tired  wheels,  gave  for  a  velocity  of  10  miles  per  hour/=  0.24 ;  for  20 
miles  per  hour/—  0.19  ;  for  30,  40,  50,  and  60  miles  per  hour  /  =  0.164, 
0.14,  0.116,  and  0.074,  respectively.  Even  smaller  values  (about  half)  for/ 
have  been  found  for  steel  tires  on  steel  rails. 


A 

V 

A 

0----/N 

FIG.  170. 


FRICTION.  169 

generated  by  revolving  00  about  ON,  the  block  A  will  not 
slip  on  B,  and  the  tangential  resistance  of  B  is  simply  P  sin 
/3 ;  but  if  ft  is  >  q>,  this  tangential  resistance  being  onlyfN 
and  <  P  sin  /?,  .A  will  begin  to  slip,  with  an  acceleration. 

162.  Problems  in  Sliding  Friction. — In  the  following  prob- 
lems/* is  supposed  known  at  points  where  rubbing  occurs,  or 
is  impending.  As  to  the  pressure  N  to  which  the  friction  is 
due,  it  is  generally  to  be  considered  unknown  until  determined 
by  the  conditions  of  the  problem.  Sometimes  it  may  be  an 
advantage  ^o  deal  with  the  single  unknown  force  P  (resultant 
of  NandifN)  acting  in  a  line  making  the  known  angle  <p  with 
the  normal  (on  the  side  away  from  the  motion). 

PROBLEM  1. — Required  the  value  of  the  weight  P,  Fig.  171, 
the  slightest  addition  to  which  will  cause  motion  of  the  hori- 
zontal rod  OB,  resting  on  rough  planes  at  45°.  The  weight 
G  of  the  rod  may  be  applied  at  the 
middle.  Consider  the  rod  free ;  at 
each  point  of  contact  there  is  an  un- 
known JW  and  a  friction  due  to  it 
fN\  the  tension  in  the  cord  will  be 
=  P,  since  there  is  no  acceleration 


and   no  friction   at  pulley.     Notice  FIG.  m. 

the  direction  of  the  frictions,  both  opposing  the  impending 
motion.  [The  student  should  not  rush  to  the  conclusion  that 
.TV  and  NI  are  equal,  and  are  the  same  as  would  be  produced  by 
the  components  of  G  if  the  latter  were  transferred  to  A  and 
resolved  along  AO  and  AB  ;  but  should  await  the  legitimate 
results  deduced  by  algebra,  from  the  equations  of  condition 
for  the  equilibrium  of  a  system  of  forces  in  a  plane.  Few 
problems  in  Mechanics  are  so  simple  as  to  admit  of  an  'imme- 
diate mental  solution  on  inspection  ;  and  guess-work  should  be 
carefully  avoided.] 

Taking  an  origin  and  two  axes  as  in  figure,  we  have  (eqs. 
(2),  §  36),  denoting  the  sine  of  45°  by  m, 

N-P  =  0;.     .     (1) 


____  fNa  +  Na  -  Gb  =  0.  .     .     (3) 


170 


MECHANICS    OF   ENGINEERING. 


The  three  unknowns  P,  JV,  and  Nl  can  now  be  found. 
Divide  (3)  by  a,  remembering  that  J  :  a  =  m,  and  solve  for 
N\  substitute  it  in  (2)  and  Nl  also  becomes  known ;  while  P 
is  then  found  from  (1)  and  is 


P  = 


/V2 


G. 


PROBLEM  2. — Fig.  172.     A  rod,  centre  of  gravity  at  middle, 
leans  against  a  rough  wall,  and  rests  on  an  equally  rough  floor; 
how  small  may  the  angle  a  become  before  it 
~-   P2        sliPs  •     Let  a  =  the  half-length.     The  figure 
Fr ~         shows  the  rod  free,  and  following  the  sugges- 
tion  of   §  162,   a   single   unknown  force  Pl 
making  a  known  angle  q>  (whose   tan  =tf) 
with  the  normal  DE,  is  put  in  at  D,  leaning 
away  from  the  direction  of  the  impending 
motion,  instead  of  an  N  and  fN\  similarly 
FIG.  ITS.  PI  acts  at  C.     The  present  system  consisting 

of  but  three  forces,  the  most  direct  method  of  finding  «,  with- 
out introducing  the  other  two  unknowns  Pl  and  P3  at  all,  is 
to  use  the  principle  that  if  three  forces  balance,  their  lines 
of  action  must  intersect  in  a  point.  That  is,  Pt  must  inter- 
sect the  vertical  containing  G,  the  weight,  in  the  same  point 
as  P,,  viz.,  A. 

Now  EA,  and  also  BC^  =  a  cos  «, 

.*.  ED  =  a  cos  a  cot  <p        and         AB  =  a  cos  a  tan  (p. 
But  DF,  which  =  20  sin  a,  =  DE '—  AB ; 

.'.  2#  sin  a  =  a  cos  a  [cot  cp  —  tan  cp].      .     .     (1) 

Dividing  by  cos  «,  and  noting  that  tan  q>  =f=  1  -f-  cot  99, 
we  obtain  for  the  required  value  of  a 


and  finally,     tan  a  =  cot 


after  some  trigonometrical  reduction.     That  is,  a  is  the  com- 
plement of  double  the  angle  of  friction. 


FRICTION. 


171 


PROBLEM  3. — Fig.  173.  Given  the  resistance  Q,  acting 
parallel  to  the  fixed  guide  C*  the  angle  <*,  and  the  (equal)  co- 
efficients of  friction  at  the  rubbing  surfaces,  required  the 


FIG.  173. 


FIG.  174. 


amount  of  the  horizontal  force  JP,  at  the  head  of  the  block  A 
(or  wedge\  to  overcome  Q  and  the  frictions.  D  is  fixed,  and 
ab  is  perpendicular  to  cd.  Here  we  have  four  unknowns,  viz., 
P,  and  the  three  pressures  N,  N»  and  Nm  between  the  blocks. 
Consider  A  and  B  as  free  bodies,  separately  (see  Fig.  174),  re- 
membering Newton's  law  of  action  and  reaction.  The  full 

O 

values  (e.g.,/W)  of  the  frictions  are  put  in,  since  we  suppose 
a  slow  uniform  motion  taking  place. 
For  A,  2X=  0  and  2Y=  0  give 


rl  —  NCOS  a  -f/ZTsin  a  —  P  sin  a  =  0  ;  .  .  (1) 
^^sin  a  +fJVGosa  —  P  cos  a  =  0.  .  .  (2) 
For  B,  ^Xand  2  T  give 

Q  —  Ni  +/-#»  =  0  ;  ....  (3)      and      Nt  —fN^  =  0.  ...  (4) 

Solve  (4)  for  N^  and  substitute  in  (3),  whence 

-/')=# :     •       •       (5) 


Solve  (2)  for  N,  substitute  the  result  in  (1),  as  also  the  value 
of  -ZV^  from  (5),  and  the  resulting  equation  contains  but  one  un- 
known, P.  Solving  for  J3,  putting  for  brevity 


sn 


a  —  m     and     cos  a  — 


a  = 


we  have 
or 


P  = 


(n  .  cos  a  -\-  m  .  sin  or)(l  — /a)' 


172  MECHANICS    OF   ENGINEERING. 

Numerical  Example  of  Problem  3.— If  Q  =  120  Ibs.,  / 
=  0.20  (an  abstract  number,  and  /.  the  same  in  any  system  of 
units),  while  a  =  14°,  whose  sine  =  0.240  and  cosine  —  .970, 
then 

m  =  0.2 X. 97  +  0.24  =  0.43  and  n  =  .97  —  .2X-24 -=  0.92, 
whence  P  =  0.64$  =  76.8  Ibs. 

While  the  wedge  moves  2  inches  P  does  the  work  (or  exerts 
an  energy)  of  2  X  76.80  =  153.6  in.-lbs.  =  12.8  ft.-lbs. 

For  a  distance  of  2  inches  described  by  the  wedge  horizon- 
tally, the  block  JS  (and  /.  the  resistance  Q)  has  been  moved 
through  a  distance  =  2  X  sin  14°  =  0.48  in.  along  the  guide 
<7,  and  hence  the  work  of  120  X  0.48  =  57.6  in.-lbs.  has  been 
done  upon  Q.  Therefore  for  the  supposed  portion  of  the 
motion  153.6  —  57.6  —  96.0  in.-lbs.  of  work  has  been  lost  in 
friction  (converted  into  heat). 

It  is  noticeable  in  eq.  (6),  that  -if /"should  =  1.00,  P  =  oc  ; 
and  that  if  a  .=  90°.  P  =  Q.  and  there  is  no  friction  (the 
weights  of  the  blocks  have  been  neglected). 

PROBLEM  4.  Numerical. — With  what  minimum  pressure 
P  should  the  pulley  A  be  held  against  B.  which  it  drives  by 
fi  fl  .x  i-i  A  "  frictional  gearing,"  to  transmit  2  H.P.; 
4  ;•  js\~  )  if  a  —  45°,  f  for  impending  (relative) 
..  N  motion,  i.e.,  for  impending  slipping  = 

FIG.  175.  0.40,  and  the  velocity  of  the  pulley-rim 

is  9  ft.  per  second  ? 

The  limit-value  of  the  tangential  "  grip" 

T  =         %fN=  2  X  0.40  X  P  sin  45°, 
2  H.  P.  =  2  X  550  =  1100  ft.-lbs.  per  second. 

Putting  T  X  9  ft.  =  1100,  we  have* 

2  X  0.40  X  I/?  X  P  X  9  =  1100  ;  .-.  P  =  215  Ibs. 

PROBLEM  6. — A  block  of  weight  G  lies  on  a  rough  plane, 
inclined  an  angle  ft  from  the  horizontal ;  find  the  pull  P,  mak- 
ing an  angle  a  with  the  first  plane,  which  will  maintain  a  uni- 
form motion  up  the  plane. 

*  In  this  problem  the  student  should  note  that,  in  general,  when  a  is  not  45°, 
we  have  N  =  %P  -s-  cos  a  (since  in  such  a  case  the  parallelogram  of  forces  is 
not  a  square). 


FRICTION.  173 

PROBLEM  7. — Same  as  6,  except  that  the  pull  P  is  to  permit 
a  uniform  motion  down  the  plane. 

PROBLEM  8. — The  thrust  of  a  screw-propeller  is  15  tons. 
The  ring  against  which  it  is  exerted  has  a  mean  radius  of  8 
inches,  the  shaft  makes  one  revolution  per  second,  and/*  =  0.06. 
Required  the  H.  P.  lost  in  friction  from  this  cause. 

Ans.  13.7  H.  P. 

163.  The  Bent-Lever  with  Friction.  Worn  Bearing. — Fig. 
176.  Neglect  the  weight  of  the  lever,  and  suppose  the  plumb- 
er-block so  worn  that  there  is 
contact  along  one  element  only  of 
the  shaft.  Given  the  amount  and 
line  of  action  of  the  resistance  72, 
and  the  line  of  action  of  P,  re- 
quired the  amount  of  the  latter  for 
impending  slipping  in  the  direction 
of  the  dotted  arrow.  As  P  grad- 
ually increases,  the  shaft  of  the 
lever  (or  gear-wheel)  rolls  on  its  FIG. 

bearing  until  the  line  of  contact  has  reached  some  position  A, 
when  rolling  ceases  and  slipping  begins.  To  find  A,  and  the 
value  of  P,  note  that  the  total  action  of  the  bearing  upon  the 
lever  is  some  force  jP,,  applied  at  A  and  making  a  known 
angle  cp  (f  —  tan  <p)  with  the  normal  A  C.  Pl  must  be  equal 
and  opposite  to  the  resultant  of  the  known  ft  and  the  unknown 
P,  and  hence  graphically  (a  graphic  is  much  simpler  here  than 
an  analytical  solution)  if  we  describe  about  C  a  circle  of  radius 
=  r  sin  <p,  r  being  the  radius  of  shaft  (or  gudgeon),  and  draw 
a  tangent  to  it  from  D,  we  determine  DA  as  the  line  of  action 
of  Pr  If  DG  is  made  =  R,  to  scale,  and  (redrawn  parallel 
to  D  .  .  .  P,  P  is  determined,  being  =  DE,  while  P,  =  DF. 

If  the  known  force  ft  is  capable  of  acting  as  a  working  force, 
by  drawing  the  other  tangent  DB  from  D  to  the  "  friction- 
circle,"  we  have  P  =  DH,  and  P1  =  DK^  for  impending 
rotation  in  an  opposite  direction. 

If  ft  and  P  are  the  tooth -pressures  upon  two  spur:wheels, 
keyed  upon  the  same  shaft  and  nearly  in  the  same  plane,  the 


174  MECHANICS    OF   ENGINEERING. 

same  constructions  hold  good,  and  for  a  continuous  uniform 
motion,  since  the  friction  =  P1  sin  <p, 
the  work  lost  in  friction 


=  [P  sin 
per  revolution, 

It  is  to  be  remarked,  that  without  friction  Pl  would  pass 
through  Gj  and  that  the  moments  of  It  and  P  would  balance 
about  C  (for  rest  or  uniform  rotation) ;  whereas  with  friction 
they  balance  about  the  proper  tangent-point  of  the  friction- 
circle. 

Another  way  of  stating  this  is  as  follows :  So  long  as  the 
resultant  of  P  and  R  falls  within  the  "  dead-angle"  BDA^ 
motion  is  impossible  in  either  direction. 

If  the  weight  of  the  lever  is  considered,  the  resultant  of  it 
and  the  force  R  can  be  substituted  for  the  latter  in  the  fore- 
going. 

164.  Bent-Lever  with  Friction.  Triangular  Bearing. — Like 
the  preceding,  the  gudgeon  is  much  exaggerated  in  the  figure 

(177).  For  impending  rotation  in 
direction  of  the  force  jP,  the  total 
actions  at  A1  and  A9  must  lie  in 
known  directions,  making  angles  =  cp 
with  the  respective  normals,  and  in- 
clined away  from  the  slipping.  Join 
the  intersections  D  and  L.  Since 
the  resultant  of  P  and  R  at  D  must 
act  along  DL  to  balance  that  of  Pl 
and  PV  having  given  one  force,  say 
FIG.  177.  7?5  We  easily  find  P  =  DE,  while 

PI  and  PS  =  ZJfand  LN  respectively,  LO  having  been  made 
=  DF,  and  the  parallelogram  completed. 

(If  the  direction  of  impending  rotation  is  reversed,  the  change 
in  the  construction  is  obvious.)  If  P9  =  0,  the  case  reduces 
to  that  in  Fig.  176 ;  if  the  construction  gives  P9  negative,  the 
supposed  contact  at  A^  is  not  realized,  and  the  angle  A.iCAl 
should  be  increased,  or  shifted,  until  /\  is  positive. 

As  before,  P  and  R  may  be  the  tooth- pressures   on  two 


FKICTION. 


175 


spur-wheels  nearly  in  the  same  plane  and  on  the  same  shaft ; 
if  so,  then,  for  a  uniform  rotation, 
Work  lost  in  fric.  per  revol.  =  \_P^  sin  <p  -f-  Pt  sin  <p]%rtr. 

165.  Axle-Friction. — The  two  foregoing  articles  are  intro- 
ductory to  the  subject  of  axle -friction.     When  the  bearing 'is 
new,  or  nearly  so,  the  elements  of  the  axle  which  are  in  contact 
with  the  bearing  are  infinite  in  number,  thus  giving  an  infinite 
number  of  unknown  forces  similar  to  P1  and  P^  of  the  last 
paragraph,  each  making  an  angle  q>  with  its  normal.     Refined 
theories  as  to  the  law  of  distribution  of  these  pressures  are  of 
little  use,   considering  the   uncertainties  as  to   the  value   of 
tf(  =  tan  q>) ;  hence  for  practical  purposes  axle-friction  may  be 
written 

F=fB, 

in  which  f  is  a  coefficient  of  axle-friction  derivable  from 
experiments  with  axles,  and  R  the  resultant  pressure  on  the 
bearing.  In  some  cases  R  may  be  partly  due  to  the  tightness 
of  the  bolts  with  which  the  cap  of  the  bearing  is  fastened. 

As  before,  the  work  lost  in  overcoming  axle-friction  per 
revolution  is  ^f'R^Ttr,  in  which  r  is  the  radius  of  the  axle. 
f,  like/1,  is  an  abstract  number.  As  in  Fig.  176,  a  "  friction- 
circle,"  of  radius  =f'r,  may  be  considered  as  subtending  the 
"  dead-angle." 

166.  Experiments  with   Axle-Friction. — Prominent  among 
recent  experiments  have  been  those 

of  Professor  Thurston  (1872-73), 
who  invented  a  special  instrument 
for  that  purpose,  shown  (in  princi- 
ple only)  in  Fig.  178.  By  means  of 
an  internal  spring,  the  amount  of 
whose  compression  is  read  on  a  scale, 
a  weighted  bar  or  pendulum  is  caused 
to  exert  pressure  on  a  projecting  axle 
from  which  it  is  suspended.  The 

i  •£..—       .^_....^i 

axle  is  made  to  rotate  at  any  desired  FIG.  ITS. 

velocity  by  some  source  of  power,  the  axle-friction  causing 


176  MECHANICS    OF   ENGINEERING. 

the  pendulum  to  remain  at  rest  at  some  angle  of  deviation 
from  the  vertical.  The  figure  shows  the  pendulum  free,  the 
action  of  gravity  upon  it  being  #,  that  of  the  axle  consisting 
of  the  two  pressures,*  each  =  7?,  and  of  the  two  frictions  (each 
being  F  —f  R\  due  to  them.  Taking  moments  about  C,  we 
have  for  equilibrium 

2f£r  =  GFb, ' 

in  which  all  the  quantities  except  f  are  known  or  observed. 
The  temperature  of  the  bearing  is  also  noted,  with  reference 
to  its  effect  on  the  lubricant  employed.  Thus  the  instrument 
covers  a  wide  range  of  relations. 

General  Morin's  experiments  as  interpreted  by  Weisbach 
give  the  following  practical  results : 

!|-  0.054  for  well-sustained 
J       lubrication; 
/  I  0.07  to  .08  for  ordinary 

I      lubrication. 

By  "  pressure  per  square  inch  on  the  bearing"  is  commonly 
meant  the  quotient  of  the  total  pressure  m  Ibs.  by  the  area  in 
square  inches  obtained  by  multiplying  the  width  of  the  axle  by 
the  length  of  bearing  (this  length  is  quite  commonly  four  times 
the  diameter)  ;  call  it^>,  and  the  velocity  of  rubbing  in  feet  per 
minute,  v.  Then,  according  to  Kankine,  to  prevent  overheat- 
ing, we  should  have 

p(v  +  20)  <  44800  .  .  .  (not  homog.). 

Still,  in  marine-engine  bearings  pv  alone  often  reaches  60,000, 
as  also  in  some  locomotives  (Cotterill).  Good  practice  keeps 
P  within  the  limit  of  800  (Ibs.  per  sq.  in.)  for  other  metals 
than  steel  (Thurston),  for  which  1200  is  sometimes  allowed. 

With  v  =  200  (feet  per  min.)  Professor  Thurston  found  that 
for  ordinary  lubricants  p  should  not  exceed  values  ranging 
from  30  to  75  (Ibs.  per  sq.  in.). 

The  product  pv  is  obviously  proportional  to  the  power  ex- 
pended in  wearing  the  rubbing  surfaces,  per  unit  of  area. 

*  The  weight  G  being  small  compared  with  the  compressive  force 
the  spring,  each  pressure  is  practically  equal  to  R. 


FRICTION.  177 

167.  Friction-Wheels. — A  single  example  of  their  use  will 
be  given,  with  some  approximations  to  avoid  complexity.    Fig. 
179.     G  is  the  weight  of  a  heavy  wheel,  P1  is  a  known  vertical 
resistance  (tooth-pressure),  and  P  an 
unknown     vertical    working    force, 
whose  value  is  to  be  determined  to 
maintain   a  uniform   rotation.     The 
utility  of  the  friction-wheels  is  also 
to  be  shown.     The  resultant  of  P^ 
6r,  and  P  is  a  vertical  force  R,  pass- 
ing nearly  through  the  centre  C  of 
the  main  axle  which  rolls  on  the  four 
friction- wheels.      R,  resolved  along          ^---.a 
CA  and  CB,  produces  (nearly)  equal  FIG.  179. 

pressures,  each  being  N  =  R  -f-  2  cos  at,  at  the  two  axles  of 
the  friction- wheels,  which  rub  against  their  fixed  plumber- 
blocks.  R  =  P  -f-  P1  -\-  G»  and  .*.  contains  the  unknown  P, 
but  approximately  =  G  -\-  2P15  i.e.,  is  nearly  the  same  (in  this 
case)  whether  friction-wheels  are  employed  or  not. 

When  G  makes  one  revolution,  the  friction  f'N  at  each  axle 
Cl  is  overcome  through  a  distance  —  (rl  :  a^  %7tr,  and 

Work  lost  per  revol.  )  1 

with  (  =  2  f'NT^nr  =  ^  -^ 

,  .    .         T     ,          (          J      a.  a,  cos 

friction-wheels,       J 

Whereas,  if  C  revolved  in  a  fixed  bearing, 

Work  lost  per  revol.  ) 

without  V  =fR%7tr. 

friction-wheels,        ) 

Apparently,  then,  there  is  a  saving  of  work  in  the  ratio  /•,: 
al  cos  a,  but  strictly  the  H  is  not  quite  the  same  in  the  two  cases ; 
for  with  friction-wheels  the  force  P  is  less  than  without,  and  R 
depends  on  P  as  well  as  on  the  known  G  and  Pv  By  dimin- 
ishing the  ratio  rl  :  al9  and  the  angle  a,  the  saving  is  increased. 
If  a  were  so  large  that  cos  a  <  r1 :  a^  there  would  be  no  saving, 
but  the  reverse. 

As  to  the  value  of  P  to  maintain  uniform  rotation,  we  have 
12 


178  MECHANICS    OF   ENGINEERING. 

for  equilibrium  of  moments  about  O9  with  friction-wheels  (con> 
sidering  the  large  wheel  and  &x\.efree\ 

Pi  =  PA  +  27V,    ......    (1) 

in  which  T  is  the  tangential  action,  or  "  grip,"  between  one 
pair  of  friction-wheels  and  the  axle  C  which  rolls  upon  them. 
T  would  not  equal  fN  unless  slipping  took  place  or  were  im- 
pending at  E,  but  is  known  by  considering  a  pair  of  friction- 
wheels  free,  when  2  (Pa)  about  Ol  gives 


which  in  (1)  gives  finally 


a,  cos 
Without  friction-wheels,  we  would  have 


(3) 


The  last  term  in  (2)  is  seen  to  be  less  than  that  in  (3)  (unless 
a  is  too  large),  in  the  same  ratio  as  already  found  for  the  saving 
of  work,  supposing  the  J?'s  equal. 

If  Pl  were  on  the  same  side  of  C  as  jP,  it  would  be  of  an 
opposite  direction,  and  the  pressure  R  would  be  'diminished. 
Again,  if  P  were  horizontal,  R  would  not  be  vertical,  and  the 
friction-wheel  axles  would  not  bear  equal  pressures.  Since  P 
depends  on  P^  G,  and  the  frictions,  while  the  friction  depends 
on  R,  and  R  on  P^  G,  and  P,  an  exact  analysis  is  quite 
complex,  and  is  not  warranted  by  its  practical  utility. 

Example.  —  If  an  empty  vertical  water-wheel  weighs  25,000 
Ibs.,  required  the  force  P  to  be  applied  at  its  circumference  to 
maintain  a  uniform  motion,  with  a  =  15  ft.,  and  r  =  5  inches. 
Here  Pl  =  0,  and  R  =  G  (nearly  ;  neglecting  the  influence  of 
P  on  R\  i.e.,  R  =  25,000  Ibs. 

First,  without  friction-wheels  (adopting  the  foot-pound-  sec- 
ond system  of  units),  withy7  =  .07  (abstract  number).  From 
eq.  (3)  we  have 

p  =  o  +  0.07  X  25,000  X  (A  •*•  15)  =  4§.6  Ibs. 


FRICTION.  179 

The  work  lost  in  friction  per  revolution  is 
fRZnr  =  0.07  X  25,000  X  2  X  3.14  X  TV  =  4580  ft.-lbs. 

/Secondly,  with  friction-wheels,  in  which  rl  :  al  =  -J  and 
cos  a  =  0.80  (i.e.,  a  =  36°).  From  eq.  (2) 

P  =  0  +  | .  Y  X  48.6  =  only  12.15  Ibs., 
while  the  work  lost  per  revolution 

=  | .  ±£  X  4580  =  1145  ft.-lbs. 

Of  course  with  friction- wheels  the  wheel  is  not  so  steady  as 
without. 

In  this  example  the  force  P  has  been  simply  enough  to 
overcome  friction.  In  case  the  wheel  is  in  actual  use,  P  is  the 
weight  of  water  actually  in  the  buckets  at  any  instant,  and  does 
the  work  of  overcoming  Plt  the  resistance  of  the  mill  machinery, 
and  also  the  friction.  By  placing  Pl  pointing  upward  on  the 
same  side  of  0  as  P,  and  making  5,  nearly  ==  &,  R  will  =  G 
nearly,  just  as  when  the  wheel  is  running  empty;  and  the 
foregoing  numerical  results  will  still  hold  good  for  practical 
purposes. 

168.  Friction  of  Pivots. — In  the  case  of  a  vertical  shaft  or 
axle,  and  sometimes  in  other  cases,  the  extremity  requires  sup- 
port against  a  thrust  along  the  axis  of  the  axle  or  pivot.  If 
the  end  of  the  pivot  is  flat  and  also  the  surface 
against  which  it  rubs,  we  may  consider  the 
pressure,  and  therefore  the  friction,  as  uniform 
over  the  surface.  With  a  flat  circular  pivot, 
then,  Fig.  180,  the  frictions  on  a  small  sector 
of  the  circle  form  a  system  of  parallel  forces 
whose  resultant  is  equal  to  their  sum,  and  is 
applied  a  distance  of  %r  from  the  centre.  Hence  the  sum  of 
the  moments  of  all  the  frictions  about  the  centre  =fR%r,  in 
which  R  is  the  axial  pressure.  Therefore  a  force  P  necessary 
to  overcome  the  friction  with  uniform  rotation  must  have  a 
moment 

Pa  =/fffr, 


180  MECHANICS   OF   ENGINEERING. 

and  the  work  lost  in  friction  per  revolution  is 

(1) 


As  the  pivot  and  step  become  worn,  the  resultant  friction* 
in  the  small  sectors  probably  approach  the  centre  ;  for  the 
greatest  wear  occurs  first  near  the  outer  edge,  since  there  the 
product^  is  greatest  (see  §  166).  Hence  for  \r  we  may  more 
reasonably  put  ^r. 

Example.  —  A  vertical  flat-ended  pivot  presses  its  step  with 
a  force  of  12  tons,  is  6  inches  in  diameter,  and  makes  40  revolu- 
tions per  minute.  Required  the  H.  P.  absorbed  by  the  friction. 
Supposing  the  pjvot  and  step  new,  and  f  for  good  lubrication 
=  0.07,  we  have,  from  eq.  (1)  (foot-lb.  -second)  < 

Work  lost  per  revolution 

=  .07  X  24,000  X  6.28  X  %  .  i  =  1758.4  f  t.-lbs., 
and  .'.  work  per  second 

=  1758.4  x  H  =  1172.2  ft.-lbs., 

which  -r-  550  gives  2.13  H.P.  absorbed  in  friction.     If  ordi- 
nary axle-friction  also  occurs  its  effect  must  be  added. 

If  the  flat-ended  pivot  is  hollow  >,  with  radii  r,  and  r^  we  may 
put  ^(rl  +  rj  instead  of  the  %r  of  the  preceding. 

It  is  obvious  that  the  smaller  the  lever-arm  given  to  the 
resultant  friction  in  each  sector  of  the  rubbing  surface  the 
smaller  the  power  lost  in  friction.  Hence  pivots  should  be 
made  as  small  as  possible,  consistently  with  strength. 

For  a  conical  pivot  and  step,  Fig.  181,  the  resultant  friction 
in  each  sector  of  the  conical  bearing  surface  has 
a  lever-arm  =  -fr,  about  the  axis  A,  and  a  value 
>  than  for  a  flat-ended  pivot  ;  for,  on  account 
of  the  wedge-like  action  of  the  bodies,  the 
pressure  causing  friction  is  greater.  The  sum  of 
the  moments  of  these  resultant  frictions  about 
A  is  the  same  as  if  only  two  elements  of  the 
cone  received  pressure  (each  =  N  =  ^R  -r-  sin  a).  Hence  the 


FKICTION. 


181 


'moment  of  friction  of  the  pivot,  i.e.,  the  moment  of  the  force 
necessary  to  maintain  uniform  rotation,  is 


p*  = 


*  sin  a  3 
and  work  lost  per  revolution       =  ^ 


sin  a 


By  making  rl  small  enough,  these  values  may  be  made  less 
than  those  for  a  flat-ended  pivot  of  the  same  diameter  =  2r. 

In  Schiele's  "anti-friction"  pivots  the  outline  is  designed 
according  to  the  following  theory  for  securing  uniform  vertical 
wear.  Let  p  —  the  pressure  per 
horizontal  unit  of  area  (i.e., 
=  H  -r-  horizontal  projection  of 
the  actual  rubbing  surface)  ; 
this  is  assumed  constant.  Let 
the  unit  of  area  be  small,  for 
algebraic  simplicity.  The  fric-  FIO.  182. 

tion'on  the  rubbing  surface,  whose  horizontal  pro  jection  =  unity, 
is  •=.  fN  •=.  f(p  -r-  sin  a)  (see  Fig.  182;  the  horizontal  com- 
ponent of  p  is  annulled  by  a  corresponding  one  opposite).  The 
work  per  revolution  in  producing  wear  on  this  area  ~  fN%ny. 
But  the  vertical  depth  of  wear  per  revolution  is  to  be  the  same 
at  all  parts  of  the  surface  ;  and  this  implies  that  the  same 
volume  of  material  is  worn  away  under  each  horizontal  unit  of 

area.     Hence  /2T2  Try,  i.e.,/.        2  Try,  is  to  be  constant  for  all 

values  of  y  ;  and  since 
as  the  law  of  the  curve, 


and  2?r  are  constant,  we  must  have, 


y 


-,  i.e.,  the  tangent  BC  =  the  same  at  all  points. 


sin  OL 

This  curve  is  called  the  " tractrix"  Schiele's  pivots  give  a 
very  uniform  wear  at  high  speeds.  The  smoothness  of  wear 
prevents  leakage  in  the  case  of  cocks  and  faucets. 

169.  Normal  Pressure  of  Belting. — When  a  perfectly  flexible 
cord,  or  belt,  is  stretched  over  a  smooth  cylinder,  both  at  rest, 


182 


MECHANICS    OF   ENGINEERING. 


the  action  between  them  is  normal  at  every  point.     As  to  its 
•\\]    s    amount, p,  per  linear  unit  of  arc,  the  fol- 
~~"**'  lowing  will  determine.     Consider  a  semi- 
circle of  the  cord  free,  neglecting  its  weight. 
Fig.  183.      The  forces  holding  it  in  equilib- 
rium are  the  tensions  at  the  two  ends  (these 
are  equal,  manifestly,  the   cylinder    being 
smooth ;  for  they  are  the  only  two  forces 
'/TT"1"   "*  having  moments  about  C,  and  each  has  the 
FIG.  183.  same  lever-arm),  and  the  normal  pressures, 

which  are  infinite  in  number,  but  have  an  intensity,  p,  per 
linear  unit,  which  must  be  constant  along  the  curve  since  S  is 
the  same  at  all  points.  The  normal  pressure  on  a  single  ele- 
ment, ds,  of  the  cord  is  =  pds,  and  its  X  component  = 
pds  cos  6  =:  prd6  cos  0.  Hence  2X=  0  gives 

1-jTT 

>  6d6  —  2#  =  0,  i.e.,  rp\        sin  6  =  28: 
'  *L-* 


--        =          or        =. 


(1) 


170.  Belt  on  Rough  Cylinder.     Impending  Slipping.— If  fric- 
tion is  possible  between  the  two  bodies,  the  tension  may  vary 
the  arc  of  contact,  so  that  p  also  varies,  and  consequently 


FIG.  184. 


the  friction  on  an  element  ds  being  =fpds  =f(S~r-  r)ds,  also 
varies-  If  slipping  is  impending,  the  law  of  variation  of  the 
tension  8  may  be  found,  as  follows :  Fig.  184,  in  which  the 


FRICTION.  183 

impending  slipping  is  toward  the  left,  shows  the  cord  free. 
For  any  element,  ds,  of  the  cord,  we  have,  putting  2  (moms. 
about  O)  =  0  (Fig.  185), 

(S+  dS)r  =  Sr  +  dFr  ;  i.e.,  dF=  dS, 

or  (see  above)  dS  =f(S  -f-  r)ds. 

But  ds  =  rdO  ;  hence,  after  transforming, 


(i) 


In  (1)  the  two  variables  0  and  /S  are  separated  ;  (1)  is  there- 
fore ready  for  integration. 


.     (2) 

Or,  by  inversion,  S,ef°-  —  8W    .......     (3) 

e,  denoting  the  Naperian  base,  =  2.71828  -{-;  «  of  course  is  in 
TT-measure. 

Since  Sn  evidently  increases  very  rapidly  as  a  becomes 
larger,  S0  remaining  the  same,  we  have  the  explanation  of  the 
well-known  fact  that  a  comparatively  small  tension,  $0,  exerted 
by  a  man,  is  able  to  prevent  the  slipping  of  a  rope  around  a 
pile-head,  when  the  further  end  is  under  the  great  tension  Sn 
due  to  the  stopping  of  a  moving  steamer.  For  example,  with 
f  =  £,  we  have  (Weisbach) 

for  a  —  \  turn,    or  a  —  \n,  Sn  =  1.69#0  ; 
=  i  turn,    or  a  =  TC,    8n  =  2.85#0  ; 
=  1  turn,    or  a  =  ZTT,  Sn  —  8.12#0  ; 
=  2  turns,  or  a  =  ITT,  Sn  =  65.94x90  ; 
=  4  turns,  or  a  =  STT,  Sn  =  4348.56x90. 

If  slipping  actually  occurs,  we  must  use  a  value  of  /"for  fric- 
tion of  motion. 

Example.  —  A  leather  belt  drives  an  iron  pulley,  covering 
one  half  the  circumference.     What  is  the  limiting  value  of  the 


184 


MECHANICS   OF  ENGINEERING. 


ratio  of  /8n  (tension  on  driving-side)  to  #0  (tension  on  follow- 
ing side)  if  the  belt  is  not  to  slip,  taking  the  low  value  of 
f  •=.  0.25  for  leather  on  iron  ? 

We  have  given  fa.  =  0.25  X  X  =  .7854,  which  by  eq.  (2)  is 
the  Naperian  log.  of  (Sn  :  S9)  when  slipping  occurs.  Hence  the 
common  log.  of  (Sn :  £.)  =  0.7854  X  0.43429  =  0.34109  ;  i.e., 
if 

(Si :  £0)  =  2.193,  say  2.2, 
the  belt  will  (barely)  slip  (for/=  0.25). 

(0.43429  is  the  modulus  of  the  common  system  of  loga* 
rithms,  and  =  1 :  2.30258.  See  example  in  §48.) 

At  very  high  speeds  the  relation^?  =  S  -f-  r  (in  §  169)  is  not 
strictly  true,  since  the  tensions  at  the  two  ends  of  an  element 
ds  are  partly  employed  in  furnishing  the  necessary  deviating 
force  to  keep  the  element  of  the  cord  in  its  circular  path,  the 
remainder  producing  normal  pressure. 

171.  Transmission  of  Power  by  Belting  or  Wire  Rope. — In  the 
simple  design  in  Fig.  186,  it  is  required  to  find  the  motive 
weight  G,  necessary  to  overcome  the  given  resistance  H  at  a 


4 


FIG.  186. 


uniform   velocity  =  -y,;   also   the   proper   stationary    tension 
weight  GQ  to  prevent  slipping  of  the  belt  on  its  pulleys,  and 
the  amount  of  power,  Z,  transmitted. 
In  other  words, 

\  **>  a>  r>  a»  r"  a  =  *  f  or  both 
I  «j5  and/  for  b 


Given  : 


d/  for  both  pulleys  ;  ( 

-n       •    j  .  {  L  ;  #,  to  furnish  L  ;  GQ  for  no  slip  ;  v  the  velocity 

'  (  of  G  ;  v'  that  of  belt  ;  and  the  tensions  in  belt. 


FRICTION. 


185 


Neglecting  axle-friction  and  the  rigidity  of  the  belting,  the 
power  transmitted  is  that  required  to  overcome  R  through  a 
distance  =  vl  every  second,  i.e., 

L  =  Rv* (1) 

Since  (if  the  belts  do  not  slip) 

a  :  r::v'  :  v,    and 


we  have 


v   =  —v, 


j 
and 


:/•,:: 


=  --  v.. 
a  r    1 


(2) 


Neglecting  the  mass  of  the  belt,  and  assuming  that  each  pul- 
ley revolves  on  a  gravity-axis,  we  obtain  the  following,  by  con- 
sidering the  free  bodies  in  Fig.  187 : 


CA  free) 


2  (moms.)  =  0  in  A  free  gives  Rr^  =  (Sn  —  Sofa  ; .    (3) 
2  (moms.)  =  0  in  B  free  gives  Gr  =  (Sn  —  S0)a ;  .    (4) 


r     a. 


whence  we  readily  find  ( 

Evidently  R  and  G  are  inversely  proportional  to  their  velo- 
cities vl  and  v  ;  see  (2).  This  ought  to  be  true,  since  in  Fig. 
186  G  is  the  only  working-force,  R  the  only  resistance,  and 
the  motions  are  uniform ;  hence  (from  eq.  (XVI.),  §  142) 

Gv  -  Rv,  =  0. 
2X  =  0,  for  B  and  truck  free,  gives 


while,  for  impending  slip, 


=  Sef". 


(6) 


186  MECHANICS   OF   ENGINEERING. 

By  elimination  between  (4),  (5),  and  (6),  we  obtain 

Ll 
-  1 


r     ef^Ll_L    ef'  +  l 
~  '    f'  -     -    '  '  ~^~^       ...     (7) 


L        ef« 
ana  Sn  =  -,  .  ^—  T  ..........    (8) 

Hence  G0  and  Sn  vary  directly  as  the  power  transmitted  and 
inversely  as  the  velocity  of  the  belt.  For  safety  G0  should  be 
made  >  the  above  value  in  (7)  ;  corresponding  values  of  the 
two  tensions  may  then  be  found  from  (5),  and  from  the  rela- 
tion (see  §  150) 

(Sn-S0)v'  =  Z,  .......    (60) 

These  new  values  of  the  tensions  will  be  found  to  satisfy  the 
condition  of  no  slip,  viz., 


For  leather  on  iron,  e^  —  2.2  (see  example  in  §  170),  as  a 
*  low  value.  The  belt  should  be  made  strong  enough  to  with- 
stand Sn  safely. 

As  the  belt  is  more  tightly  stretched,  and  hence  elongated, 
on  the  driving  than  on  the  following  side,  it  "creeps"  back- 
ward on  the  driving  and  forward  on  the  driven  pulley,  so  that 
the  former  moves  slightly  faster  than  the  latter.  The  loss  of 
work  due  to  this  cause  does  not  exceed  2  per  cent  with  ordinary 
belting  (Cotterill). 

In  the  foregoing  it  is  evident  that  the  sum  of  the  tensions  in 
the  two  sides  =  GQ,  i.e.,  is  the  same,  whether  the  power  is 
being  transmitted  or  not  ;  and  this  is  found  to  be  true,  both  in 
theory  and  by  experiment,  when  a  tension-weight  is  not  used, 
viz.,  when  an  initial  tension  8  is  produced  in  the  whole  belt 
before  transmitting  the  power,  then  after  turning  on  the  latter 
the  sum  of  the  two  tensions  (driving  and  following)  always 
=  2$,  since  one  side  elongates  as  much  as  the  other  contracts  ; 
it  being  understood  that  the  pulley-axles  preserve  a  constant 
distance  apart. 

172.  Rolling  Friction.  —  The  few  experiments  which  have 
been  made  to  determine  the  resistance  offered  by  a  level  road- 


FRICTION.  187 

way  to  the  uniform  motion  of  a  roller  or  wheel  rolling  upon  it 
corroborate  approximately  the  following  theory.  The  word 
friction  is  hardly  appropriate  in  this  connection  (except  when 
the  roadway  is  perfectly  elastic,  as  will  be  seen),  but  is  sanctioned 
by  usage. 

First,  let  the  roadway  or  track  be  compressible,  but  inelastic, 
G  the  weight  of  the  roller  and  its  load,  and  P  the  horizontal 
force  necessary  to  preserve  a  uniform  motion  ^ — "NT"? 
(both  of  translation  and  rotation).  The  track 
(or  roller  itself)  being  compressed  just  in 
front,  and  not  reacting  symmetrically  from 
behind,  its  resultant  pressure  against  the 
roller  is  not  at  0  vertically  under  the  centre, 
but  some  small  distance,  OD  =  J,  in  front.  (The  successive 
crushing  of  small  projecting  particles  has  the  same  effect.) 
Since  for  this  case  of  motion  the  forces  have  the  same  relations 
as  if  balanced  (see  §  124),  we  may  put  2  moms,  about  D  =  0, 

.-.Pr=  Gb',    or,     P  =  \Q.  •     .     .     .     (1) 

Coulomb  found  for 

Rollers  of  lignum-vitse  on  an  oak  track,  5  =  0.0189  inches; 
Rollers  of  elm  on  an  oak  track,  I  =  0.0320  inches. 

Weisbach's  experiments  give,  for  cast-iron  wheels  20  inches  in 
diameter  on  cast-iron  rails, 

I  =  0.0183  inches ; 
and  Rittinger,  for  the  same,  I  =  0.0193  inches. 

Pambour  gives,  for  iron  railroad  wheels  39.4  inches  in  diameter, 

I  =  0.0196  to 

0.0216  inches. 

According  to  the  foregoing  theory,  P,  the  "  rolling  friction" 
(seeeq.  (1)),  is  directly  proportional  to  G,  and  inversely  to  the 
radius,  if  b  is  constant.  The  experiments  of  General  Morin  and 
othersconfirm  this,  while  those  of  Dupuit,  Poiree.  and  Sauvage 
indicate  it  to  be  proportional  directly  to  G,  and  inversely  to  the 
square  root  of  the  radius. 


188  MECHANICS    OF   ENGINEERING. 

Although  b  is  a  distance  to  be  expressed  in  linear  units,  and 
not  an  abstract  number  like  the  /"and  f  for  sliding  and  axle- 
friction,  it  is  sometimes  called  a  "  coefficient  of  rolling  fric- 
tion." In  eq.  (1),  J  and  r  should  be  expressed  in  the  same 
unit. 

Of  course  if  P  is  applied  at  the  top  of  the  roller  its  lever- 
arm  about  D  is  2r  instead  of  r,  with  a  corresponding  change 
in  eq.  (1). 

With  ordinary  railroad  cars  the  resistance  due  to  axle  and 
rolling  frictions  combined  is  about  8  Ibs.  per  ton  of  weight  on 
a  level  track.  For  wagons  on  macadamized  roads  5  =  J  inch, 
but  on  soft  ground  from  2  to  3  inches. 

Secondly,  when  the  roadway  is  perfectly  elastic.  This  is 
chiefly  of  theoretic  interest,  since  at  first  sight  no  force  would 
be  considered  necessary  to  maintain  a  uniform  rolling  motion. 
But,  as  the  material  of  the  roadway  is  compressed  under  the 
roller  its  surface  is  first  elongated  and  then  recovers  its  former 
state ;  hence  some  rubbing  and  consequent  sliding  friction  must 


FIG.  189. 


occur.  Fig.  189  gives  an  exaggerated  view  of  the  circum- 
stances, P  being  the  horizontal  force  applied  at  the  centre 
necessary  to  maintain  a  uniform  motion.  The  roadway  (rub- 
ber for  instance)  is  heaped  up  both  in  front  and  behind  the 
roller,  0  being  the  point  of  greatest  pressure  arid  elongation 
of  the  surface.  The  forces  acting  are  G,  P,  the  normal 
pressures,  and  the  frictions  due  to  them,  and  must  form  a 
balanced  system.  Hence,  since  G  and  P,  and  also  the  normal 
pressures,  pass  through  (7,  the  resultant  of  the  frictions  must 
also  pass  through  C\  therefore  the  frictions,  or  tangential 
actions,  on  the  roller  must  be  some  forward  and  some  backward 


FRICTION.  189 


(and  not  all  in  one  direction,  as  seems  to  be  asserted  on  p.  260 
of  Cotterill's  Applied  Mechanics,  where  Professor  Reynolds' 
explanation  is  cited).  The  resultant  action  of  the  roadway 
upon  the  roller  acts,  then,  through  some  point  Z>,  a  distance 
OD  =  b  ahead  of  Oj  and  in  the  direction  J)C,  and  we  have  as 
before,  with  D  as  a  centre  of  moments, 


or     P  =  - 
r 


If  rolling  friction  is  encountered  above  as 
well  as  below  the  rollers,  Fig.  190,  the 
student  may  easily  prove,  by  considering 
three  separate  free  bodies,  that  for  uniform 
motion 


where  £  and  b1  are  the  respective  "  coefficients  of  rolling  fric- 
tion" for  the  upper  and  lower  contacts. 

Example  1.  —  If  it  is  found  that  a  train  of  cars  will  move 
uniformly  down  an  incline  of  1  in  200,  gravity  being  the  only 
working  force,  and  friction  (both  rolling  and  axle)  the  only 
resistance,  required  the  coefficient,  y,  of  axle-friction,  the 
diameter  of  all  the  wheels  being  2r  =  30  inches,  that  of  the 
journals  %a  =  3  inches,  taking  ~b  =  0.02  inch  for  the  rolling 
friction.  Let  us  use  equation  (XYI.)  (§  142),  noting  that  while 
the  train  moves  a  distance  s  measured  on  the  incline,  its  weight 

G  does  the  work  G  ^-^  s,  the  rolling  friction  -  G  (at*  the  axles) 

has  been  overcome  through  the  distance  s,  and  the  axle-friction 


(total)  through  the  (relative)  distance  -  s  in  the  journal  boxes; 
whence,  the  change  in  kinetic  energy  being  zero, 

15o«—!>—  £ 

Gs  cancels  out,  the"  ratios  l>  :  r  and  a  :  r  are  =  y^-g-  and  fa 
respectively  (being  ratios  or  abstract  numbers  they  have  the 

*  That  is,  the  ideal  resistance,  at  centre  of  axles  and  ||  to  the  incline,  equiv- 
alent to  actual  rolling  resistance. 


190    '  MECHANICS    OF   ENGINEERING. 

same  numerical  values,  whether  the  inch  or  foot  is  used),  anc| 
solving,  we  have 

f  =  0.05  —  0.0133  =  0.036. 

Example  2. — How  many  pounds  of  tractive  effort  per  ton 
of  load  would  the  train  in  Example  1  require  for  uniform  mo- 
tion on  a  level  track?  Ans.  10  Ibs. 

173.  Eailroad  Brakes.* — During  the  uniform  motion  of  a 
railroad  car  the  tangential  action  between  the  track  and  each 
wheel  is  small.  Thus,  in  Example  1,  just  cited,  if  ten  cars  of 
eight  wheels  each  make  up  the  train,  each  car  weighing  20  tons, 
the  backward  tangential  action  of  the  rails  upon  each  wheel  is 
only  25  Ibs.  When  the  brakes  are  applied  to  stop  the  train 
this  action  is  much  increased,  and  is  the  only  agency  by  which 
the  rails  can  retard  the  train,  directly  or  indirectly :  directly, 
when  the  pressure  of  the  brakes  is  so  great  as  to  prevent  the 
wheels  from  turning,  thereby  causing  them  to  "skid"  (i.e., 
slide)  on  the  rails ;  indirectly,  when  the  brake-pressure  is  of 
such  a  value  as  still  to  permit  perfect  rolling  of  the  wheel,  in 
which  case  the  rubbing  (and  heating)  occurs  between  the  brake 
and  wheel,  and  the  tangential  action  of  the  rail  has  a  value 
equal  to  or  less  than  the  friction  of  rest.  In  the  first  case, 
then  (skidding),  the  retarding  influence  of  the  rails  is  il\Q  fric- 
tion of  motion  between  rail  and  wheel ;  in  the  second,  a  force 
which  may  be  made  as  great  as  the  friction  of rest  between  rail 
and  wheel.  Hence,  aside  from  the  fact  that  skidding  produces 
objectionable  flat  places  on  the  wheel-tread,  the  brakes  are 
more  effective  if  so  applied  that  skidding  is  impending,  but 
not  actually  produced  ;  for  the  friction  of  rest  is  usually  greater 
than  that  of  actual  slipping  (§  160).  This  has  been  proved 
experimentally  in  England.  The  retarding  effect  of  axle  and 
rolling  friction  has  been  neglected  in  the  above  theory. 

Example  1. — A  twenty-ton  car  with  an  initial  velocity  of  80 
feet  per  second  (nearly  a  mile  a  minute)  is  to  be  stopped  on  a 
level  within  1000  feet ;  required  the  necessary  friction  on  each 
of  the  eight  wheels. 

Supposing  the  wheels  not  to  skid,  the  friction  will  occur 

*  See  foot-note  on  p.  168,  as  to  the  diminution  of  the  coefficient /with 
speed. 


FRICTION,  191 

between  the  brakes  and  wheels,  and  is  overcome  through  the 
(relative)  distance  1000  feet.  Eq.  (XYL),  §  142,  gives  (foot- 
Ib.-second  system) 

0-8^x1000  =  0-1 


from  which  F  (  —  friction  at  circumference  of  each  wheel) 
=  496  Ibs. 

Example  2.—  Suppose  skidding  to  be  impending  in  the  fore- 
going, and  the  coefficient  of  friction  of  rest  (i.e.,  impending 
slipping)  between  rail  and  wheel  to  be  f  =z  0.20.  In  what 
distance  will  the  car  be  stopped  ?  Am.  496  ft. 

Example  3.  —  Suppose  the  car  in  Example  1  to  be  on  an  up- 
grade of  60  feet  to  the  mile.  (In  applying  eq.  (XYI.)  here, 
the  weight  20  tons  will  enter  as  a  resistance.)  Ans.  439  Ibs. 

Example  4.  —  In  Example  3,  consider  all  four  resistances, 
viz.,  gravity,  rolling  friction,  and  brake  and  axle  frictions,  the 
distance  being  1000  ft.,  and  F  the  unknown  quantity. 

(Take  the  wheel  dimensions  of  p.  189.)          Ans.  414  Ibs0 

174.  Estimation  of  Engine  and  Machinery  Friction.  —  Accord- 
ing to  Professor  Cotterill,  2  convenient  way  of  estimating  the 
work  lost  in  friction  in  a  steam-engine  and  machinery  driven 
by  it  is  the  following  : 

Let  pm  =  mean  effective  steam-pressure  per  unit  of  area  of 
piston,  and  conceive  this  composed  of  three  por- 
tions, viz., 
p0  =  the  necessary  pressure  to  drive  the  engine  alone  un- 

loaded, at  the  proper  speed  ; 
p'm  =  pressure  necessary  to  overcome  the  resistance  caused 

by  the  useful  work  of  the  machines  ; 

cp'm  =  pressure  necessary  t,n  overcome  the  friction  of  the 
machinery,  and  that  of  the  engine  over  and  above* 
its  friction  when  unloaded.  This  is  about  15$  of 
p'm  (i.e.,  e  =  0.15),  except  in  large  engines,  and 
then  rather  less. 

That  is,  by  formula,  F  being  the  piston-area  and  I  the  length 
of  stroke,  the  work  per  stroke  is  thus  distributed  : 


*  Recent  experiments  (1888^  by  Prof.  Thurston  show  that  this  surplus 

engine-friction  is  practically  zero 


MECHANICS   OP  ENGINEERING. 

£>g  is  "  from  1  to  1^,  or  in  marine  engines  2  Ibs.  or  more  per 
square  inch." 

175.  Anomalies  in  Friction. — Experiment  has  shown  consid- 
erable deviation  under  certain  circumstances  from  the  laws  of 
friction,  as  stated  in  §157  for  sliding  friction.     At  pressures 
below  J  Ib.  per  sq.  inch  the  coefficient  f  increases  when  the 
pressure  decreases,  while  above  500  Ibs.  (Rennie,  with  iron  and 
steel)  it  increases  with  the  pressure.    With  high  velocities,  how- 
ever, above  10  ft.  per  second,  f  is  much  smaller  as  the  velocity 
increases  (Bochet,  1858);    (and  Galton,  1878). 

As  for  axle-friction,  experiments  instituted  by  the  Society  of 
Mechanical  Engineers  in  England  (see  the  London  Engineer 
for  March  7  and  21,  1884)  gave  values  for  f  less  than  -j-J-^ 
when  a  "  bath"  of  the  lubricant  was  employed.  These  values 
diminished  with  increase  of  pressure,  and  increased  with  the 
velocity  (see  below,  Hirn's  statement). 

Professor  Cotterill  says,  "  It  cannot  be  doubted  that  for 
values  of  pv  (see  §  166)  >  5000  the  coefficient  of  friction  of 
well-lubricated  bearings  of  good  construction  diminishes  with 
the  pressure,  and  may  be  much  less  than  the  value  at  low  speeds 
as  determined  by  Morin"  (p.  259  of  his  Applied  Mechanics). 

Professor  Thurston's  experiments  confirmed  those  of  Hirn  as 
to  the  following  relation  :  "The  friction  of  lubricated  surfaces 
is  nearly  proportional  to  the  square  root  of  the  area  and  pres- 
sure." Hirn  also  maintained  that,  "in  ordinary  machinery, 
friction  varies  as  the  square  root  of  the  velocity." 

176.  Rigidity  of  Ropes. — If  a  rope  or  wire  cable  passes  over 
a  pulley  or  sheave,  a  force  P  is  required  on  one  side  greater 
than  the  resistance  Q  on  the  other  for  uniform  motion,  aside 
from   axle-friction.     Since  in   a   given   time  both  P   and    Q 
describe  the  same  space  «,  if  P  is  >  Q,  then  Ps\s  >  Qs,  i.e., 
the  work  done  by  P  is  >  than  that  expended  upon   Q.     This 
is  because  some  of  the  work  Ps  has  been  expended  in  bending 
the  stiff  rope  or  cable,  and  in  overcoming  friction  between  the 
strands,  both  where  the  rope  passes  upon  and  where  it  leaves 


FRICTION. 


the  pulley.  "With  hemp  ropes,  Fig.  191,  the  material  being 
nearly  inelastic,  the  energy  spent  in  bending  it  on  at  D  is 
nearly  all  lost,  and  energy  must  also  be  spent  in  straightening 


Fio.  191. 

it  at  E\  but  with  a  wire  rope  or  cable  some  of  this  energy  is 
restored  by  the  elasticity  of  the  material.  The  energy  spent 
in  friction  or  rubbing  of  strands,  however,  is  lost  in  both  cases. 
The  figure  shows  geometrically  why  P  must  be  >  Q  for  a 
uniform  motion,  for  the  lever-arm,  #,  of  P  is  evidently  <  b 
that  of  Q.  If  axle-friction  is  also  considered,  we  must  have 


r  being  the  radius  of  the  journal. 

Experiments  with  cordage  have  been  made  by  Prony,  Cou- 
lomb, Eytelwein,  and  Weisbach,  with  considerable  variation  in 
the  results  and  formulae  proposed.  (See  Coxe's  translation  of 
vol.  i.,  Weisbach's  Mechanics.) 

With  pulleys  of  large  diameter  the  effect  of  rigidity  is  very 
slight.  For  instance,  Weisbach  gives  an  example  of  a  pulley 
five  feet  in  diameter,  with  which.  Q  being  =  1200  Ibs.,  P 
=  1219.  A  wire  rope  f  in.  in  diameter  was  used.  Of  this 
difference,  19  Ibs.,  only  5  Ibs.  was  due  to  rigidity,  the  remainder, 
14  Ibs.,  being  caused  by  axle-friction.  When  a  hemp-rope  1.6 
inches  in  diameter  was  substituted  for  the  wire  one,  P—Q=27 
Ibs.,  of  which  12  Ibs.  was  due  to  the  rigidity.  Hence  in  one 
case  the  loss  of  work  was  less  than  -J  of  \%.  in  the  other  abonr 
\%,  caused  by  the  rigidity.  For  very  small  sheaves  and  thick 
ropes  the  loss  is  probably  much  greater. 
13 


194  MECHANICS    OF   ENGINEERING. 

177,  Miscellaneous  Examples. — Example  1.  The  end  of  a 
shaft  12  inches  in  diameter  and  making  50  revolutions  per  min- 
ute exerts  against  its  bearing  an  axial  pressure  of  10  tons  and 
a  lateral  pressure  of  40  tons.  With  f  =ff  =  0.05,  required 
the  H.  P.  lost  in  friction.  Ans.  22.2  H.  F. 

Example  2. — A  leather  belt  passes  over  a  vertical  pulley, 
covering  half  its  circumference.  One  end  is  held  by  a  spring 
balance,  which  reads  10  Ibs.  while  the  other  end  sustains  a 
weight  of  20  Ibs.,  the  pulley  making  100  revolutions  per  min- 
ute. Required  the  coefficient  of  friction,  and  the  H.P.  spent 
in  overcoming  the  friction.  Also  suppose  the  pulley  turned 
in  the  other  direction,  the  weight  remaining  the  same.  The 
diameter  of  the  pulley  is  18  inches.  ,  (f  =  0.22  ; 

?'  (0.142  and  .284  H.  P. 

Example  3. — A  grindstone  with  a  radius  of  gyration  =  12" 
inches  has  been  revolving  at  120  revolutions  per  minute,  and 
at  a  given  instant  is  left  to  the  influence  of  gravity  and  axle 
friction.  The  axles  are  1-J-  inches  in  diameter,  and  the  wheel 
makes  160  revolutions  in  coming  to  rest.  Required  the  coeffi- 
cient of  axle-friction.  Ans.  c/'=  0.389. 

Example  4. — A  board  A,  weight  2  Ibs.,  rests  horizontally  on 
another  B\  coefficient  of  friction  of  rest  between  them  being 
f  =  0.30.  B  is  now  moved  horizontally  with  a  uniformly 
accelerated  motion,  the  acceleration  being  =  15  feet  per  "  square 
second ;"  will  A  keep  company  with  it,  or  not  \  Ans. 


PAKT    III. 


STRENGTH  OF  MATERIALS. 

[OR  MECHANICS  OF  MATERIALS]. 
CHAPTEE  I. 

ELEMENTARY   STRESSES  AND  STRAINS. 

178.  Deformation  of  Solid  Bodies. — In  the  preceding  por- 
tions of  this  work,  what  was  called  technically  a  "  rigid 
body,"  was  supposed  incapable  of  changing  its  form,  i.e., 
the  positions  of  its  particles  relatively  to  each  other,  under 
the  action  of  any  forces  to  be  brought  upon  it.  This  sup- 
position was  made  because  the  change  of  form  which  must 
actually  occur  does  not  appreciably  alter  the  distances, 
angles,  etc.,  measured  in  any  one  body,  among  most  of 
the  pieces  of  a  properly  designed  structure  or  machine. 
To  show  how  the  individual  pieces  of  such  constructions 
should  be  designed  to,  avoid  undesirable  deformation  or 
injury  is  the  object  of  this  division  of  Mechanics  of  En- 
gineering, viz.,  the  Strength  of  Materials. 


D 
FIG.  192.    §  178. 


As  perhaps  tue  simplest  instance  of  the  deformation  or 
distortion  of  a  solid,  let  us  consider  the  case  of  a  prismatic 
rod  in  a  state  of  tension,  Fig.  192  (link  of  a  surveyor's 


196  MECHANICS  OF   ENGINEERING. 

chain,  e.g.).  The  pull  at  each  end  is  P,  and  the  body  is 
said  to  be  under  a  tension  of  P  (Ibs.,  tons,  or  other  unit), 
not  2P.  Let  ABGD  be  the  end  view  of  an  elementary 
parallelepiped,  originally  of  square  section  and  with  faces 
at  45°  with  the  axis  of  the  prism.  It  is  now  deformed,  the 
four  faces  perpendicular  to  the  paper  being  longer*  than 
before,  while  the  angles  BAD  and  BCD,  originally  right 
angles,  are  now  smaller  by  a  certain  amount  d,  ABC  and 
ADC  larger  by  an  equal  amount  d.  The  element  is  said 
to  be  in  a  state  of  strain,  viz.:  the  elongation  of  its  edges 
(parallel  to  paper)  is  called  a  tensile  strain,  while  the  alter- 
ation in  the  angles  between  its  faces  is  called  a  shearing 
strain,  or  angular  distortion  (sometimes  also  called  a  slid- 
ing, or  tangential,  strain,  since  BG  has  been  made  to  slide, 
relatively  to  AD,  and  thereby  caused  the  change  of  angle). 
[This  use  of  the  word  strain,  to  signify  change  of  form  and 
not  the  force  producing  it,  is  of  recent  adoption  among 
many,  though  not  all,  technical  writers.] 

179.  Strains.  Two  Kinds  Only. — Just  as  a  curved  line  may 
be  considered  to  be  made  up  of  small  straight-line  ele- 
ments, so  the  substance  of  any  solid  body  may  be  consid- 
ered to  be  made  up  of  small  contiguous  parallelepipeds, 
whose  angles  are  each  90°  before  the  body  is  subjected  to 
the  action  of  forces,  but  which  are  not  necessarily  cubes. 
A  line  of  such  elements  forming  an  elementary  prism  is 
spmetimes  called  a,f,bre,  but  this  does  not  necessarily  imply 
a  fibrous  nature  in  the  material  in  question.  The  system 
of  imaginary  cutting  surfaces  by  which  the  body  is  thus 
subdivided  need  not  consist  entirely  of  planes  ;  in  the  sub- 
ject of  Torsion,  for  instance,  the  parallelopipedical  ele- 
ments considered  lie  in  concentric  cylindrical  shells,  cut 
both  by  transverse  and  radial  planes. 

Since  these  elements  are  taken  so  small  that  the  only 
possible  changes  of  form  in  any  one  of  them,  as  induced 
by  a  system  of  external  forces  acting  on  the  body,  are 

*  When  a  is  nearly  0°  (or  90°)  BG  and  AD  (or  AB  and  DC)  are  shorter 
than  before,  on  account  of  lateral  contraction;  see  §  193. 


ELEMENTARY   STRESSES,    ETC.  197 

elongations  or  contractions  of  its  edges,  and  alteration  of 
its  angles,  there  are  but  two  kinds  of  strain,  elongation 
(contraction,  if  negative)  and  shearing. 

180.  Distributed  Forces  or  Stresses.  —  In  the  matter  preced- 
ing this  chapter  it  has  sufficed  for  practical  purposes  to 
consider  a  force  as  applied  at  a  point  of  a  body,  but  in 
reality  it  must  be  distributed  over  a  definite  area  ;  for 
otherwise  the  material  would  be  subjected  to  an  infinite 
force  per  unit  of  area.  (Forces  like  gravity,  magnetic  at- 
traction, etc.,  we  have  already  treated  as  distributed  over 
the  mass  of  a  body,  but  reference  is  now  had  particularly 
to  the  pressure  of  one  body  against  another,  or  the  action 
of  one  portion  of  the  body  on  the  remainder.)  For  in- 
stance, sufficient  surface  must  be  provided  between  the 
end  of  a  loaded  beam  and  the  pier  on  which  it  rests  to 
avoid  injury  to  either.  Again,  too  small  a  wire  must  not 
be  used  to  sustain  a  given  load,  or  the  tension  per  unit 
of  area  of  its  cross  section  becomes  sufficient  to  rupture 
it. 

Stress  is  distributed  force,  and  its  intensity  at  any  point 
of  the  area  is 

•    .    .        a) 


where  dF  is  a  small  area  containing  the  point  and  dP  the 
force  coming  upon  that  area.  If  equal  dP's  (all  parallel) 
act  on  equal  dF'soi  a  plane  surface,  the  stress  is  said  to 
be  of  uniform  intensity,  which  is  then 


(2) 


where  P=-  total  force  and  F  the  total  area  over  which  it 
acts.  The  steam  pressure  on  a  piston  is  an  example  of 
stress  of  uniform  intensity, 


198  MECHANICS   OF   ENGINEERING. 

For  example,  if  a  force  P==  28800  Ibs,  is  uniformly  dis- 
tributed over  a  plane  area  of  F=  72  sq.  inches,  or  ^  of  a 
sq.  foot,  the  intensity  of  the  stress  is 


72 

(or  p  =  28800-^=57600  Ibs.  per  sq.  foot,  or  ^=14400-:' 
2^=28.8  tons  per  sq.  ft.,  etc.). 

181.  Stresses  on  an  Element;  of  Two  Kinds  Only.  —  When  a 
solid  body  of  any  material  is  in  equilibrium  under  a  sys- 
tem of  forces  which  do  not  rupture  it,  not  only  is  its  shape 
altered  (i.e.  its  elements  are  strained),  and  stresses  pro- 
duced on  those  planes  on  which  the  forces  act,  but  other 
stresses  also  are  induced  on  some  or  all  internal  surfaces 
which  separate  element  from  element,  (over  and  above  the 
forces  with  which  the  elements  may  have  acted  on  each 
other  before  the  application  of  the  external  stresses  or 
"  applied  forces  ").  So  long  as  the  whole  solid  is  the  "free 
body  "  under  consideration,  these  internal  stresses,  being 
the  forces  with  which  the  portion  on  one  side  of  an  imag- 
inary cutting  plane  acts  on  the  portion  on  the  other  side, 
do  not  appear  in  any  equation  of  equilibrium  (for  if  intro- 
duced they  would  cancel  out);  but  if  we  consider  free  a 
portion  only,  some  or  all  of  whose  bounding  surfaces  are 
cutting  planes  of  the  original  body,  the  stresses  existing 
on  these  planes  are  brought  into  the  equations  of  equilib- 
rium. 

Similarly,  if  a  single  element  of  the  body  is  treated  by 
itself,  the  stresses  on  all  six  of  its  faces,  together  with  its 
weight,  form  a  balanced  system  of  forces,  the  body  being 
supposed  at  rest. 

i 
!K 


FIG.  193. 


ELEMENTARY   STRESSES,    ETC.  199 

As  an  example  of  internal  stress,  consider  again  the  case 
of  a  rod  in  tension  ;  Fig.  193  shows  the  whole  rod  (or  eye- 
bar)  free,  the  forces  P  being  the  pressures  of  the  pins  in 
the  eyes,  and  causing  external  stress  (compression  here) 
on  the  surfaces  of  contact.  Conceive  a  right  section  made 
through  US,  far  enough  from  the  eye,  (7,  that  we  may  con- 
sider the  internal  stress  to  be  uniform  *  in  this  section,  and 
consider  the  portion  MSG  as  a  free  body,  in  Fig.  194.  The 
stresses  on  US,  now  one  of  the  bounding  surfaces  of  the 
free  body,  must  be  parallel  to  P,  i.e.,  normal  to  US; 
(otherwise  they  would  have  components  perpendicular  to 
P,  which  is  precluded  by  the  necessity  of  2Y  being  =  0, 
and  the  supposition  of  uniformity.)  Let  F  =  the  sec- 


FIG.  194. 


PIG.  195. 


tional  area  US,  and  p  =  the  stress  per  unit  of  area ;  then 

IX=  0  gives  P=  Fp,  i.e.,  p=  ?  .        .  (2) 

F 

The  state  of  internal  stress,  then,  is  such  that  on  planes 
perpendicular  to  the  axis  of  the  bar  the  stress  is  tensile  and 
normal  (to  those  planes).  Since  if  a  section  were  made 
oblique  to  the  axis  of  the  bar,  the  stress  would  still  be 
parallel  to  the  axis  for  reasons  as  above,  it  is  evident  that 
on  an  oblique  section,  the  stress  has  components  both  nor- 
mal and  tangential  to  the  section,  the  normal  component 
being  a  tension. 

*  As  will  be  shown  later  (§  295)  the  Hue  of  the  two  P's  in  Fig.  193  must 
pass  through  the  centre  of  gravity  of  the  cross-section  RS  (plane  figure)  of 
the  bar.  for  the  stress  to  be  uniform  over  the  section. 


200 


MECHANICS   OF   ENGINEERING. 


The  presence  of  the  tangential  or  shearing  stress  in  ob- 
lique sections  is  rendered  evident  by  considering  that  if  an 
oblique  dove-tail  joint  were  cut  in  the  rod,  Fig.  195,  the 
shearing  stress  on  its  surfaces  may  be  sufficient  to  over- 
come friction  and  cause  sliding  along  the  oblique  plane. 

If  a  short  prismatic  block  is  under  the  com pressive^ ac- 
tion of  two  forces,  each  =  P  and  applied  centrally  in  one~ 
base,  we  may  show  that  the  state  of  internal  stress  is  the 
same  as  that  of  the  rod  under  tension,  except  that  the  nor- 
mal stresses  are  of  contrary  sign,  i.e.,  compressive  instead 
of  tensile,  and  that  the  shearing  stresses  (or  tendency  to 
slide)  on  oblique  planes  are  opposite  in  direction  to  those 
in  the  rod. 

Since  the  resultant  stress  on  a  given  internal  plane  of  a 
body  is  fully  represented,  by  its  normal  and  tangential 
components,  we  are  therefore  justified  in  considering  but 
two  kinds  of  internal  stress,  normal  or  direct,  and  tang  en- 
tial  or  shearing. 

182.  Stress  on  Oblique  Section  of  Rod  in  Tension. — Consider 
free  a  small  cubic  element  whose 
edge  =a  in  length;  it  has  two 
faces  parallel  to  the  paper,  being 
taken  near  the  middle  of  the  rod 
in  Fig.  192.  Let  the  angle  which 
the  face  AB,  Fig.  196,  makes  with 
the  axis  of  the  rod  be  =  a.  This 
angle,  for  our  present  purpose,  is 
considered  to  remain  the  same 
while  the  two  forces  P  are  acting, 
as  before  their  action.  The  re- 
sultant stress  on  the  face  AB  hav- 
ing an  intensity  p=P-±F,  (see  eq. 
2)  per  unit  of  transverse  section 
of  rod,  is  =  p  (a  sin  a)  a.  Hence 
its  component  normal  to  AB  is 
pa2  sin2  a ;  and  the  tangential  or  shearing  component  along 


PIG.  197. 


ELEMENTARY    STRESSES,    ETC.  201 

sin  a  cos  a.      Dividing  by  the  area,  a2,  we  have 
the  following : 

For  a  rod  in  simple  tension  we  have,  on  a  plane  making 
an  angle,  a,  with  the  axis  : 

a  Normal  Stress  =p  sin2  a  per  unit  of  area     .         .     (1) 
and  a  Shearing  Stress  =p  sin  a  cos  a  per  unit  of  area    .     (2) 

"  Unit  of  area  "  here  refers  to  the  oblique  plane  in  ques- 
tion, while  p  denotes  the  normal  stress  per  unit  of  area  of 
a  transverse  section,  i.e.,  when  «=90°,  Fig.  194. 

The  stresses  on  CD  are  the  same  in  value  as  on  AB9 
while  for  BG  and  AD  we  substitute  90° — a  for  «.  Fig. 
197  shows  these  normal  and  shearing  stresses,  and  also, 
much  exaggerated,  the  strains  or  change  of  form  of  the 
element  (see  Fig.  192). 

182a.  Relation  between  Stress  and  Strain. — Experiment 
shows  that  so  long  as  the  stresses  are  of  such  moderate 
value  that  the  piece  recovers  its  original  form  completely 
when  the  external  forces  which  induce  the  stresses  are  re- 
moved, the  following  is  true  and  is  known  as  Hooke's  Law 
(stress  proportional  to  strain).  As  the  forces  P  in  Fig. 
193  (rod  in  tension)  are  gradually  increased,  the  elonga- 
tion, or  additional  length,  of  RK  increases  in  the  same 
ratio  as  the  normal  stress,  p,  on  the  sections  US  and  KN^ 
per  unit  of  area  [§  191]. 

As  for  the  distorting  effect  of  shearing  stresses,  consider 
in  Fig.  197  that  since 

p  sin  a  cos  a  =  p  cos  (90° — «)  sin  (90° — a) 

the  shearing  stress  per  unit  of  area  is  of  equal  value  on  aU 
four  of  the  faces  (perpendicular  to  paper)  in  the  elementary 
block,  and  is  evidently  accountable  for  the  shearing  strain, 
i.e.,  for  the  angular  distortion,  or  difference,  3,  between 
90°  and  the  present  value  of  each  of  the  four  angles.  Ac- 
cording to  Hooke's  Law  then,  as  P  increases  within 
limit  mentioned  above,  3  varies  proportionally  to 

p  sin  a  cos  a,  i.e.  to  the  stress. 


202  MECHANICS   OF  EXGINEEKING. 

182b.  Example. — Supposing  the  rod  in  question  were  of 
a  kind  of  wood  in  which  a  shearing  stress  of  200  Ibs.  per 
sq.  inch  along  the  grain,  or  a  normal  stress  of  400  Ibs.  per 
sq.  inch,  perpendicular  to  a  fibre-plane  will  produce  rup- 
ture, required  the  value  of  a  the  angle  which  the  grain 
must  make  with  the  axis  that,  as  P  increases,  the  danger 
of  rupture  from  each  source  may  be  the  same.  This  re- 
quires that  200:400::jp  sin  a  cos  a:p  sin2#,  i.e.  tan.  a  must 
«=2.000.-.«=63^°.  If  the  cross  section  of  the  rod  is  2  sq. 
inches,  the  force  P  at  each  end  necessary  to  produce  rup- 
ture of  either  kind,  when  «=63^°,  is  found  by  putting 
p  sin  a  cos  a=  200. \p  =500.0  Ibs.  per  sq.  inch.  Whence,  since 
p=P+-F,  P=1000  Ibs.  (Units,  inch  and  pound.) 

183.  Elasticity  is  the  name  given  to  the  property  which 
most  materials  have,  to  a  certain  extent,  of  regaining  their 
original  form  when  the  external  forces  are  removed.  If 
the  state  of  stress  exceeds  a  certain  stage,  called  the  Elastic 
Limit,  the  recovery  of  original  form  on  the  part  of  the  ele- 
ments is  only  partial,  the  permanent  deformation  being 
called  the  Set. 

Although  theoretically  the  elastic  limit  is  a  perfectly  defi- 
nite stage  of  stress,  experimentally  it  is  somewhat  indefi- 
nite, and  is  generally  considered  to  be  reached  when  the 
permanent  set  becomes  well  marked  as  the  stresses  are  in- 
creased and  the  test  piece  is  given  ample  time  for  recovery 
in  the  intervals  of  rest. 

The  Safe  Limit  of  stress,  taken  well  within  the  elastio 
limit,  determines  the  working  strength  or  safe  load  of  the 
piece  under  consideration.  E.g.,  the  tables  of  safe  loads 
of  the  rolled  wrought  iron  beams,  for  floors,  of  the  New 
Jersey  Steel  and  Iron  Co.,  at  Trenton,  are  computed  on 
the  theory  that  the  greatest  normal  stress  (tension  or  com- 
pression) occurring  on  any  internal  plane  shall  not  exceed 
12,000  Ibs.  per  sq.  inch ;  nor  the  greatest  shearing  stress 
4,000  Ibs.  per  sq.  inch.  (See  §270a  on  p.  319.) 


ELEMENTARY   STRESSES,    ETC.  203 

The  tntimate  Limit  is  reached  when  rupture  occurs. 

184,  The  Modulus  of  Elasticity  (sometimes  called  co-efficient 
of  elasticity)  is  the  number  obtained  by  dividing  the  stress 
per  unit  of  area  by  the  corresponding  relative  strain. 

Thus,  a  rod  of  wrought  iron  y2  sq.  inch  sectional  area 
being  subjected  to  a  tension  of  2^  tons  =5,000  Ibs.,  it  is 
found  that  a  length  which  was  six  feet  before  tension  is 
=  6.002  ft.  during  tension.  The  relative  longitudinal  strain 
or  elongation  is  then=  (0.002)4-  6=1  :  3,000  and  the  corres- 
ponding stress  (being  the  normal  stress  on  a  transverse 
plane)  has  an  intensity  of 

pt=P~F=  5,000-^-  ^=10,000  Ibs.,  per  sq.  inch. 

Hence  by  definition  the  modulus  of  elasticity  is  (for  ten- 
sion) 

-r  3-^=30,000,000^8.  per  sq.  inch,  (the 


3 

sub-script  "t"  refers  to  tension). 

It  will  be  noticed  that  since  e  is  an  abstract  number,  Et 
is  of  the  same  quality  as  pt,  i.e.,  Ibs.  per  sq.  inch,  or  one  di- 
mension of  force  divided  by  two  dimensions  of  length. 
(In  the  subject  of  strength  of  materials  the  inch  is  the 
most  convenient  English  linear  unit,  when  the  pound  is 
the  unit  of  force  ;  sometimes  the  foot  and  ton  are  used  to- 
gether.) 

The  foregoing  would  be  called  the  modulus  of  elasticity 
of  ivrought  iron  in  tension  in  the  direction  of  the  fibre,  as 
given  by  the  experiment  quoted.  But  by  Hooke's  Law  p 
and  e  vary  together,  for  a  given  direction  in  a  given  ma- 
terial, hence  within  the  elastic  limit  E  is  constant  for  a  given 
direction  in  a  given  material.  Experiment  confirms  this 
approximately. 

Similarly,  the  modulus  of  elasticity  for  compression  E(, 


204  MECHANICS   OF   ENGINEERING. 

t 

in  a  given  direction  in  a  given  material  may  be  determined 
by  experiments  on  short  blocks,  or  on  rods  confined  lat- 
erally to  prevent  flexure. 

As  to  the  modulus  of  elasticity  for  shearing,  E^  we 
divide  the  shearing  stress  per  unit  of  area  in  the  given 
direction  by  d  (in  TT  measure)  the  corresponding  angular 
.strain  or  distortion  ;  e  g.,  for  an  angular  distortion  of  0.10° 
or  d  =  .00174,  and  a  shearing  stress  of  15,660  Ibs.  per  sq. 
inch,  we  have  E  =  ^  =  9,000,000  Ibs.  per  sq.  inch. 

Unless  otherwise  specified,  by  modulus  of  elasticity  will 
be  meant  a  value  derived  from  experiments  conducted 
within  the  elastic  limit,  and  this,  whether  for  normal  stress 
or  for  shearing,  is  approximately  constant  for  a  given  di- 
rection in  a  given  substance.* 

185.  Isotropes. — This  name  is  given  to  materials  which 
are  homogenous  as  regards  their  elastic  properties.     In 
such  a  material  the  moduli  of  elasticity  are  individually 
the  same  for  all  directions.     E.g.,  a  rod  of  rubber  cut  out 
of  a  large  mass  will  exhibit  the  same  elastic  behavior  when 
subjected  to  tension,  whatever  its  original  position  in  the 
mass.     Fibrous  materials  like  wood  and  wrought  iron  are 
not  isotropic  ;  the  direction  of  grain  in  the  former   must 
always  be  considered.     The  "  piling  "  and  welding  of  nu- 
merous small  pieces  of  iron  prevent  the  resultant  forging 
from  being  isotropic. 

186.  Resilience  refers  to  the  potential  energy  stored  in  a 
body  held  under  external  forces  in  a  state  of  stress  which 
does  not  pass  the  elastic  limit.     On  its  release  from  con- 
straint, by  virtue  of  its  elasticity  it  can  perform  a  certain 
amount  of  work  called  the  resilience,  depending  in  amount 
upon  the  circumstances  of  each  case  and  the  nature  of  the 
material.     See  §  148. 

187.  General  Properties  of  Materials. — In  view  of  some  defi- 
nitions already  made  we  may  say  that  a  material  is  ductile 

*  The  moduli,  or  "co-efficients,"  of  elasticity  as  used  by  physicists  are  well  explained 
in  Stewart  and  Gee's  Practical  Physics,  Vol.  I.,  pp.  164,  etc.  Their  "co-efficient  of 
rigidity"  is  our  Es. 


ELEMENTARY   STRESSES,    ETC.  205 

when  the  ultimate  limit  is  far  removed  from  the  elastic 
limit ;  that  it  is  brittle  like  glass  and  cast  iron,  when  those 
limits  are  near  together.  A  small  modulus  of  elasticity 
means  that  a  comparatively  small  force  is  necessary  to 
produce  a  given  change  of  form,  and  vice  versa,  but  implies 
little  or  nothing  concerning  the  stress  or  strain  at  the 
elastic  limit ;  thus  Weisbach  gives  E^  Ibs.  per  sq.  inch  for 
wrought  iron  =  28,000,000=  double  the  E<.  for  cast  iron 
while  the  compressive  stresses  at  the  elastic  limit  are  the 
same  for  both  materials  (nearly). 

188.  General  Problem  of  Internal  Stress. — This,  as  treated 
in  the  mathematical  Theory  of  Elasticity,  developed  by 
Lame,  Clapeyron  and  Poisson,  may  be  stated  as  follows : 

Given  the  original  form  of  a  body  when  free  from  stress, 
and  certain  co-efficients  depending  on  its  elastic  proper- 
ties ;  required  the  new  position,  the  altered  shape,  and  the  in- 
tensity of  the  stress  on  each  of  the  six  faces,  of  every  element 
of  the  body,  when  a  given  balanced  system  of  forces  is  applied 
to  the  body. 

Solutions,  by  this  theory,  of  certain  problems  of  the  na- 
ture just  given  involve  elaborate,  intricate,  and  bulky 
analysis ;  but  for  practical  purposes  Navier's  theories 
(1838)  and  others  of  more  recent  date,  are  sufficiently  exact, 
when  their  moduli  are  properly  determined  by  experiments 
covering  a  wide  range  of  cases  and  materials.  These  will 
be  given  in  the  present  work,  and  are  comparatively  sim- 
ple. In  some  cases  graphic  will  be  preferred  to  analytic 
methods  as  more  simple  and  direct,  and  indeed  for  some 
problems  they  are  the  only  methods  yet  discovered  for  ob- 
taining solutions.  Again,  experiment  is  relied  on  almost 
exclusively  in  dealing  with  bodies  of  certain  forms  under 
peculiar  systems  of  forces,  empirical  formulae  being  based 
on  the  experiments  made ;  e.g.,  the  collapsing  of  boilei 
flues,  and  in  some  degree  the  flexure  of  long  columns. 


206  MECHANICS   OF   ENGINEERING. 

189.  Classification  of  Cases, — Although  in  almost  any  case 
whatever  of  the  deformation  of  a  solid  body  by  a  balanced 
system  of  forces  acting  on  it,  normal  and  shearing  stresses 
are  both  developed  in  every  element  which  is  affected  at 
all  (according  to  the  plane  section  considered,)  still,  cases 
where  the  body  is  prismatic,  and  the  external  system  con- 
sists of  two  equal  and  opposite  forces,  one  at  each  end  of 
the  piece  and  directed  away  from  each  other,  are  commonly 
called  cases  of  Tension;  (Fig.  192);  if  the  piece  is  a  short 
prism  with  the  same  two  terminal  forces  directed  toward 
each  other,  the  case  is  said  to  be  one  of  Compression ;  a  case 
similar  to  the  last,  but  where  the  prism  is   quite  long 
("  long  column  "),  is  a  case  of  Flexure  or  bending,  as  are  also 
most  cases  where  the  "  applied  forces  "  (i.e.,  the  external 
forces),  are  not  directed  along  the  axis  of  the  piece.   Rivet- 
ed joints  and  "  pin-connections  "  present  cases  of  Shearing; 
a  twisted  shaft  one  of  Torsion.     When  the  gravity  forces 
due  to  the  weights  of  the  elements  are  also  considered,  a 
combination  of  two  or  more  of  the  foregoing  general  cases 
may  occur. 

In  each  case,  as  treated,  the  principal  objects  aimed  at 
are,  so  to  design  the  piece  or  its  loading  that  the  greatest 
stress,  in  whatever  element  it  may  occur,  shall  not  exceed 
a  safe  value  ;  and  sometimes,  furthermore,  to  prevent  too 
great  deformation  on  the  part  of  the  piece.  The  first  ob- 
ject is  to  provide  sufficient  strength;  the  second  sufficient 
stiffness. 

190,  Temperature  Stresses. — If  a  piece  is  under  such  con- 
straint that  it  is  not  free  to  change  its  form  with  changes 
of  temperature,  external  forces  are  induced,  the  stresses 
produced  by  which  axe  called  temperature  stresses. 


TENSION. 


207 


TENSION. 

191.  Hooke's  Law  by  Experiment. — As  a  typical  experiment 
in  the  tension  of  a  long  rod  of  ductile  metal  such  as 
wrought  iron  and  the  mild  steels,  the  following  table  is  quot- 
ed from  Prof.  Cotterill's  "  Applied  Mechanics."  The  experi- 
ment is  old,  made  by  Hodgkinson  for  an  English  Kailway 
Commission,  but  well  adapted  to  the  purpose.  From  the 
great  length  of  the  rod,  which  was  of  wrought  iron  and 
0.517  in.  in  diameter,  the  portion  whose  elongation  was 
observed  being  49  ft.  2  in.  long,  the  small  increase  in  length 
below  the  elastic  limit  was  readily  measured.  The  succes- 
sive loads  were  of  such  a  value  that  the  tensile  stress 
p=P~T-F,  or  normal  stress  per  sq.  in.  in  the  transverse 
section,  was  made  to  increase  by  equal  increments  of  2657.5 
Ibs.  per  sq.  in.,  its  initial  value.  After  each  application  of 
load  the  elongation  was  measured,  and  after  the  removal 
of  the  load,  the  permanent  set,  if  any. 

Table  of  elongations  of  a  wrought  iron  rod,  of  a  lengtt=49  ft.  2  in. 


p 

; 

Al 

t^M 

/ 

Load,  (Ibs.  per 
square  inch.) 

Elongation, 
(inches.) 

Increment 
of 
Elongation. 

«,  the  relative 
elongation,  (ab- 
stract number.) 

Permanent 
Set, 
(inches.) 

1X2667.5 

.0485 

.0485 

0.000082 

2X    " 

.1095 

.061 

.000186 

8X    " 

.1675 

.058 

.000283 

0.0019 

4X    " 

.224 

.0565 

.000379 

.002 

6X    ** 

.2805 

.0565 

.000475 

.00* 

0X    " 

.887 

.0565 

.000570 

.003 

TX    - 

.393 

.056 

.004 

8X    " 

.462 

.059 

.000706 

.0078 

9X    M 

.5155 

.0635 

.0195 

MX    " 

.598 

0825 

.049 

11X    " 

.760 

162 

.1545 

1«X    " 

1.810 

-550 

.667 

etc. 

208 


MECHANICS   OP   ENGINEERING. 


Referring  now  to  Fig.  198,  the  notation  is  evident.  P 
is  the  total  load  in  any  experiment,  F  the  cross  section  of 
the  rod  ;  hence  the  normal  stress  on  the  transverse  section 
is  p=P+F.  When  the  loads  are  increased  by  equal  in- 
crements, the  corresponding  increments  of  the  elongation 
J  should  also  be  equal  if  Hooke's  law  is  true.  It  will  be 
noticed  in  the  table  that  this  is  very  nearly  true  up  to  the 
8th  loading,  i.e.,  that  AX,  the  difference  between  two  con- 
secutive values  of  ^,  is  nearly  constant.  In  other  words  the 
proposition  holds  good  : 


if  P  and  Pi  are  any  two  loads  below  the  8th,  and  >l  and  Ji 
the  corresponding  elongations. 

The  permanent  set  is  just  perceptible  at  the  3d  load,  and 
increases  rapidly  after  the  8th,  as  also  the  increment  of 
elongation.  Hence  at  the  8th  load,  which  produces  a  ten- 
sile stress  on  the  cross  section  of  p=  8x2667.5=  21340.0 
Ibs.  per  sq.  inch,  the  elastic  limit  is  reached. 

As  to  the  state  of  stress  of  the  individual  elements,  if 

we  conceive  such  sub -division 
of  the  rod  that  four  edges  of 
each  element  are  parallel  to  the 
axis  of  the  rod,  we  find  that  it 
is  in  equilibrium  between  two 
normal  stresses  on  its  end  faces 
(Fig.  199)  of  a  value  =pdF= 
(P-^F)dF  where  dF  is  the  hor- 
izontal section  of  the  element. 
If  dx  was  the  original  length, 
and  dA  the  elongation  produced  by  pdFt  we  shall  have, 
since  all  the  cfo's  of  the  length  are  equally  elongated  at  the 
same  time, 


FIG.  19& 


^ 

dx    T 


TENSION.  209 

where  1=  total  (original)  length.  But  dX-r-dx  is  the  rela- 
tive elongation  e,  and  by  definition  (§  184)  the  Modulus  of 
Elasticity  for  Tension,  E^ 


a) 


Eq.  (1)  enables  us  to  solve  problems  involving  the  elonga- 
tion of  a  prism  under  tension,  so  long  as  the  elastic  limit 
is  not  surpassed. 

The  values  of  E^  computed  from  experiments  like  those 
just  cited  should  be  the  same  for  any  load  under  the  elas- 
tic limit,  if  Hooke's  law  were  accurately  obeyed,  but  in 
reality  they  differ  somewhat,  especially  if  the  material 
lacks  homogeneity.  In  the  present  instance  (see  Table) 
we  have  from  the 

2d  Exper.  ^=^-4-6=28,680,000  Ibs.  per  sq.  in. 

5th     "        #t=  "    =28,009,000 

8th    "        Et=  "    =27,848,000        «          " 

If  similar  computations  were  made  beyond  the  elastic 
limit,  i.e.,  beyond  the  8th  Exper.,  the  result  would  be  much 
smaller,  showing  the  material  to  be  yielding  much  more 
readily. 

192,  Strain  Diagrams. — If  we  plot  the  stresses  per  sq.  inch 
(p)  as  ordinates  of  a  curve,  and  the  corresponding  relative 
elongations  (e)  as  abscissas,  we  obtain  a  useful  graphic  re- 
presentation of  the  results  of  experiment. 

Thus,  the  table  of  experiments  just  cited  being  utilized 
in  this  way,  we  obtain  on  paper  a  series  of  points  through 
which  a  smooth  curve  may  be  drawn,  viz. :  OBC  Fig.  200, 
foi  wrought  iron.  Any  convenient  scales  may  be  used  for 
p  and  e ;  and  experiments  having  been  made  on  other 
metals  in  tension  and  the  results  plotted  to  the  samt>  scales 


210 


MECHANICS  OF   ENGINEERING. 


as  before  for  p  and  e,  we  have  the  means  of  comparing  their 
tensile  properties.  Fig.  200  shows  two  other  curves,  rep- 
resenting (roughly)  the  average  behavior  of  steel  and  cast 
iron.  At  the  respective  elastic  limits  J5,  B •',  and  B",  it  will 
be  noticed  that  the  curve  for  wrought  iron  makes  a  sudden 
turn  from  the  vertical,  while  those  of  the  others  curve  away 
more  gradually ;  that  the  curve  for  steel  lies  nearer  the 
vertical  axis  than  the  others,  which  indicates  a  higher 
value  for  Et ;  and  that  the  ordinates  BA't  B'A't  and  B'  A  * 
(respectively  21,000,  9,000,  and  30,000  Ibs.  per  sq.  inch)  in- 


Pie.  200. 

dicate  the  tensile  stress  at  the  elastic  limit.  These  latter 
quantities  will  be  called  the  moduli  of  tenacity  at  elastic 
limit  for  the  respective  materials.  [On  a  true  scale  the 
point  C  would  be  much  further  to  the  right  than  here 
shown.  Only  one  half  of  the  curve  for  steel  is  given,  for 
want  of  space.] 

Within  the  elastic  limit  the  curves  are  nearly  straight 
(proving  Hooke's  law)  and  if  a,  a',  and  a"  are  the  angles 
made  by  these  straight  portions  with  the  axis  of  X  (i.e., 
of  e),  we  shall  have 

(Et  for  w.  iron) :  ( J5?t  o.  iron) :  '(E,  steel) : :  tan  a :  tan  a' :  tan  «" 


TENSION.  213 

as  a  graphic  relation  between  their  moduli  of  elasticity 
/since 


e 


Beyond  the  elastic  limit  HLQ  wrought  iron  rod  shows  large 
increments  of  elongation  for  small  increments  of  stress, 
i.e.,  the  curve  becomes  nearly  parallel  to  the  horizontal 
axis,  until  rupture  occurs  at  a  stress  of  53,000  Ibs.  per  sq. 
inch  of  original  sectional  area  (afc  rupture  this  area  is  some- 
what reduced,  especially  in  the  immediate  neighborhood 
of  the  section  of  rupture  ;  see  next  article)  and  after  a  rel- 
ative elongation  e=  about  0.30,  or  30%.  (The  preceding 
table  shows  only  a  portion  of  the  results.)  The  curve 
for  steel  shows  a  much  higher  breaking  stress  (100,000 
Ibs.  per  sq.  in.)  than  the  wrought  iron,  but  the  total 
elongation  is  smaller,  e=-  about  10%.  This  is  an  average 
curve ;  tool  steels  give  an  elongation  at  rupture  of  about 
4  to  5%,  while  soft  steels  resemble  wrought  iron  in  their 
ductility,  giving  an  extreme  elongation  of  from  10  to  20%. 
Their  breaking  stresses  range  from  70,000  to  150,000  Ibs. 
or  more  per  sq.  inch.  Cast  iront  being  comparatively  brit- 
tle, reaches  at  rupture  an  elongation  of  only  3  or  4  tenths 
of  one  per  cent.,  the  rupturing  stress  being  about  18,000 
Ibs.  per  sq.  inch.  The  elastic  limit  is  rather  ill  defined  in 
the  case  of  this  metal ;  and  the  proportion  of  carbon  and 
the  mode  of  manufacture  have  much  influence  on  its  be- 
havior under  test. 

193.  Lateral  Contraction. — In  the  stretching  of  prisms  of 
nearly  all  kinds  of  material,  accompanying  the  elongation 
of  length  is  found  also  a  diminution  of  width  whose  rela- 
tive amount  in  the  case  of  the  three  metals  just  treated  ia 
about  y£  or  ^  of  the  relative  elongation  (within  elastic 
limit).  Thus,  in  the  third  experiment  in  the  table  of  §  191, 
this  relative  lateral  contraction  or  decrease  of  diameter 
*=  Y^  to  y±  of  e,  i.e.,  about  0.00008.  In  the  case  of  cast 
iron  and  hard  steels  contraction  is  not  noticeable  ex- 


212  MECHANICS  OF  ENGINEERING 

oept  by  very  delicate  measurements,  both  within  and  with- 
out the  elastic  limit ;  but  the  more  ductile  metals,  as 
wrought  iron  and  the  soft  steels,  when  stretched  beyond 
the  elastic  limit  show  this  feature  of  their  deformation 
in  a  very  marked  degree.  Fig.  201  shows  by  dotted  lines 
the  original  contour  of  a  wrought  iron  rod,  while  the  con- 
tinuous lines  indicate  that  at  rupture.  At  the  cross  section 
of  rupture,  whose  position  is  determined  by  some 
local  weakness,  the  drawing  out  is  peculiarly 
pronounced. 

The  contraction  of  area  thus  produced  is  some- 
times as  great  as  50  or  60%  at  the  fracture. 

194.  "Flow  of  Solids." — When  the  change  in  re- 
lative position  of  the  elements  of  a  solid  is  ex- 
treme, as  occurs  in  the  making  of  lead  pipe, 
drawing  of  wire,  the  stretching  of  a  rod  of  duc- 
tile metal  as  in  the  preceding  article,  we  have 
Pie.  201.     instances  of  what  is  called  the  Flow  of  Solids,  in- 
teresting  experiments    on    which    have    been    made   by 
Tresca, 

195.  Moduli  of  Tenacity. — The  tensile  stress  per  square 
inch  (of  original  sectional  area)  required  to  rupture  a 
prism  of  a  given  material  will  be  denoted  by  T  and  called 
the  modulus  of  ultimate  tenacity  ;  similarly,  the  modulus  of 
safe  tenacity,  or  greatest  safe  tensile  stress  on  an  element, 
by  T' ;  while  the  tensile  stress  at  elastic  limit  may  be 
called  T".  The  ratio  of  T  to  T  '  is  not  fixed  in  practice 
but  depends  upon  circumstances  (from  ^  to  %). 

Hence,  if  a  prism  of  any  material  sustains  a  total  pull 
or  load  P,  and  has  a  sectional  area  =  F,  we  have 

P=FT  for  the  ultimate  or  breaking  load.  \ 

P'=FT'   "      "    safe  load.  >•     .    .     (2) 

P'=FT""      "     load  at  elastic  limit.  ) 

«* 

Of  course  T'  should  always  be  less  than  T". 


TENSION.  213 

196.  Resilience  of  a  Stretched  Prism. — Fig.  202.  In  the 
gradual  stretching  of  a  prism,  fixed  at  one  extremity,  the 
value  of  the  tensile  force  P  at  the  other  necessarily  de- 
pends on  the  elongation  ^  at  each  stage  of  the  lengthening, 
according  to  the  relation  [eq.  (1)  of  §  191.] 


within  the  elastic  limit.  (If  we  place  a  weight  G  on  the 
flanges  of  the  unstretched  prism  and  then  leave 
it  to  the  action  of  gravity  and  the  elastic  action 
of  the  prism,  the  weight  begins  to  sink,  meeting 
an  increasing  pressure  P,  proportional  to  A,  from 
the  flanges).  Suppose  the  stretching  to  continue 
until  P  reaches  some  value  P"  (at  elastic  limit 
|A  say),  and  A  a  value  A".  Then  the  work  done  so 
far  is 

27=  mean  force  X  space  =j£  P"  X"    .    .    (4) 

But  from  (2)  P  =  FT  ,  and  (see  §§  184  and  191) 


.-.  (4)  becomes  U=*/2  T  e".  Fl=tf  T"  s"  V     .    .      (5) 

where  V  is  the  volume  of  the  prism.  The  quantity  y%  T"e"9 
or  work  done  in  stretching  to  the  elastic  limit  a  cubic 
inch  (or  other  unit  of  volume)  of  the  given  material,  Weis- 
bach  calls  the  Modulus  of  Resilience  for  tension.  From  (5) 
it  appears  that  the  amounts  of  work  done  in  stretching  to 
the  elastic  limit  prisms  of  the  same  material  but  of  differ- 
ent dimensions  are  proportional  to  their  volumes  simply. 
The  quantity  ^T"e"  is  graphically  represented  by  the 
area  of  one  of  the  triangles  such  as  OAB,  OA"B"  in  Fig. 
200 ;  for  (in  the  curve  for  wrought  iron  for  instance)  tht^ 
modulus  of  tenacity  at  elastic  limit  is  represented  by  A'B, 
and  e"  (i.e.,  e  for  elastic  limit)  by  OA'.  The  remainder  of 


214  MECHANICS   OP  ENGINEERING. 

the  area  OBG  included  between  the  curve  and  the  hori- 
zontal axis,  i.e.,  from  B  to  (7,  represents  the  work  done  in 
stretching  a  cubic  unit  from  the  elastic  limit  to  the  point 
of  rupture,  for  each  vertical  strip  having  an  altitude  =p 
and  a  width  =  de,  has  an  area  =pde,  i.e.,  the  work  done  by 
the  stress  p  on  one  face  of  a  cubic  unit  through  the  dis- 
tance de,  or  increment  of  elongation. 

If  a  weight  or  load  =  G  be  "  suddenly  "applied  to  stretch 
the  prism,  i.e.,  placed  on  the  flanges,  barely  touching 
them,  and  then  allowed  to  fall,  when  it  comes  to  rest  again 
it  has  fallen  through  a  height  ^l?  and  experiences  at  this 
instant  some  pressure  P:  from  the  flanges;  PI=?  The 
work  GXl  has  been  entirely  expended  in  stretching  the 
prism,  none  in  changing  the  kinetic  energy  of  G,  which 
=0  at  both  beginning  and  end  of  the  distance  ^, 


Since  Pi=26r,  i.e.,  is  >  G,  the  weight  does  not  remain  in 
this  position  but  is  pulled  upward  by  the  elasticity  of  the 
prism.  In  fact,  the  motion  is  harmonic  (see  §§  59  and 
138).  Theoretically,  the  elastic  limit  not  being  passed,  the 
oscillations  should  continue  indefinitely. 

Hence  a  load  G  "  suddenly  applied  "  occasions  double  the 
tension  it  would  if  compelled  to  sink  gradually  by  a  sup- 
port underneath,  which  is  not  removed  until  the  tension  is 
just  =  G,  oscillation  being  thus  prevented. 

If  the  weight  G  sinks  through  a  height  =h  before  strik- 
ing the  flanges,  Fig.  202,  we  shall  have  similarly,  within 
slastic  limit,  if  ^=  greatest  elongation,  (the  mass  of  rod 
being  small  compared  with  that  of  G). 

G(h+*l)=}4P1ll        ....    (6) 

If  the  elastic  limit  is  to  be  just  reached  we  have  from  eqs. 
(5)  and  (6),  neglecting  ^  compared  with  h, 

"F         .        .        .        (7) 


TENSION. 


215 


an  equation  of  condition  that  the  prism  shall  not  be  in- 
jured. 

Example. — If  a  steel  prism  have  a  sectional  area  of  \£ 
gq.  inch  and  a  length  1=10  ft.  =120  inches,  what  is  the 
greatest  allowable  height  of  fall  of  a  weight  of  200  Ibs.; 
that  the  final  tensile  stress  induced  may  not  exceed  T"  — 
30,000  Ibs.  per  sq.  inch,  if  e"  =.002  ?  From  (7),  using  the 
inch  and  pound,  we  have 


•197,  Stretching  of  a  Prism  by  Its  Own  Weight. — In  the  case 
of  a  very  long  prism  such  as  a  mining- 
pump  rod,  its  weight  must  be  taken  into 
account  as  well  as  that  of  the  terminal 
load  P ,  see  Fig.  203.  At  (a.)  the  prism 
is  shown  in  its  unstrained  condition ;  at 
(b)  strained  by  the  load  Pl  and  its  own 
weight.  Let  the  cross  section  be  =F,  the 
heaviness  of  the  prism  =f.  Then  the  rela- 
tive extension  of  any  element  at  a  distance 


L 


- 


FIG.  203. 


x  from  o  is  * 


dx 


FEt 


(1) 


[See  eq.  (1)  §  191) ;  since  P^+Fyx  is  the  load  hanging  upon 
the  cross  section  at  that  locality.  Equal  dx'a,  therefore, 
are  unequally  elongated,  x  varying  from  0  to  I.  The  total 
elongation  is 


=~  Cl 
FEJo 


FEt 


. 


Gl 


Le.,  A=  the  amount  due  to  Plt  plus  an  extension  which 
half  the  weight  of  the  prism  would  produce,  hung  at  the 
lower  extremity. 

Pl 

*  in  A  = put  dX  for  A,  dx  for  I,  and  (Pl  +  yFx)  for  P. 


216  MECHANICS   OF   ENGINEERING. 

The  foregoing  relates  to  the  deformation  of  the  piece, 
and  is  therefore  a  problem  of  stiffness.  As  to  the  strength 
of  the  prism,  the  relative  elongation  e=dX-r-dx  [see  eq.  (1)], 
which  is  variable,  must  nowhere  exceed  a  safe  value  e'= 
T'+EL  (from  eq.  (1)  §  191,  putting  P=FT  ,  and  ;=/). 
Now  the  greatest  value  of  the  ratio  dX  :  dx,  by  inspecting 
eq.  (1),  is  seen  to  be  at  the  upper  end  where  x—l.  The 
proper  cross  section  F,  for  a  given  load  Plt  is  thus  found. 


Putting  ™  *ave  f  =  -        (2) 


198.  Solid  of  Uniform  Strength  in  Tension,  or  hanging  body 
of  minimum  material  supporting  its  own 
weight  and  a  terminal  load  PL.  Let  it  be  a 
solid  of  revolution.  If  every  cross-section 
F  at  a  distance  =x  from  the  lower  extrem- 
ity, bears  its  safe  load  FT',  every  element 
of  the  body  is  doing  full  duty,  and  its  form 
p,  is  the  most  economical  of  material. 

The   lowest  section  must  have  an  area 
».  204.         p^=p^rji^  sjnce  p^  js  its  safe  loado 

204.     Consider  any  horizontal  lamina ;  its  weight  is 

(f=  heaviness  of  the  material,  supposed  homogenous),  and 

its  lower  base  F  must  have  PI+  6r  for  its  safe  load,  i.e. 

in  which  G  denotes  the  weight  of  the  portion  of  the  solid 
below  F.     Similarly  for  the  upper  base  F+dF,  we  have 


.        .        (2) 
By  subtraction  we  obtain 


e. 


TENSION.  217 

in  which  the  two  variables  x  and  F  are  separated.     By  in> 
tegration  we  now  have 


(4) 


from  which  .F  may  be  computed  for  any  value  of  x. 

The  weight  of  the  portion  below  any  F  is  found  from  (1) 
and  (4)  ;  i.e. 


while  the  total  extension  A  will  be 

.......    (6) 


the  relative  elongation  dX-^dx  being  the  same  for  every  dx 
and  bearing  the  same  ratio  to  e"  (at  elastic  limit),  as  T' 
does  to  T". 

199.  Tensile  Stresses  Induced  by  Temperature.  —  If  the  two 
ends  of  a  prism  are  immovably  fixed,  when  under  no  strain 
and  at  a  temperature  t,  and  the  temperature  is  then  low- 
ered to  a  value  t',  the  body  suffers  a  tension  proportional 
to  the  fall  in  temperature  (within  elastic  limit).  If  for  a 
rise  or  fall  of  1°  Fahr.  (or  Cent.)  a  unit  of  length  of  the 
material  would  change  in  length  by  an  amount  YJ  (called 
the  co-efficient  of  expansion)  a  length  =1  would  be  con- 
tracted an  amount  X=rll(t-t'}  during  the  given  fall  of  tem- 
perature if  one  end  were  free.  Hence,  if  this  contraction 
is  prevented  by  fixing  both  ends,  the  rod  must  be  under  a 
tension  P,  equal  in  value  to  the  force  which  would  be 


218  MECHANICS   OF  ENGINEERING. 

necessary  to  produce  the  elongation  /,  just  stated,  under 
ordinary  circumstances  at  the  lower  temperature. 

From  eq.  (1)  §191,  therefore,  we  have  for  this  tension 
due  to  fall  of  temperature 


P= 


For  1°  Cent,  we  may  write 

For  Cast  iron  y  =  .0000111 ; 
"  Wrought  iron  =  .0000120 ; 
"  Steel  =  .0000108  to  .0000114 ; 

«     Copper          rj  =  .0000172 ; 
«    Zinc  7  =  .0000300. 



COMPRESSION    OF    SHORT    BLOCKS. 

200,  Short  and  Long  Columns. — In  a  prism  in  tension,  its 
own  weight  being  neglected,  all  the  elements  between  tht 
localities  of  application  of  the  pair  of  external  forces  pro- 
ducing the  stretching  are  in  the  same  state  of  stress,  if  the 
external  forces  act  axially  (excepting  the  few  elements  in  the 
immediate  neighborhood  of  the  forces;  these  suffering 
local  stresses  dependent  on  the  manner  of  application  of 
the  external  forces),  and  the  prism  may  be  of  any  length 
without  vitiating  this  statement.  But  if  the  two  external 
forces  are  directed  toward  each  other  the  intervening  ele- 
ments will  not  all  be  in  the  same  state  of  compressive 
stress  unless  the  prism  is  comparatively  short  (or  unless 
numerous  points  of  lateral  support  are  provided).  A  long 
prism  will  buckle  out  sideways,  thus  even  inducing  tensile 
stress,  in  some  cases,  in  the  elements  on  the  convex  side. 

Hence  the  distinction  between  short  blocks  and  long 
columns.  Under  compression  the  former  yield  by  crush- 
ing or  splitting,  while  the  latter  give  way  by  flexure  (i.e. 
bending).  Long  columns,  then  will  be  treated  separately 


COMPRESSION   OF    SHORT   BLOCKS.  219 

in  ft  subsequent  chapter.  In  the  present  section  the  blocks 
treated  being  about  three  or  four  times  as  long  as  wide, 
all  the  elements  will  be  considered  as  being  under  equal 
compressive  stresses  at  the  same  time. 

201.  Notation  for  Compression.  —  By  using  a  subscript  c, 
we  may  write 

Ec=  Modulus  of  Elasticity;*  i.e.  the  quotient  of  the 
compressive  stress  per  unit  of  area  divided  by  the  relative 
shortening  ;  also 

C=  Modulus  of  crushing  ;  i.e.  the  force  per  unit  of  sec- 
tional area  necessary  to  rupture  the  block  by  crushing  ; 

C'=  Modulus  of  safe  compression,  a  safe  compressive 
stress  per  unit  of  area  ;  and 

C"=  Modulus  of  compression  at  elastic  limit. 

For  the  absolute  and  relative  shortening  in  length  we 
may  still  use  A  and  e,  respectively,  and  within  the  elastic 
limit  may  write  equations  similar  to  those  for  tension,  F 
being  the  sectional  area  of  the  block  and  P  one  of  the  ter- 
minal forces,  while  p  =  compressive  stress  per  unit  of  area 
of  F,  viz.: 


within  the  elastic  limit. 
Also  for  a  short  block 

Crushing  force  =FC  \ 

Compressive  force  at  elastic  limit  =FC"     >   .     (21/ 
Safe  compressive  force  —FC' 

202.  Remarks  on  Crushing.  —  As  in  §  182  for  a  tensile 
stress,  so  for  a  compressive  stress  we  may  prove  th-at  a 

*[NOTE.  —  It  must  be  remembered  that  the  modulus  of  elasticity, 
whether  for  normal  or  shearing  stresses,  is  a  number  indicative  of  stiff- 
ness, not  of  strength,  and  has  no  relation  to  the  elastic  limit  (except 
that  experiments  to  determine  it  must  not  pass  that  limit).] 


220  MECHANICS   OF  ENGINEERING. 

shearing  stress  =p  sin  a  cos  a  is  produced  on  planes  at  an 
angle  a  with  the  axis  of  the  short  block,  p  being  the  com- 
pression per  unit  of  area  of  transverse  section.  Accord- 
ingly it-  is  found  that  short  blocks  of  many  comparatively 
brittle  materials  yield  by  shearing  on  planes  making  an 
angle  of  about  45°  with  the  axis,  the  expression  p  sin  a 
cos  a  reaching  a  maximum,  for  a=45°  ;  that  is,  wedge- 
shaped  pieces  are  forced  out  from  the  sides.  Hence  the 
necessity  of  making  the  block  three  or  four  times  as  long 
as  wide,  since  otherwise  the  friction  on  the  ends  would 
cause  the  piece  to  show  a  greater  resistance  by  hindering 
this  lateral  motion.  Crushing  by  splitting  into  pieces 
parallel  to  the  axis  sometimes  occurs. 

Blocks  of  ductile  material,  however,  yield  by  swelling 
out,  or  bulging,  laterally,  resembling  plastic  bodies  some- 
what in  this  respect. 

The  elastic  limit  is  more  difficult  to  locate  than  in  ten- 
sion, but  seems  to  have  a  position  corresponding  to  that 
in  tension,  in  the  case  of  wrought  iron  and  steel.  With 
cast  iron,  however,  the  relative  compression  at  elastic 
limit  is  about  double  the  relative  extension  (at  elastic 
limit  in  tension),  but  the  force  producing  it  is  also  double. 
For  all  three  metals  it  is  found  that  Ec=Et  quite  nearly, 
so  that  the  single  symbol  E  may  be  used  for  both. 


EXAMPLES  IN  TENSION  AND  COMPRESSION. 

203.  Tables  for  Tension  and  Compression.  —  The  round  num- 
bers of  the  following  tables  are  to  be  taken  as  rude  averages 
only,  for  use  in  the  numerical  examples  following.  (The 
scope  and  design  of  the  present  work  admit  of  nothing 
more.  For  abundant  detail  of  the  results  of  the  more  im- 
portant experiments  of  late  years,  the  student  is  referred 
to  the  recent  works*  of  Profs.  Thurston,  Burr,  Lanza,  and 
Wood).  Another  column  might  have  been  added  giving 
the  Modulus  of  Kesilience  in  each  case,  viz.:  y2  e"T" 

rp»2\ 

(which  also   =__  }  ;     see  §  196.     e  is  an  abstract  num- 
2x£  / 


*  Other  valuable  and  recent  works  (1897  and  later)  are  "Materials  of 
Construction,"  by  Prof.  J.  B.  Johnson,  Unwin's  "  Testing  of  Materials," 
and  Martens'  work  of  same  title. 


EXAMPLES  IN  TENSION  AND   COMPRESSION.         221 


ber,  and  =^-f-Z,  while  Elt  T",  and  T  are  given  in  pounds 
per  square  inch: 

TABLE   OF  THE   MODULI,    ETC.,    OF  MATERIALS   IN  TENSION, 


e" 

£ 

Et 

fT<ll 

T 

Material. 

(Elastic  limit.) 

At  Rupture. 

Mod.  of  Blast. 

Elastic  limit. 

Rupture. 

abst.  number. 

abst.  number. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in. 

bs.  per  sq.  in. 

Soft  Steel, 

.00200 

.2500 

26,000,000 

50,000 

80,000 

Hard  Steel, 

.00200 

.0500 

40,000,000 

90,000 

130,000 

Cast  Iron, 
Wro't  Iron, 

Brass, 

.00066 
.00080 

.00100 

.0020 
.2500 

14,000,000 
28,000,000 

10,000,000 

9,000 
22,000 

(   7,000 
\     to 
(19,000 

18,000 
(45,000 
\    to 
(60,000 
16,000 
to 
50,000 

Glass, 

9,000,000 

3,500 

Wood,    with 
the  fibres, 

I       .00200 
\         to 
(        .01100 

.0070 
to 
.0150 

200,000 
to 
2,000,000 

3,000 
to 
19,000 

6,000 
to 
28,000 

Hemp  rope, 

7,000 

[N.B.— Expressed  in  kilograms   per  square  centim.,  Et,  T  and  T"    would  be  nu 
merically  about  Vi4  as  large  as  above,  while  e  and  e"  would  be  unchanged.] 

TABLE  OF  MODULI,  ETC.;  COMPRESSION  OF  SHORT  BLOCKS. 


Material. 

e" 

Elastic  limit. 

e 

At  lupture. 

Mod.  of  Blast. 

C" 

Elastic  limit. 

C 
Rupture. 

abst.  number 

abst.  number. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in. 

Soft  Steel, 

0.00100 

30,000,000 

30,000 

Hard  Steel, 

0.00120 

0.3000 

40,000,000 

50,000' 

200,000 

Cast  Iron, 

0.00150 

14,000,000 

20,000 

90,000 

Wro't  Iron, 

0.00080 

0.3000 

28,000,000 

24,000 

40,000 

31ass, 

20,000 

Granite, 
Sandstone, 
Brick, 

(See 
§213a 

10,000 
5,000 
3,000 

Wood,  with 
the  fibres, 

(   0.0100 
•]      to 
0.0400 

350,000 
to 
2,000,000 

2,000 
to 
10,000 

Portland     1 
Cement,  f 

(§  213a)      ' 

4,000 

222  MECHANICS   OF   ENGINEERING. 

204,  Examples.  No.  1.  —  A  bar  of  tool  steel,  of  sectional 
area  =0.097  sq.  inches,  is  ruptured  by  a  tensile  force  of 
14,000  Ibs.  A  portion  of  its  length,  originally  y2  a  foot, 
is  now  found  to  have  a  length  of  0.532  ft.  Required  T, 
and  e  at  rupture.  Using  the  inch  and  pound  as  units  (as 
in  the  foregoing  tables)  we  have  r=li|?=  144326  Ibs.  per 
sq.  in.;  (eq.  (2)  §  195)  ;  while 

e=(0.532—  0.5)xl2-4-(0.50xl2)=0.064. 

EXAMPLE  2.  —  Tensile  test  of  a  bar  of  "  Hay  Steel  "  for 
the  Glasgow  Bridge,  Missouri.  The  portion  measured  was 
originally  3.21  ft.  long  and  2.09  in.  X  1.10  in.  in  section. 
At  the  elastic  limit  P  was  124,200  Ibs.,  and  the  elongation 
was  0.064  ins.  Required  Ett  T",  and  e"  (for  elastic  limit). 


t"=     =  =.00165  at  elastic  limit. 

i  O..21  X  12 

^"=124,200-^(2.09  x  1.10)=  54,000  Ibs.  per  sq.  in. 

lbs- 


Nearly  the  same  result  for  Et  would  probably  have  been 
obtained  for  values  of  p  and  e  below  the  elastic  limit. 

The  Modulus  of  Resilience  of  the  above  steel  (see  §  196) 
would  be  y2  e"  T"=  44.82  inch-pounds  of  work  per  cubic 
inch  of  metal,  so  that  the  whole  work  expended  in  stretch- 
ing to  the  elastic  limit  the  portion  above  cited  is 

U=  y2  e"  T"  7=3968.  inch  -lbs. 

An  equal  amount  of  work  will  be  done  by  the  rod  in  re- 
covering its  original  length. 

EXAMPLE  3.  —  A  hard  steel  rod  of  ^  sq.  in.  section  and 
20  ft.  long  is  under  no  stress  at  a  temperature  of  130" 


EXAMPLES   IN    TENSION   AND   COMPRESSION.        223 

Cent.,  and  is  provided  with  flanges  so  that  the  slightest 
tjontraction  of  length  will  tend  to  bring  two  walls  nearer 
together.  If  the  resistance  to  this  motion  is  10  tons  how 
low  must  the  temperature  fall  to  cause  any  motion  ?  ^  be- 
ing =.0000110  (Cent,  scale).  From  §  199  we  have,  ex- 
pressing P  in  Ibs.  and  F  in  sq.  inches,  since  Et=  40,000,000 
Ibs.  per  sq.  inch, 

10X2,000  =  40,000,000  X  #  X(130-f )  x  0.000011 ;  whence 
^=39.0°  Centigrade. 

EXAMPLE  4. — If  the  ends  of  an  iron  beam  bearing  5  tons 
at  its  middle  rest  upon  stone  piers,  required  the  necessary 
bearing  surface  at  each  pier,  putting  G1  for  stone  =200 
Ibs.  per  sq.  inch.  25  sq.  in.,  Ans. 

EXAMPLE  5. — How  long  must  a  wrought  iron  wire*  be, 
supported  vertically  at  its  upper  end,  to  break  with  ita 
own  weight  ?  216,000  inches,  Ans. 

EXAMPLE  6. — One  voussoir  (or  block)  of  an  arch -ring 
presses  its  neighbor  with  a  force  of  50  tons,  the  joint  hav- 
ing a  surface  of  5  sq.  feet ;  required  the  compression  per 
sq.  inch.  138.8  Ibs.  per  sq.  in.,  Ans. 

205.  Factor  of  Safety. — When,  as  in  the  case  of  stone,  the 
value  of  the  stress  at  the  elastic  limit  is  of  very  uncertain 
determination  by  experiment,  it  is  customary  to  refer  the 
value  of  the  safe  stress  to  that  of  the  ultimate  by  making 
it  the  w'th  portion  of  the  latter,  n  is  called  a  factor  of 
safety,  and  should  be  taken  large  enough  to  make  the  safe 
stress  come  within  the  elastic  limit.  For  stone,  n  should 
not  be  less  than  10,  i.e.  C'^C+n',  (see  Ex.  6,  just  given). 


206.  Practical  Notes. — It  was  discovered  independently  by 
'Commander  Beardslee  and  Prof.  Thurston,  in  1873,  that 
if  wrought  iron  rods  were  strained  considerably  beyond 
the  elastic  limit  and  allowed  to  remain  free  from  stress 

*  Take  T  =  60,000  Ibs.  per  square  inch. 


MECHANICS   OF   ENGINEERING. 

for  at  least  one  day  thereafter,  a  second  test  would  show 
higher  limits  both  elastic  and  ultimate. 

When  articles  of  cast  iron  are  imbedded  in  oxide  of  iron 
and  subjected  to  a  red  heat  for  some  days,  the  metal  loses 
most  of  its  carbon,  and  is  thus  nearly  converted  into 
wrought  iron,  lacking,  however,  the  property  of  welding. 
Being  malleable,  it  is  called  malleable  cast  iron. 

Chrome  steel  (iron  and  chromium)  and  tungsten  steel  pos- 
sess peculiar  hardness,  fitting  them  for  cutting  tools,  rock 
drills,  picks,  etc. 

By  fatigue  of  metals  we  understand  the  fact,  recently  dis- 
covered by  Wohler  in  experiments  made  for  the  Prussian 
Government,  that  rupture  may  be  produced  by  causing  the 
stress  on  the  elements  to  vary  repeatedly  between  two 
limiting  values,  the  highest  of  which  may  be  considerably 
below  T  (or  (7),  the  number  of  repetitions  necessary  to 
produce  rupture  being  dependent  both  on  the  range  of 
variation  and  the  higher  value. 

For  example,  in  the  case  of  Phoenix  iron  in  tension, 
rupture  was  produced  by  causing  the  stress  to  vary  from 
)  to  52,800  Ibs.  per  sq.  inch,  800  times ;  also,  from  0  tc 
14,000  Ibs.  per  sq.  inch  240,853  times  ;  while  4,000,000  va- 
riations between  26,400  and  48,400  per  sq.  inch  did  not 
cause  rupture.  Many  other  experiments  were  made  and 
the  following  conclusions  drawn  (among  others): 

Unlimited  repetitions  of  variations  of  stress  (Ibs.  per 
0q.  in.)  between  the  limits  given  below  will  not  injure  the 
metal  (Prof.  Burr's  Materials  of  Engineering). 

Wrought  iron    I  From  17>600  ComP'  to  17'6°°  Tension' 
1    (      «  0  to  33,000 

f  From  30,800  Comp.  to  30,800  Tension, 
Axle  Cast  Steel  J      "  0  to  52,800 

(     "       38500  Tens,    to  88,000        " 
(See  p,  832  for  an  addendum  to  this  paragraph.) 


SHEARING. 

SHEARING. 


225 


207.  Eivets, — The  angular  distortion  called  shearing 
strain  in  the  elements  of  a  body,  is  specially  to  be  provided 
for  in  the  case  of  rivets  joining  two  or  more  plates.  This 
distortion  is  shown,  in  Figs.  205  and  206,  in  the  elements 
near  "he  plane  of  contact  of  the  plates,  much  exaggerated* 


-— lT« 


FIG.  205. 


FIG.  206. 


In  Fig.  205  (a  lap-joint)  the  rivet  is  said  to  be  in  single 
shear ;  in  Fig.  206  in  double  shear.  If  P  is  just  great 
enough  to  shear  off  the  rivet,  the  modulus  of  ultimate  shear- 
ing, which  may  be  called  S,  (being  the  shearing  force  per 
unit  of  section  when  rupture  occurs)  is 


(1) 


in  which  F=  the  cross  section  of  the  rivet,  its  diameter 
being  =d.  For  safety  a  value  S'=  %  to  ^  of  S  should 
be  taken  for  metal,  in  order  to  be  within  the  elastic  limit. 

As  the  width  of  the  plate  is  diminished  by  the  rivet 
hole  the  remaining  sectional  area  of  the  plate  should  be 
ample  to  sustain  the  tension  P,  or  2P,  (according  to  the 
plate  considered,  see  Fig.  206),  P  being  the  safe  shearing 
force  for  the  rivet.  Also  the  thickness  t  of  the  plate 
should  be  such  that  the  side  of  the  hole  shall  be  secure^ 
against  crushing ;  P  must  not  be  >  C'td,  Fig.  205. 

Again,  the  distance  a,  Fig.  205,  should  be  such  as  to 
prevent  the  tearing  or  shearing  out  of  the  part  of  the 
plate  between  the  rivet  and  edge  of  the  plate. 


226  MECHANICS  OF   ENGINEERING. 

For  economy  of  material  the  seam  or  joint  should  be 
no  more  liable  to  rupture  by  one  than  by  another,  of  the 


FIG.  207. 

four  modes  just  mentioned.  The  relations  which  must 
then  subsist  will  be  illustrated  in  the  case  of  the  "  butt- 
joint  "  with  two  cover-plates,  Fig.  207.  Let  the  dimen- 
sions be  denoted  as  in  the  figure  and  the  total  tensile  force 
on  the  joint  be  =  Q.  Each  rivet  (see  also  Fig.  206)  is  ex- 
posed in  each  of  two  of  its  sections  to  a  shear  of  /2  Q, 
hence  for  safety  against  shearing  of  rivets  we  put 


(1) 


Along  one  row  of  rivets  in  the  main  plate  the  sectional 
area  for  resisting  tension  is  reduced  to  (b  —  3d)tlt  hence  for 
safety  against  rupture  of  that  plate  by  the  tension  Q,  we 
put 


(2) 


Equations  (1)  and  (2)  suffice  to  determine  d  for  the  rivets 
and  #!  for  the  main  plates,  Q  and  b  being  given;  but  the 
values  thus  obtained  should  also  be  examined  with  refer- 
ence to  the  compression  in  the  side  of  the  rivet  hole,  i.e., 
y6  Q  must  not  be  >  C'tLd.  [The  distance  a,  Fig.  205,  to  the 
edge  of  the  plate  is  recommended  by  different  authorities 
to  be  from  d  to  3d.] 

Similarly,  for  the  cover  -plate  we  must  have 


.        .        .    (3) 
< 

and  }2Q  not  >  C'td. 


SFTEAKING.  227 

If  the  rivets  do  not  fit  their  holes  closely,  a  large  margin 
should  be  allowed  in  practice.  Again,  in  boiler  work,  the 
pitch,  or  distance  between  centers  of  two  consecutive  rivets 
may  need  to  be  smaller,  to  make  the  joint  steam-tight,  than 
would  be  required  for  strength  alone. 

208.  Shearing  Distortion.  —  The  change  of  form  in  an  ele- 
ment due  to  shearing  is  an  angular  deformation  and  will 
be  measured  in  7r-measure.  This  angular  change  or  dif- 
ference between  the  value  of  the  corner  angle  during  strain 
and  I^TT,  its  value  before  strain,  will  be  called  3,  and  is 
proportional  (within  elastic  limit)  to  the  shearing  stress 
per  unit  of  area,  J9S,  existing  on  all  the  four  faces  whose 
angles  with  each  other  have  been  changed. 

Fig.  208.  (See  §  181).  By  §  184  the  Modulus  of  Shearing 
Elasticity  is  the  quotient  obtained  by  dividing  pK  by  d  ;  i.e. 
(elastic  limit  not  passed), 


or  inversely,  3=ps-trE&  .....      (1)' 

The  value  of  E&  for  different  substances  is  most  easily 
determined  by  experiments  on  torsion 
in  which  shearing  is  the  most  promi- 
nent stress.  (This  prominence  depends 
on  the  position  of  the  bounding  planes 
of  the  element  considered  ;  e.g.,  in  Fig. 
208,  if  another  element  were  considered 
within  the  one  there  shown  and  with 
PIG.  208.  its  planes  at  45°  with  those  of  the  first, 

we  should  find  tension  alone  on  one  pair  of  opposite  faces, 
compression  alone  on  the  other  pair.)  It  will  be  noticed 
that  shearing  stress  cannot  be  present  on  two  opposite 
faces  only,  but  exists  also  on  another  pair  of  faces  (those 
perpendicular  to  the  stress  on  the  first),  forming  a  couple 
of  equal  and  opposite  moment  to  the  first,  this  being 
necessary  for  the  equilibrium  of  the  element,  even  when 


228 


MECHANICS   OF   ENGINEERING. 


tensile  or  compressive  stresses  are  also  present  on  the 
faces  considered. 

209.  Shearing  Stress  is  Always  of  the  Same  Intensity  on  the 
Four  Faces  of  an  Element. — (By  intensity  is  meant  per  unit 
of  area ;  and  the  four  faces  referred  to  are  those  perpen- 
dicular to  the  paper  in  Fig.  208,  the  shearing  stress  being 
parallel  to  the  paper.) 

Let  dx  and  dz  be  the  width  and  height  of  the  element 
in  Fig.  208,  while  dy  is  its  thickness  perpendicular  to  the 
paper.  Let  the  intensity  of  the  shear  on  the  right  hand 
face  be  =q^  that  on  the  top  face  =ps.  Then  for  the  ele- 
ment aw  a  free  body,  taking  moments  about  the  axis  0  per- 
pendicular to  paper,  we  have 

gs  dz  dy  X  dx — pa  dx  dy  x  dz  =0  /.  gs  =ps 

{dx  and  dz  being  the  respective  lever  arms  of  the  forces 
gs  dz  dy  and  ps  dx  dy.) 

Even  if  there  were  also  tensions  (or  compressions)  on 
one  or  both  pairs  of  faces  their  moments  about  0  would 
balance  (or  fail  to  do  so  by  a  differential  of  a  higlier  order) 
independently  of  the  shears,  and  the  above  result  would 
still  hold. 

210.  Table  of  Moduli  for  Shearing. 


S" 

E. 

8" 

S 

Material. 

i.e.  6  at  elastic 
limit. 

Mod.  of  Elasticity 
for  Shearing. 

(Elastic  limit.) 

(Rupture.) 

arc  in  w-measure. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in 

Ibs.  per  sq.  in. 

Soft  Steel, 

9,000,000 

70,000 

Hard  Steel, 

0.0032 

14,000,000 

45,000 

90,000 

Cast  Iron, 

0.0021 

7,000,000 

15,000 

30,000 

Wrought  Iron, 

0.0022 

9,000,000 

20,000 

50,000 

Brass, 

5,000,000 

Glass, 

Wood,  across  J 
fibre,            1 

1,500 
to 
8,000 

Wood,  along  j 
fibre,             1 

500 
to 

:,soo 

SHEARING. 


229 


As  in  the  tables  for  tension  and  compression,  the  above 
values  are  averages.  The  true  values  may  differ  from 
these  as  much  as  30  per  cent,  in  particular  cases,  accord' 
ing  to  the  quality  of  the  specimen. 

211.  Punching  rivet  holes  in  plates  of  metal  requires  the 
overcoming  of  the  shearing  resistance  along  the.  convex 
surface  of  the  cylinder  punched  out.  Hence  if  d  =  diam- 
eter of  hole,  and  t  =  the  thickness  of  the  plate,  the  neces- 
sary force  for  the  punching,  the  surface  sheared  being 
F=  tnd,  is 


P=Stxd 


(2)' 


Another  example  of  shearing  action  is  the  "  stripping  " 
of  the  threads  of  a  screw,  when  the  nut  is  forced  off  lon- 
gitudinally without  turning,  and  resembles  punching  in 
its  nature. 

212.  EandEg;  Theoretical  Relation.  —  In»case  a  rod  is  in 
tension  within  the  elastic  limit,  the  relative  (linear)  lateral 
contraction  (let  this  =m)  is  so  connected  with  Ut  and  Et 
that  if  two  of  the  three  are  known  the  third  can  be  de- 
duced theoretically.  This  relation  is  proved  as  follows, 
by  Prof.  Burr.  Taking  an  elemental  cube  with  four  of  its 
faces  at  45°  with  the  axis  of  the  piece,  Fig.  209,  the  axial 
half-diagonal  AD  becomes  of  a  length  AD'=AD-\-e.AD 
under  stress,  while  the  transverse  half  diagonal  contracts 
to  a  length  B'D'=AD—m.AD.  The  angular  distortion  3 


FIG.  209.    §  212. 


>0 


FIG.  210. 


230  MECHANICS    OF   ENGINEERING. 

is  supposed  very  small  compared  with  90°  and  is  due  to 
the  shear  ps  per  unit  of  area  on  the  face  B  G  (or  BA). 
From  the  figure  we  have 


tan(45°—    )  =  ==1-^,  appro*. 


[But,  Fig.  210,  tan(45°  —  #)=!  —  2#  nearly,  where  x  is  a 
small  angle,  for,  taking  GA=  unity=  AE,  tan  AD=  AF= 
AE—EF.  _Now  approximately  EF=~E~G.^o>ndiEG^ 
£J)</2=x</2  .'.  AF=  I—2x  nearly.]  Hence 

1—  3=1—  m—  e;  or  S=m+e        .        .       (2) 

Eq.  (2)  holds  good  whatever  the  stresses  producing  the 
deformation,  but  in  the  present  case  of  a  rod  in  tension, 
if  it  is  an  isotrope,  and  if  p  =  tension  per  unit  of  area  on 
its  transverse  section,  (see  §  182,  putting  «=45°),  we  have 
Et=p+e  and  Es=(ps  on  BC)-r-3=  }4p+S.  Putting  also 
(w:e)=r,  whence  m=re,  eq.  (2)  may  finally  be  written 


•      0) 


Prof.  Bauschinger,  experimenting  with  cast  iron  rods, 
found  that  in  tension  the  ratio  m  :  e  was  =  {g,,  as  an  average, 
which  in  eq.  (3)  gives 


t=    ^nearly.       .        .        .     (4) 


His  experiments  on  the  torsion  of  cast  iron  rods  gave 
Es=  6,000,000  to  7,000,000  Ibs.  per  sq.  inch.  By  (4),  then, 
•Et  should  be  15,000,000  to  17,500,000  which  is  approxi- 
mately true  (§  203). 

Corresponding  results  may  be  obtained  for  short  blocks 
in  compression,  the  lateral  change  being  a  dilatation  in- 
stead of  a  contraction. 


SHBAKiNGk  231 

Examples  in  Shearing. — EXAMPLE  1. — Kequired  the 
proper  length,  a,  Fig.  211,  to 
guard  against  the  shearing  off, 
along  the  grain,  of  the  portion 
ab,  of  a  wooden  tie-rod,  the  force 
P  being  =  2  tons,  and  the  width 
of  the  tie  =  4  inches.  Using  a 
value  of  S1  =  100  Ibs.  per  sq.  in., 
we  put  fo/S^  4,000  cos  45° ;  i.e. 
a==(4,000x0.707)-f-(4x!00)=  7.07 
inches. 

EXAMPLE  2.— A  ^  in.  rivet  of  wrought  iron,  in  single 
shear  (see  Fig.  205)  has  an  ultimate  shearing  strength 
P==FS=}£ntfg=  i^7r(^)2x 50,000=  30,050 Ibs.  Forsafety, 
putting  #'=8,000  instead  of  S,P'=  4,800  Ibs.  is  its  safe 
shearing  strength  in  single  shear. 

The  wrought  iron  plate,  to  be  secure  against  the  side- 
crushing  in  the  hole,  should  have  a  thickness  ty  computed 
thus : 

P'=tdC'\  or  4,800=^x12,000  .-.  *=0.46  in. 

If  the  plate  were  only  0.23  in.  thick  the  safe  value  of  P 
would  be  only  y2  of  4,800. 

EXAMPLE  3. — Conversely,  given  a  lap-joint,  Fig.  205,  in 
which  the  plates  are  %  in.  thick  and  the  tensile  force  on 
the  joint  =  600  Ibs.  per  linear  inch  of  seam,  how  closely 
must  y^  inch  rivets  be  spaced  in  one  row,  putting  #'=8,000 
and  C' =12,000  Ibs.  per  sq.  in.  ?  Let  the  distance  between 
centres  of  rivets  be  =x  (in  inches),  then  the  force  upon 
each  rivet  =600x,  while  its  section  7^=0.44  sq.  in.  Having 
regard  to  the  shearing  strength  of  the  rivet  we  put  600cc= 
0.44x8,000  and  obtain  x= 5.86  in.;  but  considering  that  the 
safe  crushing  resistance  of  the  hole  is  =  %/ ^-12,000= 
2,250  Ibs.,  600x=2,250  gives  #=3.75  inches,  which  is  the 
pitch  to  be  adopted.  What  is  the  tensile  strength  of  the 
reduced  sectional  area  of  the  plate,  with  this  pitch  ? 


232  MECHANICS   OF   ENGINEERINGS. 

EXAMPLE  4. — Double  butt-joint ;  (see  Fig.  207) ;  3/8  inch 
plate;  %  in.  rivets;  Tf=  <7'= 12,000 ;  #'=8,333;  width  of 
plates=14  inches.  Will  one  row  of  rivets  be  sufficient  at 
each  side  of  joint,  if  Q=30,000  Ibs.?  The  number  of  rivets 
=  ?  Here  each  rivet  is  in  double  shear  and  has  therefore 
a  double  strength  as  regards  shear.  In  double  shear  the 
safe  strength  of  each  rivet  =2FS'=7,333  Ibs.  Now  30,000-- 
7,333=4.0  (say).  With  the  four  rivets  in  one  row  the  re- 
duced sectional  area  of  the  main  plate  is  =[14 — 4x  ^]  X3/8 
=4.12  sq.  in.,  whose  safe  tensile  strength  is  =.FT'=4.12x 
12,000=49,440  Ibs.;  which  is  >  30,000  Ibs.  .-.  main  plate  is 
safe  in  this  respect.  But  as  to  side-crushing  in  holes 
in  main  plate  we  find  that  C'^d  (i.e.  12,000  X  3/8  X  %=  3,375 
Ibs.)  is  <^Q  i.e.  <7,500  Ibs.,  the  actual  force  on  side  of 
hole.  Hence  four  rivets  in  one  row  are  too  few  unless 
thickness  of  main  plate  be  doubled.  Will  eight  in  one 
row  be  safe  ? 

213a.  (Addendum  to  §  206.)  Elasticity  of  Stone  and  Cements. 
— Experiments  by  Gen.  Gill  more  with  the  large  Watertown 
testing-machine  in  1883  resulted  as  follows  (see  p.  221  for 
notation) : 

With  cubes  of  Haverstraw  Freestone  (a  homogeneous  brown- 
stone)  from  1  in.  to  12  in.  on  the  edge,  Ec  was  found  to  be 
from  900,000  to  1,000,000  Ibs.  per  sq.  in.  approximately  ;  and 
C  about  4,000  or  5,000  Ibs.  per  sq.  in.  Cubes  of  the  same 
range  of  sizes  of  Dyckerman's  Portland  cement  gave  EG  from 
1,350,000  to  1,630,000,  and  C  from  4,000  to  7,000,  Ibs.  per  sq. 
in.  ,  Cubes  of  concrete  of  the  above  sizes,  made  with  the 
Newark  Ce.'s  Eosendale  cement,  gave  Ee  about  538,000,  while 
cubes  of  cement-mortar,  and  some  of  concrete,  both  made  with 
National  Portland  cement,  showed  E0  from  800,000  to  2,000,- 
000  Ibs.  per  sq.  in. 

The  compressibility  of  "brick  piers  12  in.  square  in  section 
and  16  in.  high  was  also  tested.  They  were  made  of  common 
North  River  brick  with  mortar  joints  f  in.  thick,  and  showed 
a  value  for  E0  of  about  300,000  or  400,000,  while  at  elastic 
limit  C"  was  on  the  average  1,000,  Ibs.  per  sq.  in. 


TORSION.  233 


CHAPTEE  IL 
TORSION. 

214.  Angle  of  Torsion  and  of  Helix.  When  a  cylindrical 
beam  or  shaft  is  subjected  to  a  twisting  or  torsional  action, 
i.  e.  when  it  is  the  means  of  holding  in  equilibrium  two 
couples  in  parallel  planes  and  of  equal  and  opposite  mo- 
ments, the  longitudinal  axis  of  symmetry  remains  straight 
— 7^^  and  the  elements  along  it  exper- 
, \ience  no  stress  (whence  it  may  be 
-^°  Jcalled  the  "line  of  no'  twist"), 
5J[t/  while  the  lines  originally  parallel  to 


1  FIG.  212.  it  assume  the  form  of  helices,  each 

element  of  which  is  distorted  in  its  angles  (originally 
right  angles),  the  amount  of  distortion  being  assumed  pro- 
portional to  the  radius  of  the  helix.  The  directions  of  the 


faces  of  any  element  were  originally  as  follows  :  two  radial, 
two  in  consecutive  transverse  sections,  and  the  other  two 
tangent  to  two  consecutive  circular  cylinders  whose  com- 
mon axis  is  that  of  the  shaft.  E.g.  in  Fig.  212  we  have 
an  unstrained  shaft,  while  in  Fig.  213  it  holds  the  two 


234  MECHANICS  OF  ENGINEERING. 

couples  (of  equal  moment  P  a  =  Q  b)  in  equilibrium.  These 
couples  act  in  parallel  planes  perpendicular  to  the  axis  of 
the  prism  and  a  distance,  I,  apart.  Assuming  that  the 
transverse  sections  remain  plane  and  parallel  during  tor- 
sion, any  surface  element,  m,  which  in  Fig.  212  was  entire- 
ly right-angled,  is  now  distorted.  Two  of  its  angles  have 
been  increased,  two  diminished,  by  an  amount  d,  the  angle 
between  the  helix  and  a  line  parallel  to  the  axis.  Suppos- 
ing m  to  be  the  most  distant  of  any  element  from  the  axis, 
this  distance  being  e,  any  other  element  at  a  distance  * 

from  the  axis  experiences  an  angular  distortion  «*«v  £ 

e 

If  now  we  draw  0  B'  parallel  to  O1  A  the  angle  B  0  B', 
=«,  is  called  the  Angle  of  Torsion,  while  d  may  be  called  the 
helix  angle;  the  former  lies  in  a  transverse  plane,  the  latter 

in  a  plane  tangent  to  the  cylinder.     Now 

• 

tan  d  =(  linear  arc  B  B'}-^!;  but  lin.  arc  B  Bl==t  30fc;  hence, 
putting  d  for  tan  d,  (3  being  small) 


(3  and  a  both  in  it  measure). 

215.  Shearing  Stress  on  the  Elements.  The  angular  distor- 
tion, or  shearing  strain,  3,  of  any  element  (bounded  as  al- 
ready described)  is  due  to  the  shearing  stresses  exerted  on 
it  by  its  neighbors  on  the  four  faces  perpendicular  to  the 

tangent  plane  of  the  cylindri- 
cal shell  in  which  the  element 
is  situated.  Consider  these 
neighboring  elements  of  an 
outside  element  removed,  and 
the  stresses  put  in  ;  the  latter 
are  accountable  for  the  dis- 
tortion of  the  element  and  so 


TORSION.  235 

hold  it  in  equilibrium.  Fig.  214  shows  this  element 
"free."  Within  the  elastic  limit  d  is  known  to  be  propor- 
tional to  ps>  the  shearing  stress  per  unit  of  area  on  the 
faces  whose  relative  angular  positions  have  been  changed. 
That  is,  from  eq.  (1),  §  208,  d=ps+Es-,  whence,  see  (1)  of 
§214, 


(2) 


In  (2)  pa  and  e  both  refer  to  a  surface  element,  e  being 
the  radius  of  the  cylinder,  and  ps  the  greatest  intensity  of 
shearing  stress  existing  in  the  shaft.  Elements  lying  nearer 
the  axis  suffer  shearing  stresses  of  less  intensity  in  pro- 
portion to  their  radial  distances,  i.e.,  to  their  helix-angles. 
That  is,  the  shearing  stress  on  that  face  of  the  element 
which  forms  a  part  of  a  transverse  section  and  whose  dis- 
tance from  the  axis  is  z,  is  p,  =-  ps,  per  unit  of  area,  and 

6 

the  total  shear  on  the  face  is  pdF9  dF  being  the  area  of  the 
face. 

216.  Torsional  Strength. — We  are  now  ready  to  expose  the 
full  transverse  section  of  a  shaft  under  torsion,  to  deduce 
formulae  of  practical  utility.  Making  a  right  section  of 
the  shaft  of  Fig.  213  anywhere  between  the  two  couples 
and  considering  the  left  hand  portion  as  a  free  body,  the 
forces  holding  it  in  equilibrium  are  the  two  forces  P  of 
the  left-hand  couple  and  an  infinite  number  of  shearing 
forces,  each  tangent  to  its  circle  of  radius  z,  on  the  cross 
section  exposed  by  the  removal  of  the  right-hand  portion. 
The  cross  section  is  assumed  to  remain  plane  during  tor- 
sion, and  is  composed  of  an  infinite  number  of  dfs,  each 
being  the  area  of  an  exposed  face  of  an  element;  see 


MECHANICS  OF  E^GItfEEBLNG. 


Each  elementary  shearing  force  ••  ~|  p$Ft  and  3  is  its 
lever  arm  about  the  axis  Go  .  For  equilibrium,  2  (mom.) 
about  the  axis  Oo  must  =0  ;  i.e.  in  detail 


or,  reducing, 

P*  fz*dF=Pa  ;  or,  M.=pa  .     (3) 

C  eX  6 

Eq.  (3)  relates  to  torsional  strength,  since  it  contains  ps,  the 
greatest  shearing  stress  induced  by  the  torsional  couple, 
whose  moment  Pa  is  called  the  Moment  of  Torsion,  the 
stresses  in  the  cross  section  forming  a  couple  of  equal  and 
opposite  moment. 

/p  is  recognized  as  the  Polar  Moment  of  Inertia  of  the  cross 
section,  discussed  in  §  94  ;  e  is  the  radial  distance  of  the 
outermost  element,  and  =  the  radius  for  a  circular  shafto 

217.  Torsional  Stiflhess.  —  In  problems  involving  the  angle 
of  torsion,  or  deformation  of  the  shaft,  we  need  an  equa- 
tion connecting  Pa  and  a,  which  is  obtained  by  substitut- 
ing in  eq.  (3)  the  value  of  pa  in  eq.  (2),  whence 


(4) 


TORSION.  237 

From  this  it  appears  that  the  angle  of  torsion,  a,  is  propor- 
tional to  the  moment  of  torsion,  Pa,  within  the  elastic 
limit  ;  a  must  be  expressed  in  it-measure.  Trautwine  cites  1° 
(i.e.  o=  0.0174)  as  a  maximum  allowable  value  for  shafts. 

218.  Torsional  Resilience  is  the  work  done  in  twisting  a 
shaft  from  an  unstrained  state  until  the  elastic  limit  is 
reached  in  the  outermost  elements.     If  in  Fig.  213  we 
imagine  the  right-hand  extremity  to  be  fixed,  while  the 
other  end  is  gradually  twisted  through  an  angle        each 
force  P  of  the  couple  must  be  made  to  increase  gradually 
from  a  zero  value  up  to  the  value  P19  corresponding  to  ax. 
In  this  motion  each  end  of  the   arm  a  describes  a  spacr 
«=  ^a«i,  and  the  mean  value  of  the  force  =  %Pi  (compai 
§  196).     Hence  the  work  done  in  twisting  is 


By  the  aid  of  preceding  equations,  (5)  can  be  written 


If  for  ps  we  write  S'  (Modulus  of  safe  shearing)  we  have 
for  the  safe  resilience  of  the  shaft 


7T/  _ 

' 


If  the  torsional  elasticity  of  an  originally  unstrained  shaft 
is  to  be  the  means  of  arresting  the  motion  of  a  moving 
mass  whose  weight  is  G,  (large  compared  with  the  parts 
'intervening)  and  velocity  =vt  we  write  (§  133) 

U'-G    **• 
^—  2' 

*s  the  condition  that  the  shaft  shall  not  be  injnrecL      * 


238  MECHANICS   OF   ENGINEERING. 


Polar  Moment  of  Inertia.  —  For  a  shaft  of  circular 
cross  section  (see  §  94)  Jp=  J^nr4  ;  for  a  hollow  cylinder 
JJ,==^^(r14  —  r24)  ;  while  for  a  square  shaft  Jp=i^&4,  b  being 
the  side  of  the  square  ;  for  a  rectangular  cross-section 
sides  &  and  h,  Ip=$h(P-}-h2).  For  a  cylinder  e=r;  if  hol- 
low, e=r  ,  the  greater  radius.  For  a  square,  e 


220.  Non-Circular  Shafts.  —  If  the  cross-section  is  not  cir- 
cular it  becomes  warped,  in  torsion,  instead  of  remaining 
plane.  Hence  the  foregoing  theory  does  not  strictly  ap- 
ply. The  celebrated  investigations  of  St.  Venant,  how- 
ever, cover  many  of  these  cases.  (See  §  708  of  Thompson 
and  Tait's  Natural  Philosophy  ;  also,  Prof.  Burr's  Elas- 
ticity and  Strength  of  the  Materials  of  Engineering).  Hi» 
results  give  for  a  square  shaft  (instead  of  the 


.        .        .        .    (1) 

and  Pa=lf^ps9  instead  of  eq.  (3)  of  §  216,  ps  being  the 
greatest  shearing  stress. 

The  elements  under  greatest  shearing  strain  are  found 
at  the  middles  of  the  sides,  instead  of  at  the  corners,  when 
the  prism  is  of  square  or  rectangular  cross-section.  The 
warping  of  the  cross-section  in  such  a  case  is  easily  veri  • 
fied  by  the  student  by  twisting  a  bar  of  india-rubber  in 
his  fingers. 

221.  Transmission  of  Power.  —  Fig.  216.     Suppose  the  cog- 

wheel B  to  cause  A,  on  the 
same  shaft,  to  revolve  uni- 
formly and  overcome  a  resis- 
tance Q,  the  pressure  of  the 
teeth  of  another  cog-wheel, 
B  being  driven  by  still  another 
FIG.  SIB,  wheel.  The  shaft  AB  is  un- 


TOBSION.  239 

der  torsion,  the  moment  of  torsion  being  =  Pa=  Qb.  (Pl 
and  Qi  the  bearing  reactions  have  no  moment  about  the 
axis  of  the  shaft).  If  the  shaft  makes  u  revolutions  per 
unit-time,  the  work  transmitted  (transmitted  ;  not  expend- 
ed in  twis-ting  the  shaft  whose  angle  of  torsion  remains 
constant,  corresponding  to  Pa)  per  unit-time,  i.e.  the  Power, 
is 

L=P.Zna.u=27tuPa        .        .        .   (8) 

To  reduce  L  to  Horse  Power  (§  132),  we  divide  by  N9 
the  number  of  units  of  work  per  unit-time  constituting 
one  H.  P.  in  the  system  of  units  employed,  i.e., 

Horse  Power  =H.  P.- 

For  example  JV=33,000  ft.-lbs.  per  minute,  or  =396,000 
inch-lbs.  per  minute  ;  or  =  550  ft.-lbs.  per  second.  Usually 
the  rate  of  rotation  of  a  shaft  is  given  in  revolutions  per 
minute, 

But  eq.  (8)  happens  to  contain  Pa  the  moment  of  torsion 
acting  to  maintain  the  constant  value  of  the  angle  of  tor- 
sion, and  since  for  safety  (see  eq.  (3)  §  216)  Pa=/S"7p-7-e, 
with  Jp=  y2T^  and  e=r  for  a  solid  circular  shaft,  we  have 
for  such  a  shaft 


(Safe),  H.  P.=  .        .        .        (9) 

N 

which  is  the  safe  H.  P.,  which  the  given  shaft  can  trans^ 
mit  at  the  given  speed.  S'  may  be  made  7,000  Ibs.  per  sq. 
inch  for  wrought  iron  ;  10,000  for  steel,  and  5,000  for  cast- 
iron.  If  the  value  of  Pa  fluctuates  periodically,  as  when 
a  shaft  is  driven  by  a  connecting  rod  and  crank,  for  (H.  P.) 
we  put  mX(H.  P.),  m  being  the  ratio  of  the  maximum  to 
the  mean  torsional  moment;  m=  about  Bunder  ordi- 
nary circumstances  (Cotterill). 


240 


MECHANICS    OF   ENGINEERING. 


222.  Autographic  Testing  Machine.  —The  principle  of  Prof 
Thurston's  invention  bearing  this  name  is  shown  in  Fig 


FIG.  217. 

217.  The  test-piece  is  of  a  standard  shape  and  size,  its 
central  cylinder  being  subjected  to  torsion.  A  jaw,  carry- 
ing a  handle  (or  gear-wheel  turned  by  a  worm)  and  a  drum 
on  which  paper  is  wrapped,  takes  a  firm  hold  of  one  end 
of  the  test-piece,  whose  further  end  lies  in  another  jaw 
rigidly  connected  with  a  heavy  pendulum  carrying  a  pen- 
cil free  to  move  axially.  By  a  continuous  slow  motion  of 
the  handle  the  pendulum  is  gradually  deviated  more  and 
more  from  the  vertical,  through  the  intervention  of  the 
test-piece,  which  is  thus  subjected  to  an  increasing  tor- 
sional  moment.  The  axis  of  the  test-piece  lies  in  the  axis 
of  motion.  This  motion  of  the  pendulum  by  means  of  a 
properly  curved  guide,  WR,  causes  an  axial  (i.e.,  parallel 
to  axis  of  test-piece)  motion  of  the  pencil  A,  as  well  as  an 
angular  deviation  /?  equal  to  that  of  the  pendulum,  and 
'this  axial  distance  CF,=sf,  of  the  pencil  from  its  initial 
position  measures  the  moment  of  torsion =Pa=Pc  sin  ft. 
As  the  piece  twists,  the  drum  and  paper  move  relatively 
to  the  pencil  through  an  angle  silo  equal  to  the  angle 


TOKSIOX. 


241 


of  torsion  a  so  far  attained.  The  abscissa  so  and  ordinate 
sT  of  the  curve  thus  marked  on  the  paper,  measure, 
when  the  paper  is  unrolled,  the  values  of  a  and  Pa  through 
all  the  stages  of  the  torsion.  Fig.  218  shows  typical 


FIG.  218. 


jurves  thus  obtained.  Many  valuable  indications  are 
given  by  these  strain  diagrams  as  to  homogeneousness  of 
composition,  ductility,  etc.,  etc.  On  relaxing  the  strain 
at  any  stage  within  the  elastic  limit,  the  pencil  retraces 
its  path ;  but  if  beyond  that  limit,  a  new  path  is  taken 
called  an  "  elasticity-line,"  in  general  parallel  to  the  first 
part  of  the  line,  and  showing  the  amount  of  angular  re- 
covery, B(Jy  and  the  permanent  angular  set,  OB. 

223.  Examples  in  Torsion. — The  modulus  of  safe  shearing 
strengtn,  S',  as  given  in  §  221,  is  expressed  in  pounds  per 
square  inch ;  hence  these  two  units  should  be  adopted 
throughout  in  any  numerical  examples  where  one  of  the 
above  values  for  S*  is  used.  The  same  statement  applies 
to  the  modulus  of  shearing  elasticity,  E*,  in  the  table  of 
§  210. 

EXAMPLE  1.— Fig.  216.  With  P  =  1  ton,  a  =  3  ft.,  I  = 
10  ft.,  and  the  radius  of  the  cylindrical  shaft  r=2.5  inches, 
required  the  max.  shearing  stress  per  sq.  inch,  ps,  the 
shaft  being  of  wrought  iron.  From  eq.  (3)  §  216 


Pae  __  2,000x36x2.5 


=2,930  Ibs.  per  sq.  inch, 


which  Is  a  safe  value  for  any  ferrous  metaL 


242 


MECHANICS  OF   ENGINEERING. 


EXAMPLE  2.  —  What  H.  P.  is  the  shaft  in  Ex.  1  transmit- 
ting, if  it  makes  50  revolutions  per  minute  ?  Let  u  = 
number  of  revolutions  per  unit  of  time,  and  N  '  =  the  num- 
ber of  units  of  work  per  unit  of  time  constituting  one 
horse-power.  Then  H.  P.=Pw27ra-f-jV,  which  for  the  foot- 
pound-minute system  of  units  gives 

H.  P.=2,000x50x27rx3-i-33,000=57^  H.  P. 

EXAMPLE  3.  —  What  different  radius  should  be  given  to 
the  shaft  in  Ex.  1,  if  two  radii  at  its  extremities,  originally 
parallel,  are  to  make  an  angle  of  2°  when  the  given  moment 
of  torsion  is  acting,  the  strains  in  the  shaft  remaining  con- 
stant. From  eq.  (4)  §  217,  and  the  table  210,  with  «=1 
0.035  radians  (i.e.  TT  -measure),  and  Ip=l/2xr*,  we  have 


2,000X36X120       =17.45,.r=2.04inclies. 
^0.035x9,000,000 


(This  would  bring  about  a  different  p«,  but  still  safe.)  Ti* 
foregoing  is  an  example  in  stiffness. 

EXAMPLE  4.  —  A  working  shaft  of  steel  (solid)  is  to  trans- 
mit 4,000  H.  P.  and  make  60  rev.  per  minute,  the  maximum 
twisting  moment  being  \l/2  times  the  average;  required 
its  diameter.  d=14.74  inches.  Ans. 

EXAMPLE  5.  —  In  example  1,  pA=  2,930  Ibs.  per  square 
inch  ;  what  tensile  stress  does  this  imply  on  a  plane  at  45° 
with  the  pair  of  planes  on  which  ps  acts  ?  Fig.  219  shows 


FIG.  220. 


TOKSIOX.  243 

a  small  cube,  of  edge  =dx,  (taken  from  the  outer  helix  of 
Fig.  215,)  free  and  in  equilibrium,  the  plane  of  the  paper 
being  tangent  to  the  cylinder  ;  while  220  shows  the  portion 
BDC,  also  free,  with  the  unknown  total  tensile  stress pdx* ^2 
acting  on  the  newly  exposed  rectangle  of  area  =dxXdx^/%, 
p  being  the  unknown  stress  per  unit  of  area.  From  sym- 
metry the  stress  on  this  diagonal  plane  has  no  shearing 
component.  Putting  ^[components  normal  io  BD~\=Q, 
we  have 

pdx2^/2=2dx2pscos4:50=dx2ps^2~.:p=pa    .      (1) 

That  is,  a  normal  tensile  stress  exists  in  the  diagonal 
plane  BD  of  the  cubical  element  equal  in  intensity  to  the 
shearing  stress  on  one  of  the  faces,  i.e.,  =2,930  Ibs.  per  sq. 
in.  in  this  case. 

Similarly  in  the  plane  A  G  will  be  found  a  compressive 
stress  of  2,930  Ibs.  per  sq.  in.  If  a  plane  surface  had  been 
exposed  making  any  other  angle  than  45°  with  the  face  of 
the  cube  in  Fig.  219,  we  should  have  found  shearing  and 
normal  stresses  each  less  than  ps  per  sq.  inch.  Hence  the 
interior  dotted  cube  in  219,  if  shown  "  free  "  is  in  tension 
in  one  direction,  in  compression  in  the  other,  and  with 
no  shear,  these  normal  stresses  having  equal  intensities. 
Since  S'  is  usually  less  than  T'  or  G',  if  ps  is  made  =  S' 
the  tensile  and  compressive  actions  are  not  injurious.  It 
follows  therefore  that  when  a  cylinder  is  in  torsion  any 
helix  at  an  angle  of  45°  with  the  axis  is  a  line  of  tensile, 
or  of  compressive  stress,  according  as  it  is  a  right  or  left 
handed  helix,  or  vice  versa. 

EXAMPLE  6. — A  solid  and  a  hollow  cylindrical  shaft,  of 
equal  length,  contain  the  same  amount  of  the  same  kind 
of  metal,  the  solid  one  fitting  the  hollow  of  the  other. 

Compare  their  torsional  strengths,  used  separately. 
The  solid  shaft  has  only  ^  the  strength  of  the  hollow 
one.  Ans. 


244  MECHANICS  OF 


CHAPTER  IIL 

FLEXURE  OF  HOMOGENEOUS  PRISMS  UNDER 
PERPENDICULAR  FORCES  IN  ONE  PLANE. 

224.  Assumptions  of  the  Common  Theory  of  Flexure. — When 
a  prism  is  bent,  under  the  action  of  external  forces  per- 
pendicular to  it  and  in  the  same  plane  with  each  other,  it 
may  be  assumed  that  the  longitudinal  fibres  are  in  tension 
on  the  convex  side,  in  compression  on  the  concave  side, 
and  that  the  relative  stretching  or  contraction  of  the  ele- 
ments is  proportional  to  their  distances  from  a  plane  in- 
termediate between,  with  the  understanding  that  the  flex- 
ure is  slight  and  that  the  elastic  limit  is  not  passed  in  any 
element.  ) 

This  "  common  theory  "  is  sufficiently  exact  for  ordinary 
engineering  purposes  if  the  constants  employed  are  prop- 
erly determined  by  a  wide  range  of  experiments,  and  in- 
volves certain  assumptions  of  as  simple  a  nature  as  possi- 
ble, consistently  with  practical  facts.  These  assumptions 
are  as  follows,  (for  prisms,  and  for  solids  with  variable  cross 
sections,  when  the  cross  sections  are  similarly  situated  as 
regards  a  central  straight  axis)  and  are  approximately 
borne  out  by  experiment : 

(1.)  The  external  or  "  applied  "  forces  are  all  perpendicu- 
lar to  the  axis  of  the  piece  and  lie  in  one  plane,  which  may 
be  called  the  force-plane ;  the  force-plane  contains  the 
axis  of  the  piece  and  cuts  each  cross-section  symmetri- 
cally ; 

(2.)  The  cross-sections  remain  plane  surfaces  during 
flexure  ; 

(3.)  There  is  a  surface  (or,  rather,  sheet  of  elements) 
which  is  parallel  to  the  axis  and  perpendicular  to  the 
force-plane,  and  along  which  the  elements  of  the  solid  ex- 


FLEXURE. 


245 


perience  no  tension  nor  compression  in  an  axial  direction, 
this  being  called  the  Neutral  Surface; 

(4.)  The  projection  of  the  neutral  surface  upon  the  force 
plane  (or  a  ||  plane)  being  called  the  Neutral  Line  or  Elastic 
Curve,  the  bending  or  flexure  of  the  piece  is  so  slight  that 
an  elementary  division,  ds,  of  the  neutral  line  may  be  put 
=dx,  its  projection  on  a  line  parallel  to  the  direction  of 
the  axis  before  flexure"; 

(5.)  The  elements  of  the  body  contained  between  any 
two  consecutive  cross-sections,  whose  intersections  with 
the  neutral  surface  are  the  respective  Neutral  Axes  of  the 
sections,  experience  elongations  (or  contractions,  accord- 
ing as  they  are  situated  on  one  side  or  the  other  of  the 
neutral  surface),  in  an  axial  direction,  whose  amounts  are 
proportional  to  their  distances  from  the  neutral  axis,  and 
indicate  corresponding  tensile  or  compressive  stresses  ; 

(6.)  E^E.; 

(7.)  The  dimensions  of  the  cross-section  are  small  com- 
pared with  the  length  of  the  piece  ; 

(8.)  There  is  no  shear  perpendicular  to  the  force  plane 
on  internal  surfaces  perpendicular  to  that  plane. 

In  the  locality  where  any  one  of  the  external  forces  is 
Applied,  local  stresses  are  of  course  induced  which  demand 
separate  treatment.  These  are  not  considered  at  present. 

225.  Illustration. — Consider  the  case  of  flexure  shown  in 
Fig.  221.  The  external  forces  are  three  (neglecting  the 


FORCE  PLANE 
1  £._ 

m 

Jl  : 

p 

-hr    i- 

{ 

>t  i     id 

—  1  —              o> 

Ni     i           "  L 

j 

s~                              1  ^ 

,     i2 

r 

FIG.  221. 


246 


MECHANICS   OF   ENGINEERING. 


weight  of  the  beam),  viz.:  Plt  P2,  and  P3.     Pl  and  P3  are 
loads,  P2  the  reaction  of  the  support. 

The  force  plane  is  vertical.  NiL  is  the  neutral  line  or 
elastic  curve.  NA  is  the  neutral  axis  of  the  cross-section 
at  m ;  this  cross-section,  originally  perpendicular  to  the 
sides  of  the  prism,  is  during  flexure  "|  to  their  tangent 
planes  drawn  at  the  intersection  lines  ;  in  other  words,  the 
side  view  QNB,  of  any  cross-section  is  perpendicular  to 
the  neutral  line.  In  considering  the  whole  prism  free  we 
have  the  system  Plf  P2,  and  P3  in  equilibrium,  whence 
from  2Y—Q  we  have  P2=Px+P3,  and  from  JT(mom.  about 
O)  =  0,  P3?3==  PA.  Hence  given  P1  we  may  determine  the 
other  two  external  forces.  A  reaction  such  as  P2  is  some- 
times called  a  supporting  force.  The  elements  above  the 
neutral  surface  N^LS&re  in  tension ;  those  below  in  com- 
pression (in  an  axial  direction). 

jj 

226.  The  Elastic  Forces. — Conceive  the  beam  in  Fig.  221 
separated  into  two  parts  by  any  transverse  section  such, 
as  QAy  and  the  portion  NiON,  considered  as  a  free  body 
in  Fig.  222.  Of  this  free  body  the  surface  QAB  is  one  of 


par 


Fie.  222. 


FLEXURE.  247 

I 

the  bounding  surfaces,  but  was  originally  an  internal  sur- 
face of  the  beam  in  Fig.  221.  Hence  in  Fig.  222  we  must 
put  in  the  stresses  acting  on  all  the  dF's  or  elements  of  area 
of  QAB.  These  stresses  represent  the  actions  of  the  body 
taken  away  upon  the  body  which  is  left,  and  according  to 
assumptions  (5),  (6)  and  (8)  consist  of  normal  stresses  (ten- 
sion or  compression)  proportional  per  unit  of  area,  to  the 
distance,  z,  of  the  dF's  from  the  neutral  axis,  and  of  shear- 
ing stresses  parallel  to  the  force-plane  (which  in  most 
cases  will  be  vertical). 

The  intensity  of  this  shearing  stress  on  any  dF  varies 
with  the  position  of  the  dF  with  respect  to  the  neutral 
axis,  but  the  law  of  its  variation  will  be  investigated  later 
(§§  253  and  254).  These  stresses,  called  the  Elastic  Forces 
of  the  cross-section  exposed,  and  the  external  forces  Pl  and 
P2,  form  a  system  in  equilibrium.  We  may  therefore  ap- 
ply any  of  the  sonditions  of  equilibrium  proved  in  §  38. 

227.  The  Neutral  Axis  Contains  the  Centre  of  Gravity  of  the 
Cross-Section. — Fig.  222.  Let  e~  the  distance  of  the  outer- 
most element  of  the  cross-section  from  the  neutral  axis,  and 
the  normal  stress  per  unit  of  area  upon  it  be  =p,  whether 
tension  or  compression.  Then  by  assumptions  (5)  and  (6), 
§  224,  the  intensity  of  nprmal  stress  on  any  dF  is  =  •-.  p 
and  the  actual 

normal  stress  on  any  dF  is=  ~  pdF       .        (1) 

This  equation  is  true  for  dF'a  having  negative  «'s,  i.e. 
on  the  other  side  of  the  neutral  axis,  the  negative  value 
of  the  force  indicating  normal  stress  of  the  opposite  char- 
acter ;  for  if  the  relative  elongation  (or  contraction)  of  two 
axial  fibres  is  the  same  for  equal  z's,  one  above,  the  other 
below,  the  neutral  surface,  the  stresses  producing  the 
changes  in  length  are  also  the  same,  provided  Ut=Ec;  see  §§ 
184  and  201. 


248  MECHANICS   OF   ENG1NEEKING. 

For  this  free  body  in  equilibrium  put  SX=Q  (JTisa 
horizontal  axis).  Put  the  normal  stresses  equal  to  their 
X  components,  the  flexure  being  so  slight,  and  the  X  com- 
ponent of  the  shears  =  0  for  the  same  reason.  This  gives 
(see  eq.  (1)  ) 

Ai  pdF=  0  ;  i.e.  *L    CdFz=  0  ;  or,  £  Fz=0        (2) 
J   e  e  J  e 

IB.  which  ~z—  distance  of  the  centre  of  gravity  of  the  cross- 
section  from  tne  neutral  axis,  from  which,  though  un- 
known in  position,  the  a's  have  been  measured  (see  eq. 
(4)  §  23). 

In  eq.  (2)  neither  p-z-e  nor  F  can  be  zero  .•.  z  must  =  0 ; 
i.e.  the  neutral  axis  contains  the  centre  of  gravity.  Q.  E.  D. 
[If  the  external  forces  were  not  all  perpendicular  to  the 
beam  this  result  would  not  be  obtained,  necessarily.] 

228.  The  Shear. — The     "  total    shear,"    or    simply    the 
"  shear,"  in  the  cross-section  is  the  sum   of  th.e  vertical 
shearing  stresses  on  the  respective  dF's.     Call  this  sum 
e7,  and  we  shall  have  from  the  free  body  in  Fig.  222,  by 
putting  -£7=0  (T being  vertical) 

P2—Pl—J=Q.\J=P^-Pl        .        .         (3) 

That  is,  the  shear  equals  the  algebraic  sum  of  the  ex- 
ternal forces  acting  on  one  side  (only)  of  the  section  con- 
sidered. This  result  implies  nothing  concerning  its  mode 
of  distribution  over  the  section. 

229.  The  Moment. — By   the   "Moment    of    Flexure"  or 
simply  the  Moment,  at  any  cross- section  is  meant  the  sum 
of  the  moments  of  the  elastic  forces  of  the  section,  taking 
the  neutral  axis  as  an  axis  of  moments.     In  this  summa- 
tion the  normal  stresses  appear  alone,  the  shear  taking  no 
part,  having  no  lever  arm  about  the  neutral 

Fig.  222,  the  moment  of  flexure 


FLEXURE.  249 


This  function,     CdFtf,  of  the  cross-section  or  plane  figure 

is  the  quantity  called  Moment  of  Inertia  of  a  plane  figure, 
§  85.  For  the  free  body  in  Fig.  222,  by  putting  ^(mom.s 
about  the  neutral  axis  NA)=Q,  we  have  then 


=0,  or  in  general,    *--  =  M    .  (5) 

o  6 

in  which  M  signifies  the  sum  of  moments,*  about  the  neutral 
axis  of  the  section,  of  all  the  forces  acting  on  the  free  body 
considered,  exclusive  of  the  elastic  forces  of  the  exposed 
section  itself. 


230.  Strength  in  Flexure. — Eq.  (5)  is  available  for  solving 
problems  involving  the  Strength  of  beams  and  girders,  since 
it  contains  p,  the  greatest  normal  stress  per  unit  of  area  to 
be  found  in  the  section. 

In  the  cases  of  the  present  chapter,  where  all  the  exter- 
nal forces  are  perpendicular  to  the  prism  or  beam,  and 
have  therefore  no  components  parallel  to  the  beam,  i.e.  to 
the  axis  X,  it  is  evident  that  the  normal  stresses  in  any 
section,  as  QB  Fig.  222,  are  equivalent  to  a  couple  ;  for  the 
condition  2X=Q  falls  entirely  upon  them  and  cannot  be 
true  unless  the  resultant  of  the  tensions  is  equal,  parallel, 
and  opposite  to  that  of  the  compressions.  These  two  equal 
and  parallel  resultants,  not  being  in  the  same  line,  form  a 
couple  (§  28),  which  we  may  call  the  stress-couple.  The 
moment  of  this  couple  is  the  "  moment  of  flexure  "  p~  ,  and 
it  is  further  evident  that  the  remaining  forces  in  Fig.  222, 
viz.:  the  shear  J  and  the  external  forces  Pl  and  P2,  are 
equivalent  to  a  couple  of  equal  and  opposite  moment  to 
the  one  formed  by  the  normal  stresses. 

*  It  is  evident,  therefore,  that  M (ft.-lbs.,  or  in.-lbs.)  is  numerically  equal 
to  the  "moment  of  flexure,"  or  moment  of  the  "  stress  couple  ";  so  that 
occasionally  it  may  be  convenient  to  use  "  M"  to  denote  the  value  of  the 
latter  moment  also. 


MECHANICS   OF  ENGINEERING. 


1 


231,  Flexural  Stiffness. — The  neutral  line,  or  elastic  curve, 
containing  the  centres  of  gravity  of  all  the  sections,  was 
originally  straight ;  its  radius  of  curvature  at  any  point, 
as  N,  Fig.  222,  c7uring  flexure  may  be  introduced  as  fol- 
lows. QB  and  U'V  are  two  consecutive  cross-sections, 
originally  parallel,  but  now  inclined  so  that  the  intersec- 
tion C,  found  by  prolonging  them  sufficiently,  is  the  centre 
of  curvature  of  the  ds  (put  =dx)  which  separates  them  at 
N,  and  CG=p=?  the  radius  of  curvature  of  the  elastic 
curve  at  N.  From  the  similar  triangles  U'UGa,nd  GNCwQ 
have  dlidx  ::e: pt  in  which  dX  is  the  elongation,  V U,  of  a 
portion,  originally  =dx,  of  the  outer  fibre.  But  the  rela- 
tive elongation  e=  -y-  of  the  latter  is,  by  §184,  within  the 
dx 

elastic  limit,  =J^L*.  ~  =—  and  eq.  (5)  becomes. 


(6) 


AXI6X 


From  (6)  the  radius  of  curvature  can  be  computed.  E= 
the  value  of  E^=E^  as  ascertained  from  experiments  in 
bending. 

To  obtain  a  differential  equation  of  the  elastic  curve,  (6) 
may  be  transformed  thus,^Fig.  223.  The  curve  being  very 

flat,  consider  two  consecutive 
cfo's  with  equal  dx's ;  they  may 
be  put  =  their  dx's.  Produce 
the  first  to  intersect  the  dy  of  the 
second,  thus  cutting  off  the  d*y9 
i.e.  the  difference  between  two 
consecutive  cfa/'s.  Drawing  a  per- 
pendicular to  each  ds  at  its  left 
extremity,  the  centre  of  curva- 
ture C  is  determined  by  their  in- 
tersection, and  thus  the  radius 
of  curvature  p.  The  two  shaded 
PIG.  233.  triangles  have  their  small  angles 


FLEXURE.  ,  251 

equal,  and  d?y  is  nearly  perpendicular  to  the  prolonged 
ds ;  hence,,  considering  them  similar,  we  have 

p:dx::dx:d2y.:  - 
p 

and  hence  from  eq.  (6)  we  have 

(approx.)     ±EI^=M    .         .         (7) 
diX 

as  a  differential  equation  of  the  elastic  curve.  From  this 
the  equation  of  the  elastic  curve  may  be  found,  the  de- 
flections at  different  points  computed,  and  an  idea  thus 
formed  of  the  stiffness.  All  beams  in  the  present  chap- 
ter being  prismatic  and  homogeneous  both  E  and  /  are  the 
same  (i.e.  constant)  at  all  points  of  the  elastic  curve.  In 
using  (7)  the  axis  JTmust  be  taken  parallel  to  the  length 
of  the  beam  before  flexure,  which  must  be  slight  ;  the 
minus  sign  in  (7)  provides  for  the  case  when  d^y-^dx2  is  es- 
sentially negative. 

232.  Resilience  of  Flexure. — If  the  external  forces  are  made- 
to  increase  gradually  from  zero  up  to  certain  maximum 
values,  some  of  them  may  do  work,  by  reason  of  their 
points  of  application  moving  through  certain  distances 
due  to  the  yielding,  or  flexure,  of  the  body.  If  at  the  be- 
ginning and  also  at  the  end  of  this  operation  the  body  is 
at  rest,  this  work  has  been  expended  on  the  elastic  resis- 
tance of  the  body,  and  an  equal  amount,  called  the  work 
of  resilience  (or  springing-back),  will  be  restored  by  the 
elasticity  of  the  body,  if  released  from  the  external  forces, 
provided  the  elastic  limit  has  not  been  passed.  The  energy 
thus  temporarily  stored  is  of  the  potential  kind ;  see  §§ 
148,  180,  196  and  218. 

232a.  Distinction  Between  Simple,  and  Continuous,  Beams  (or 
"Girders"). — The  external  forces  acting  on  a  beam  consist 


252 


MECHANICS   OF  ENGINEERING. 


generally  of  the  loads  and  the  "  reactions  "  of  the  sup- 
ports. If  the  beam  is  horizontal  and  rests  on  two  supports 
only,  the  reactions  of  those  supports  are  easily  found  by 
elementary  statics  [§  36]  alone,  without  calling  into  ac- 
count the  theory  of  flexure,  and  the  beam  is  said  to  be  a 
Simple  Beam,  or  girder ;  whereas  if  it  is  in  contact  with 
more  than  two  supports,  being  "  continuous,"  therefore, 
over  some  of  them,  it  is  a  Continuous  Girder  (§  271).  The 
remainder  of  this  chapter  will  deal  only  with  simple 
beams. 


ELASTIC  CURVES. 

233.  Case  I.  Horizontal  Prismatic  Beam,  [Supported  at  Both 
Ends,  With  a  Central  Load,  Weight  of  Beam  Neglected. — Fig. 
224  First  considering  the  whole  beam  free,  we  find  each 


-w 


FIG.  224.    §  233. 

reaction  to  be  =^£P.  AOB  is  the  neutral  line  ;  required 
the  equation  of  the  portion  OB  referred  to  0  as  an  origin, 
and  to  the  tangent  line  through  0  as  the  axis  of  X.  To 
do  this  consider  as  free  the  portion  mB  between  any  sec- 
tion m  on  the  right  of  0  and  the  near  support,  in  Fig. 
225.  The  forces  holding  this  free  body  in  equilibrium 


FIG.  825. 


FIG.  226. 


ELASTIC   CURVES.  253 


are  the  one  external  force  ^P,  and  the  elastic  forces  act- 
ing on  the  exposed  surface.  The  latter  consist  of  Jt  the 
shear,  and  the  tensions  and  compressions  represented  in 
the  figure  by  their  equivalent  "  stress-couple."  Selecting 
N9  the  neutral  axis  of  ra,  as  an  axis  of  moments  (that  J 
may  not  appear  in  the  moment  equation)  and  putting 
2*  (mom)  =0  we  have 


Fig.  226  shows  the  elastic  curve  OB  in  its  purely  geomet- 
rical aspect,  much  exaggerated.  For  axes  and  origin  as  in 
figure  d?y-r-dx*  is  positive. 

Eq.  (1)  gives  the  second  as-derivative  of  y  equal  to  a 
function  of  x.  Hence  the  first  as-derivative  of  y  will  be 
equal  to  the  as-anti-derivative  of  that  function,  plus  a  con- 
stant, C.  (By  anti  -derivative  is  meant  the  converse  of  de- 
rivative, sometimes  called  integral  though  not  in  the  sense 
of  summation).  Hence  from  (1)  we  have  (El  being  a  con- 
stant factor  remaining  undisturbed) 


(2)'  is  an  equation  between  two  variables  dy+dx  and  x,  and 
holds  good  for  any  point  between  0  and  B\  dy-^-dx  de- 
noting the  tang,  of  «,  the  slope,  or  angle  between  the  tan- 
gent line  and  X.  At  0  the  slope  is  zero,  and  x  also  zero  ; 
nence  at  0  (2)'  becomes 

#7x0=0—  0+0 

which  enables  us  to  determine  the  constant  C,  whose  value 
must  be  the  same  at  0  as  for  all  points  of  the  curve. 

Hence  C=0  and  (2)'  becomes 


254  MECHANICS  OF  ENGINEERING. 


from  which  the  slope,  tan.  «,  (or  simply  «,  IP.  ^--measure  j 
since  the  angle  is  small)  may  be  found  at  any  point.  Thus 
at  B  we  have  x=^l  and  dy-s-dx=al)  and 

1       PI? 
•••"^IT  ST 

Again,  taking  the  a?-anti-derivative  of  both  members  of  eq. 
(2)  we  have 


and  since  at  0  both  x  and  y  are  zero,  <7'  is  zero.    Hence 
the  equation  of  the  elastic  curve  OB  is 


To  compute  the  deflection  of  0  from  the  right  line  join 
ing  A  and  B  in  Fig.  224,  i.e.  B  K,  =d,  we  put  a?=  j£i  in  (3), 
being  then  =d,  and  obtain 


Eq.  (3)  does  not  admit  of  negative  values  for  x  ;  for  if 

the  free  body  of  Fig.  225  extended  to  the  left  of  0,  the  ex- 
ternal forces  acting  would  be  P,  downward,  at  0  ;  and  y2P% 
upward,  at  B,  instead  of  the  latter  alone  ;  thus  altering 
the  form  of  eq.  (1).  From  symmetry,  however,  we  know 
that  the  curve  AO,  Fig.  224,  is  symmetrical  with  OB  about 
the  vertical  through  Q. 


ELASTIC    CURVES.  255 

233a.  Load  Suddenly  Applied.  —  Eq.  (4)  gives  the  deflection 
d  corresponding  to  the  force  or  pressure  P  applied  at  the 
middle  of  the  beam,  and  is  seen  to  be  proportional  to  it. 
If  a  load  G  hangs  at  rest  from  the  middle  of  the  beam, 
P=  G  ;  but  if  the  load  G,  being  initially  placed  at  rest 
upon  the  unbent  beam,  is  suddenly  released  from  the  ex- 
ternal constraint  necessary  to  hold  it  there,  it  sinks  and 
deflects  the  beam,  the  pressure  P  actually  felt  by  the  beam 
varying  with  the  deflection  as  the  load  sinks.  What  is 
the  ultimate  deflection  dm  ?  Let  Pm—  the  pressure  be- 
tween the  load  and  the  beam  at  the  instant  of  maximum 
deflection.  The  work  so  far  done  in  bending  the  beam 
=  J^Pmdm.  The  potential  energy  given  up  by  the  load 
=  Gdm,  while  the  initial  and  final  kinetic  energies  are  both 
nothing. 

...  Gdm=y2Pmdm        .        .          (5) 

That  is,  Pm=26r.  Since  at  this  instant  the  load  is  sub- 
jected to  an  upward  force  of  ZG  and  to  a  downward  force 
of  only  G  (gravity)  it  immediately  begins  an  upward  mo- 
tion, reaching  the  point  whence  the  motion  began,  and 
thus  the  oscillation  continues.  We  here  suppose  the  elas- 
ticity of  the  beam  unimpaired.  This  is  called  the  "  sud- 
den "  application  of  a  load,  and  produces,  as  shown  above, 
double  the  pressure  on  the  beam  which  it  does  when  grad- 
ually applied,  and  a  double  deflection.  The  work  done 
by  the  beam  in  raising  the  weight  again  is  called  its  re- 
silience. 

Similarly,  if  the  weight  G  is  allowed  to  fall  on  the  mid- 
dle of  the  beam  from  a  height  h,  we  shall  have 


,  or  approx.,        , 
and  hence,  since  (4)  gives  dm  in  terms  of  P 


256  MECHANICS   OF  ENGINEERING. 

This  theory  supposes  the  mass  of  the  beam  small  com- 
pared with  the  falling  weight. 

234.  Case  II  Horizontal  Prismatic  Beam,  Supported  at  Both 
Ends.  Bearing  a  Single  Eccentric  Load.  Weight.  of  Beam  Neg- 

lected.— Fig.  227.  The  reactions 
of  the  points  of  support,  P0  and 
PU  are  easily  found  by  consider- 
ing the  whole  beam  free,  and  put- 
ting first  ^(mom.^  =0,  whence  P\ 
=Pl-r-llf  and  then  ^(mom.)B=0, 
Fie  227.  whence  P0=P(lr-J)+ll.  PQ  and 

PI  will  now  be  treated  as  known  quantities. 

The  elastic  curves  OC  and  CB,  though  having  a  common 
tangent  line  at  C  (and  hence  the  same  slope  «c),  and  a  com- 
mon ordinate  at  C,  have  separate  equations  and  are  both 
referred  to  the  same  origin  and  axes,  as  shown  in  the 
figure.  The  slope  at  0,  «o,  and  that  at  B,al9  are  unknown 
constants,  to  be  determined  in  the  progress  of  the  work. 

Equation  of  OC.  —  Considering  as  free  a  portion  of  the 
beam  extending  from  B  to  a  section  made  anywhere  on 
0(7,  x  and  y  being  the  co-ordinates  of  the  neutral  axis  of 
that  section,  we  conceive  the  elastic  forces  put  in  on  the 
exposed  surface,  as  in  the  preceding  problem,  and  put 
^(mom.  about  neutral  axis  of  the  section)  =0  which  gives 
(remembering  that  here  dfy-z-dx2  is  negative.) 


,        .        (1) 
whence,  by  taking  the  x  anti  -derivatives  of  both  members 


To  find  (7,  write  out  this  equation  for  the  point  0,  where 
dy-^-dx—OQ  and  x=Q,  and  we  have  C=EIa^',  hence  the 
equation  for  slope  is 


FLEXURE  ELASTIC  CURVES.  257 

lx-^-P^v-^+EIa,        .  (2) 
Again  taking  the  x  ant  i  -derivatives,  we  have  from  (2) 

Ely  =P  M-P*—       +EIa&+(  0'  =0)  (3) 


(at  Oboth  x  and  y  are  —0  .-.  (7'=0).  In  equations  (1),  (2), 
and  (3)  no  value  of  x  is  to  be  used  <0  or  >Z,  since  for 
points  in  CB  different  relations  apply,  thus 

Equation  of  CB.  —  Fig.  227.  Let  the  free  body  extend 
from  B  to  a  section  made  anywhere  on  (Tfi.^moms.),  as 
before,  =0,  gives 


=-Pl(ll-x)     ...        (4) 


(N.J3.  In  (4),  as  in  (1),  Eld^y-^dx2  is  written  equal  to  a  neg- 
ative quantity  because  itself  essentially  negative  ;  for  the 
curve  is  concave  to  the  axis  X  in  the  first  quadrant  of  the 
co-ordinate  axes.) 
From  (4)  we  have  in  the  ordinary  way  (cc-anti-deriv.) 


.        .        (5)' 

CLX  A 

To  determine  C",  consider  that  the  curves  CB  and  OG 
have  the  same  slope  (dy-^dx)  at  C  where  x—l\  hence  put 
x—  I  in  the  right-hand  members  of  (2)  and  of  (5)'  and 
equate  the  results.  This  gives  C"  =  ^PP-\-EIo^  and  .-. 


.        (5) 
l[l—?L-\  +G'"      .      (6)' 


258  MECHANICS   OF   ENGINEERING. 

At  (7,  where  x—lt  both  curves  have  the  same  ordinate  ; 
hence,  by  putting  x=l  in  the  right  members  of  (3)  and  (6)' 
and  equating  results,  we  obtain  C'"=  —  %PV.  .'.  (6)'  be 
comes 


as  the  Equation  of  CB,  Fig.  227.     But  OQ  is  still  an  unknown 
constant,  to  find  which  write  out  (6)  for  the  point  B  where 

x  =  l\,  and  y  =  0,  whence  we  obtain 


«!=  a  similar  form,  putting  P0  for  Plt  and  (?x  —  1)  for  L 

235.  Maximum  Deflection  in  Case  II.  —  Fig.  227.  The  or- 
dinate ym  of  the  lowest  point  is  thus  found.  Assuming 
t>  /4li,  it  will  occur  in  the  curve  0(7.  Hence  put  the 
dy-r-dx  of  that  curve,  as  expressed  in  equation  (2),  =0. 
Also  for  «o  write  its  value  from  (7),  having  putP1=Pl-:r-li3 
and  we  have 


whence    [a?  for  max.  y]= 

Now  substitute  this  value  of  x  in  (3),  also  OQ  from  (7),  and 
put  Pl=Pl  -r-li,  whence 


Max.  Deflec.=2/max=V9  - 


236.  Case  III.  Horizontal  Prismatic  Beam  Supported  at  Both 
Ends  and  Bearing  a  Uniformly  Distributed  Load  along  its  Whole 
Length.  —  (The  weight  of  the  beam  itself,  if  considered, 


FLEXURE.  ELASTIC  CUKVES. 


259 


constitutes  a  load  of  this  nature.)  Let  1=  the  length 
of  the  beam  and  w—  the  weight,  per  unit  of  length, 
of  the  loading ;  then  the  load  coming  upon  any  length  x 
will  be  =wx,  and  the  whole  load  =wL  By  hypothesis  w 
is  constant.  Fig.  228.  From  symmetry  we  know  that  the 


FIG.  228. 


reactions  at  A  and  B  are  each  =%wl,  that  the  middle  0  of 
the  neutral  line  is  its 'lowest  point,  and  the  tangent  line  at 
0  is  horizontal.  Conceiving  a  section  made  at  any  point 
m  of  the  neutral  line  at  a  distance  x  from  0,  consider  as 
free  the  portion  of  beam  on  the  right  of  m.  The  forces 
holding  this  portion  in  equilibrium  are  y£wl,  the  reaction 
at  B ;  the  elastic  forces  of  the  exposed  surface  at  m,  viz.: 
the  tensions  and  compressions,  forming  a  couple,  and  J 
the  total  sheor  ;  and  a  portion  of  the  load,  iv(l/2l—x).  The 
sum  of  the  me  ments  of  these  latter  forces  about  the  neu- 
tral axis  of  m,  is  the  same  as  that  of  their  resultant ;  (i.e., 
their  sum,  since  they  are  parallel),  and  this  resultant  acts  in 
the  middle  of  the  length  y£l — x.  Hence  the  sum  of  these 
moments  =w(l/^l — x)^(^l — x).  Now  putting  I  (mom. 
about  neutral  axis  of  m)  =0  for  this  free  body,  we  have 


(i) 


260  MECHANICS   OF   ENGINEERING. 

Taking  the  ic-anti-derivative  of  both  sides  of  (1), 

0)  (2) 


as  the  equation  of  slope.  (The  constant  is  =0  since  at  0 
both  dy-s-dx  and  x  are  =0.)  From  (2), 

^=|(^2-Y^4)+[C'=0]       .      .    (3) 

which  is  the  equation  of  the  elastic  curve  ;  throughout, 
i.e.,  it  admits  any  value  of  x  from  x=-}-}4lto  x=  —  ^l. 
This  is  an  equation  of  the  fourth  degree,  one  degree  high- 
er than  those  for  the  Curves  of  Cases  I  and  II,  where 
there  were  no  distributed  loads.  If  w  were  not  constant, 
but  proportional  to  the  ordinates  of  an  inclined  right  line, 
eq.  (3)  would  be  of  the  fifth  degree  ;  if  w  were  propor- 
tional to  the  vertical  ordinates  of  a  parabola  with  axis 
vertical,  (3)  would  be  of  the  sixth  degree  ;  and  so  on. 

By  putting  x=*4l  in  (3)  we  have  the  deflection  of  0  be- 
low the  horizontal  through  A  and  B,  viz.:  (with  W=  total 
load  =wl) 


=__ 
384     El     384        El 

237.  Case  IV.  Cantilevers,  —  A  horizontal  beam  whose  only 
support  consists  in  one  end  being  built  in  a  wall,  as  in 
Fig.  229(a),  or  supported  as  in  Fig. 
229(6)  is  sometimes-  called  a  canti- 
lever. Let  the  student  prove  that  in 
Fig.  229(a)  with  a  single  end  load  Pt 
the  deflection  of  B  below  the  tangent 
at  Ois  c?=^P^3-i-^B7;the  same  state- 
ment applies  to  Fig.  229(6),  but  the 
tangent  at  0  is  not  horizontal  if  the 
beam  was  originally  so.  It  can  also 
FIG.  m  be  proved  that  the  slope  at  B,  Fig. 

229(a)  (from  the  tangent  at  0)  is 


FLEXURE   ELASTIC   CURVES.  261 

PI2 


The  greatest  deflection  of  the  elastic  curve  from  the  right 
line  joining  AB,  in  Fig.  229(6),  is  evidently  given  by  the 
equation  for  y  max.  in  §  235,  by  writing,  instead  of  P  of 
that  equation,  the  reaction  at  0  in  Fig.  229(6).  This  assumes 
that  the  max.  deflection  occurs  between  A  and  0.  If  it 
occurs  between  0  and  B  put  (l\—l)  for  I. 

If  in  Fig.  229(a)  the  loading  is  uniformly  distributed 
along  the  beam  at  the  rate  of  w  pounds  per  linear  unit, 
the  student  may  also  prove  that  the  deflection  of  B  below 
the  tangent  at  0  is 


238,  Case  V.  Horizontal  Prismatic  Beam  Bearing  Equal  Ter- 
minal Loads  and  Supported  Symmetrically  at  Two  Points.— 
Fig.  231.  Weight  of  beam  neglected.  In  the  preceding 
cases  we  have  made  use  of  the  approximate  form  EldPy-^-dx* 
in  determining  the  forms  of  elastic  curves.  In  the  present 


FIG.  231. 


case  the  elastic  curve  from  0  to  C  is  more  directly  dealt 
with  by  employing  the  more  exact  expression  EI-^p  (see 
§  231)  for  the  moment  of  the  stress-couple  in  any  section. 
The  reactions  at  0  and  C  are  each  =P,  from  symmetry. 
Considering  free  a  portion  of  the  beam  extending  from  A 
to  any  section  ra  between  0  and  C  (Fig.  232)  we  have,  by 
putting  2  (mom.  about  neutral  axis  of  m)=0, 


fc-62  MECHANICS   OF   ENGINEERING. 

That  is,  the  radius  of  curvature  is  the  same  at  all  points 
of  00;  in  other  words  OC  is  the  arc  of  a  circle  with  the 
above  radius.  The  upward  deflection  of  F  from  the  right 
line  joining  0  and  C  can  easily  be  computed  from  a  knowl- 
edge of  this  fact.  This  is  left  to  the  student  as  also  the 
value  of  the  slope  of  the  tangent  line  at  0  (and  C).  The 
deflection  of  D  from  the  tangent  at  C=l/$PP-t-JSI,  as  IB 
Fig.  229(a), 


SAFE  LOADS  IN  FLEXURE* 

239.  Maximum  Moment.  —  As  we  examine  the  different  sec- 
tions of  a  given  beam  under  a  given  loading  we  find  differ* 
ent  values  of  p,  the  normal  stress  per  unit  of  area  in  the 
outer  element,  as  obtained  from  eq.  (5)  §  229,  viz.: 


in  which  I  is  the  "  Moment  of  Inertia  "  (§  85)  of  the  plane 
figure  formed  by  the  section,  about  its  neutral  axis,  e  the 
distance  of  the  most  distant  (or  outer)  fibre  from  the  neu- 
tral axis,  and  M  the  sum  of  the  moments,  about  this  neu- 
tral axis,  of  all  the  forces  acting  on  the  free  body  of  which 
the  section  in  question  is  one  end,  exclusive  of  the  stresses 
on  the  exposed  surface  of  that  section.  In  other  words 
M  is  the  sum  of  the  moments  of  the  forces  which  balance 
the  stresses  of  the  section,  these  moments  being  taken 
about  the  neutral  axis  of  the  section  under  examination. 
For  the  prismatic  beams  of  this  chapter  e  and  I  are  the 
same  at  all  sections,  hence  p  varies  with  M  and  becomes  a 
maximum  when  M  is  a  maximum.  In  any  given  case  the 
location  of  the  "dangerous  section"  or  section  of  maximum 
My  and  the  amount  of  that  maximum  value  may  be  deter- 
mined by  inspection  and  trial,  this  being  the  only  method 
(except  by  graphics)  if  the  external  forces  are  detached. 


FLEXURE  SAFE  LOADS.  263 

If,  however,  the  loading  is  continuous  according  to  a  de- 
finite algebraic  law  the  calculus  may  often  be  applied, 
taking  care  to  treat  separately  each  portion  of  the  beam 
between  two  consecutive  reactions  of  supports,  or  detached 
loads. 

As  a  graphical  representation  of  the  values  of  M  along 
the  beam  in  any  given  case,  these  values  may  be  conceived 
laid  off  as  vertical  ordinates  (according  to  some  definite 
scale,  e.g.  so  many  inch-lbs.  of  moment  to  the  linear  inch 
of  paper)  from  a  horizontal  axis  just  below  the  beam.  If 
the  upper  fibres  are  in  compression  in  any  portion  of  the 
beam,  so  that  that  portion  is  convex  downwards,  these  or- 
dinates will  be  laid  off  below  the  axis,  and  vice  versa  ;  for 
it  is  evident  that  at  a  section  where  M=Q,  p  also  =0,  i.e., 
the  character  of  the  normal  stress  in  the  outermost  fibre 
changes  (from  tension  to  compression,  or  vice  versa)  when 
M  changes  sign.  It  is  also  evident  from  eq.  (6)  §  231  that 
the  radius  of  curvature  changes  sign,  and  consequently  the 
curvature  is  reversed,  when  M  changes  sign.  These  mo- 
ment ordinates  form  a  Moment  Diagram,  and  the  extremities 
a  Moment  Curve. 

The  maximum  moment,  Mmt  being  found,  in  terms  of 
the  loads  and  reactions,  we  must  make  the  p  of  the  "  dan- 
gerous section,"  where  M=  Mm,  equal  to  a  safe  value  H'9 
and  thus  may  write 


(2) 


Eq.  (2)  is  available  for  finding  any  one  nnknown  quanti- 
ty, whether  it  be  a  load,  span,  or  some  one  dimension  of 
the  beam,  and  is  concerned  only  with  the  Strength,  and  not 
with  the  stiffness  of  th,e  beam.  If  it  is  satisfied  in  any 
given  case,  the  normal  stress  on  all  elements  in  all  sections 
is  known  to  be  =  or  <R',  and  the  design  is  therefore  safe 
in  that  one  respect. 

As  to  danger  arising  from  the  shearing  stresses  in  any 


264 


MECHANICS   OF   ENGINEERING. 


section,  the  consideration  of  the  latter  will  be  taken  up  in 
n  subsequent  chapter  and  will  be  found  to  be  necessary 
only  in  beams  composed  of  a  thin  web  uniting  two  flanges. 
The  total  shear,  however,  denoted  by  J,  bears  to  the  mo- 
ment Mt  an  important  relation  of  great  service  in  deter- 
mining Mm.  This  relation,  therefore,  is  presented  in  the 
next  article. 

240.    The  Shear  is  the  First    x-Derivative  of  the  Moment— 
Fig.  233.  (x  is  the  distance  of  any  section,  measured  parallel 

to  the  'beam  from  an  arbitrary 
origin).  Consider  as  free  a  ver- 
tical slice  of  the  beam  included 
between  any  two  consecutive 
vertical  sections  whose  distance 
apart  is  dx.  The  forces  acting 
are  the  elastic  forces  of  the  two 
internal  surfaces  now  laid  bare, 
and,  possibly,  a  portion,  wdx, 
of  the  loading,  which  at  this 

part  of  the  beam  has  some  intensity  =w  Jbs.  per  running 
linear  unit.  Putting  ^(mom.  about  axis  .?V')=Owe  have 
(noting  that  since  the  tensions  and  compressions  of  section 
N  form  a  couple,  the  sum  of  their  moments  about  N'  is 
just  the  same  as  about  N,) 


F        - 


: 

N 


• — dx~- 

J 

FIG.  233. 


But  £ }-~=M,  the  Moment  of  the  left  hand  section/? =M^t 

e  e 

that  of  the  right ;   whence  we   may  write,  after  dividing 
through  by  dx  and  transposing, 

dx       .      dM_jt 
T       1<e''~dx         ' 


(3) 


for  w  ^  vanishes  when  added  to  the  finite  J,  and  M' — M= 
dM=  increment  of  the  moment  corresponding  to  the  incre- 
ment, dx,  of  x.  This  proves  the  theorem. 


FLEXUKE.   SAFE  LOADS.  265 

Now  the  value  of  x  which  renders  M  a  maximum  or 
minimum  would  be  obtained  by  putting  the  derivative 
dM-~  dx  =  zero;  hence  we  may  state  as  a 

Corollary. — At  sections  where  the  moment  is  a  maximum 
or  minimum  the  shear  passes  through  the  value  zero. 

The  shear  J  at  any  section  is  easily  determined  by  con- 
sidering free  the  portion  of  beam  from  the  section  to  either 
end  of  the  beam  and  putting  JT(  vertical  components)=0. 

In  this  article  the  words  maximum  and  minimum  are 
used  in  the  same  sense  as  in  calculus ;  i.e.,  graphically, 
.  they  are  the  ordinates  of  the  moment  curve  at  points 
where  the  tangent  line  is  horizontal.  If  the  moment  curve  be 
reduced  to  a  straight  line,  or  a  series  of  straight  lines,  it 
has  no  maximum  or  minimum  in  the  strict  sense  just 
stated  ;  nevertheless  the  relation  is  still  practically  borne 
out  by  the  fact  that  at  the  sections  of  greatest  and  least 
ordinates  in  the  moment  diagram  the  shear  changes  sign 
suddenly.  This  is  best  shown  by  drawing  a  shear  diagram, 
whose  ordinates  are  laid  off  vertically  from  a  horizontal 
axis  and  under  the  respective  sections  of  the  beam.  They 
will  be  laid  off  upward  or  downward  according  as  e/is 
found  to  be  upward  or  downward,  when  the  free  body  con- 
sidered extends  from  the  section  toward  the  right. 

In  these  diagrams  the  moment  ordinates  are  set  off  on 
an  arbitrary  scale  of  so  many  inch -pounds,  or  foot-pounds, 
to  the  linear  inch  of  paper ;  the  shears  being  simply 
pounds,  or  some  other  unit  of/orce,  on  a  scale  of  so  many 
pounds  to  the  inch  of  paper.  The  scale  on  which  the 
beam  is  drawn  is  so  many  feet,  or  inches,  to  the  inch  of 
paper. 

241.  Safe  Load  at  the  Middle  of  a  Prismatic  Beam  Support- 
ed at  the  Ends. — Fig.  234.  The  reaction  at  each  support 
is  y2P.  Make  a  section  n  at  any  distance  x<±  from  J5. 
Consider  the  portion  nB  free,  putting  in  the  proper  elas- 
tic and  external  forces.  The  weight  of  beam  is  neglected. 
From  ^(mom.  about  n)=0  we  have 


266 


MECHANICS  OF  ENGINEERING. 


Evidently  M  is  proportional  to  x,  and  the  ordinates  repre« 
senting  it  will  therefore  be  limited  by  the  straight  line 


,Fio.  234. 


Hit,  forming  a  triangle  B'RA'.  From  symmetry,  another 
triangle  O'RA'  forms  the  other  half  of  the  moment  dia- 
gram. From  inspection,  the  maximum  M  is  seen  to  be  in 
the  middle  where  x=  ^l,  and  hence 


.  (1) 


Again  by  putting  J?(vert.  compons.)=0,  for  the  free  body 
nB  we  have 


and  must  point  downward  since  ~  points  upward.  Hence 
the  shear  is  constant  and  =  y2P  at  any  section  in  the  right 
hand  half.  If  n  be  taken  in  the  left  half  we  would  have, 
nB  being  free,  from  J(vert.  com.)=0, 


FLEXURE.  SAFE  LOADS.  267 

the  same  numerical  value  as  before  ;  but  e/must  point  up- 
ward, since  £  at  B  and  J  at  n  must  balance  the  downward 
P  at  A.  At  A,  then,  the  shear  changes  sign  suddenly, 
that  is,  passes  through  the  value  zero  ;  also  at  A,  M  is  a 
maximum,  thus  illustrating  the  statement  in§  240.  Notice 
the  shear  diagram  in  Fig.  234. 

To  find  the  safe  load  in  this  case  we  write  the  maximum 
value  of  the  normal  stress,  p,=  Bl,  a  safe  value,  (see  table 
in  a  subsequent  article)  and  solve  the  equation  for  P. 
But  the  maximum  value  of  p  is  in  the  outer  fibre  at  A, 
since  M  for  that  section  is  a  maximum.  Hence 


is  the  equation  for  safe  loading  in  this  case,  so  far  as  the 
normal  stresses  in  any  section  are  concerned. 

ExAMPL^.-t-If  the  beam  is  of  wood  and  has  a  rectangu- 
lar section  with  width  b=  2  in.,  height  &-=  4  in.,  while  its 
length  1=  ICfcjL,  required  the  safe  load,  if  the  greatest  nor- 
mal stress  iiMdmited  to  1,000  Ibs.  per  sq.  in.  Use  the 
pound  and  inch.  From  §  90  I=l/l2  M3=Y12x2x64=10.66 
biquad.  inches,  while  e=^s=2  in. 

,  P_4#/_  4x1,000x10.66 
~~ 


e  120x2 

242,  Safe  Load  Uniformly  Distributed  along  a  Prismatic  Beam 
Supported  at  the  Ends.—  Let  the  load  per  lineal  unit  of  the 
length  of  beam  be  =w  (this  can  be  made  to  include  the 
weight  of  the  beam  itself).  Fig.  235.  From  symmetry, 

each  reaction  =  y^wl.     For  the  free  body  nO  we  have,  put* 
ting  2(wom.  about  w)=0, 


268 


MECHANICS    OF   ENGINEERING. 


which  gives  M for  any  section  by  making  x  vary  from  0 
to  I.  Notice  that  in  this  case  the  law  of  loading  is  con- 
tinuous along  the  whole  length,  and  that  hence  the  mo- 
ment curve  is  continuous  for  the  whole  length. 


FIG.  235. 


To  find  the  shear  J",  at  n,  we  may  either  put  J?(vert.  com 
pons.)=0  for  the  free  body,  whence  J=  y?wl — wx,  and  mus 
therefore  be  downward  for  a  small  value  of  x  ;  or,  employ 
ing  §  240,  we  may  write  out  dM-r-dx,  which  gives 


the  same  as  before.  To  find  the  max.  M,  or  Mmt  put  eA*  Ov 
which  gives  x=^l.  This  indicates  a  maximum,  for  whan 
substituted  in  d2M-^-dx2t  i.e.,  in  — w,  a  negative  result  is 
obtained.  Hence  Mm  occurs  at  the  middle  of  the  beam  and 
its  value  is 


the  equation  of  safe  loading.      W=  total 

It  can  easily  be  shown  that  the  moment  curve  is  p  por. 


FLEXURE.  SAFE  LOADS. 


269 


;Ion  of  a  parabola,  whose  vertex  is  at  A"  under  the  mid- 
Jib  of  the  beam,  and  axis  vertical.  The  shear  diagram 
consists  of  ordinates  to  a  single  straight  line  inclined  to 
its  axis  and  crossing  it,  i.e.,  giving  a  zero  shear,  under  the 
middle  of  the  beam,  where  we  find  the  max.  M. 

If  a  frictionless  dove  -tail  joint  with  vertical  faces  were 
introduced  at  any  locality  in  the  beam  and  thus  divided 
the  beam  into  two  parts,  the  presence  of  J  would  be  made 
manifest  by  the  downward  slipping  of  the  left  hand  part 
on  the  right  hand  part  if  the  joint  were  on  the  right  of  the 
middle,  and  vice  versa  if  it  were  on  the  left  of  the  middle. 
This  shows  why  the  ordinates  in  the  two  halves  of  the 
shear  diagram  have  opposite  signs.  The  greatest  shear 
is  close  to  either  support  and  is  Jm= 


243.  Prismatic  Beam  Supported  at  its  Extremities  and  Loaded 

in  any  Manner.    Equation  for  Safe  Loading.  —  Fig.  236.    Given 

P         .p2          D  the  loads  Pl9  P2,  and  P3,  whose 

B         |  '        I          I       o    distances  from  the  right  sup- 

Port  are  li>  h  and  £5  ;  ,required 
the  equation  for  safe  loading  ; 
i.e.,  find  Mm  and  write  it  = 
ITI+e. 

If  the  moment  curve  were 
continuous,  i.e.,  if  M  were  a 
continuous  function  of  x  from 
end  to  end  of  the  beam,  we 
could  easily  find  Mm  by  making 
FIG.  236.  dM-:rdx=0,  i.e.,  e/=0,  and  sub- 

stitute the  resulting  value  of  x  in  the  expression  for  M. 
But  in  the  present  case  of  detached  loads,  J  is  not  zero, 
necessarily,  at  any  section  of  the  beam.  Still  there  is 
some  one  section  where  it  changes  sign,  i.e.,  passes  sud- 
denly through  the  value  zero,  and  this  will  be  the  section 
of  greatest  moment  (though  not  a  maximum  in  the  stricf; 
sense  used  in  calculus).  By  considering  any  portion  n  1 
as  free,  «7is  found  equal  to  the  Reaction  at  0  Diminished  by 
the  Loads  Occurring  Between  n  and  0.  The  reaction  at  B  is 


270  MECHANICS   OP   ENGINEERING. 


obtained  by  treating  the  whole  beam  as  free  (in  which  case 
no   elastic  forces    come  into   play)    and  putting  ^(mom. 
about  6>)=0;  while  that  at  0,=P0=Pl-i-P2-}-P3—PB 
If  n  is  taken  anywhere  between  0  and  E,  t7=Po 

E    "    F,  J=P0-Pl . 
"  "         F    "    H,  J=P0-Pl-P2 

"  "  "  "         H    tf    _B  tT:=P  —  P  —  P  —P 

This  last  value  of  e/also  =  the  reaction  at  the  other 
support,  j5.  Accordingly,  the  shear  diagram  is  seen  to 
consist  of  a  number  of  horizontal  steps.  The  relation 
J=dM-:rdx  is  such  that  the  slope  of  the  moment  curve  is 
proportional  to  the  ordinate  of  the  shear  diagram,  and 
that  for  a  sudden  change  in  the  slope  of  the  moment  curve 
there  is  a  sudden  change  in  the  shear  ordinate.  Hence  in 
the  present  instance,  J  being  constant  between  any  two 
consecutive  loads,  the  moment  curve  reduces  to  a  straight 
line  between  the  same  loads,  this  line  having  a  different 
inclination  under  each  of  the  portions  into  which  the  beam 
is  divided  by  the  loads.  Under  each  load  the  slope  of  the 
moment  curve  and  the  ordinate  of  the  shear  diagram  change 
suddenly.  In  Fig.  236  the  shear  passes  through  the  value 
zero,  i.e.,  changes  sign,  at  F;  or  algebraically  we  are  sup- 
posed to  find  that  P0—Pl  is  -f-  while  PQ—Pl~P2  is  — ,  in 
the  present  case.  Considering  FO,  then,  as  free,  we  find 
Mm  to  be 

—Pi(l2—li)  and  the  equation  for  safe  loading  is 

V 

(i.e.,  if  the  max.  M  is  at  F).  It  is  also  evident  that  the 
greatest  shear  is  equal  to  the  reaction  at  one  or  the  other 
support,  whichever  is  the  greater,  and  that  the  moment 
at  either  support  is  zero. 

The  student  should  not  confuse  the  moment  curve,  which 


FLEXURE.  SAFE   LOADS. 


271 


is  entirely  imaginary,  with  the  neutral  line  (or  elastic 
curve)  of  the  beam  itself.  The  greatest  moment  is  not 
necessarily  at  the  section  of  maximum  deflection  of  the 
neutral  line  (or  elastic  curve). 

For  the  case  in  Fig.  236  we  may  therefore  state  that  the 
max.  moment,  and  consequently  the  greatest  tension  or 
compression  in  the  outer  fibre,  will  be  found  in  the  sec- 
tion under  that  load  for  which  the  sum  of  the  loads  (in- 
cluding this  load  itself)  between  it  and  either  support  first 
equals  or  exceeds  the  reaction  of  that  support.  The 
amount  of  this  moment  is  then  obtained  by  treating  as  free 
either  of  the  two  portions  of  the  beam  into  which  this 
section  divides  the  beam. 

244,  Numerical  Example  of  the  Preceding  Article. — Fig.  237. 
Given  Plt  P2,  P3,  equal  to  y£  ton,  1  ton,  and  4  tons,  re- 


FIG.  237. 

spectively ;  ^=5  feet,  12=7  feet,  and  ^=10  feet;  while  the 
total  length  is  15  feet.  The  beam  is  of  timber,  of  rectan- 
gular cross-section,  the  horizontal  width  being  &=10 
inches,  and  the  value  of  R'  (greatest  safe  normal  stress), 
=  y2  ton  per  sq.  inch,  or  1,000  Ibs.  per  sq  inch. 


272  MECHANICS   OF  ENGINEERING. 

Required  the  proper  depth  h  lor  the  beam,  for  safe  load- 
ing- 

Solution.  —  Adopting  a  definite  system  of  units,  viz.,  the 
inch-ton-second  system,  we  must  reduce  all  distances  such 
as  I,  etc.,  to  inches,  express  all  forces  in  tons,  write  R'  =  y2 
(tons  per  sq.  inch),  and  interpret  all  results  by  the  same  sys- 
tem. Moments  will  be  in  inch-tons,  and  shears  in  tons. 
[N.  B.  In  problems  involving  the  strength  of  materials 
the  inch  is  more  convenient  as  a  linear  unit  than  the  foot, 
since  any  stress  expressed  in  Ibs.,  or  tons,  per  sq.  inch,  is 
numerically  144  times  as  small  as  if  referred  to  the  square 
foot.] 

Making  the  whole  beam  free,  we  have  from  moms,  about 
O,Pfi=^[^X  60+1x84+4x120]  =3.3  tons.-.  P0=5.5— 
3.3=2.2  tons. 

The  shear  anywhere  between  O  and  E  is  J=  P0=2.2  tons. 

"        E  and  F  is  ^=2.2—  #=1.7 

tons. 
The  shear  anywhere  between  F  and  HisJ  =2.2  —  y2  —  1  = 

0.7  tons. 
The  shear  anywhere  between  H  and  B  is  J  =  2.2  —  y2  —  1 

—  4=  —  3.3  tons. 

Since  the  shear  changes  sign  on  passing  H%  .%  the  max. 
moment  is  at  If;  whence  making  HO  free,  we  have 
M  at  H=Mm  =2.2x120—  }4x  60—  1x36=198  inch  -tons. 


For  safety  Mm  must  =  -  ,  in   which   JZ  —  y2    ton   per   sq. 

e 


inch,  e  =  y2\L  —  y2  of  unknown  depth  of  beam,  and  /,  §90,  = 

1  W,  with  6=10  inches 

»'•£•  #  •  I-  X  10^=198;  or  tf  =237.6.%  A  =15.4  inches. 

245.  Comparative  Strength  of  Rectangular  Beams.  —  For  such 
a  beam,  under  a  given  loading,  the  equation  for  safe  load- 
ing is 


=Mm  i.  e.  %  R  bh*=Mm  ....  (1) 

6 


FLEXURE.   SAFE  LOADS.  273 

whence  the  following  is  evident,  (since  for  the  same  length, 
mode  of  support,  and  distribution  of  load,  Mm  is  propor- 
tional to  the  safe  loading.) 

For  rectangular  prismatic  beams  of  the  same  length, 
same  material,  same  mode  of  support  and  same  arrange- 
ment of  load  : 

(1)  The  safe  load  is  proportional  to  the  width  of  beams 
having  the  same  depth  (h). 

(2)  The  safe  load  is  proportional  to  the  square  of  the 
depth  of  beams  having  the  same  width  (&). 

(3)  The  safe  load  is  proportional  to  the  depth  of  beams 
having  the  same  volume  (i.  e.  the  same  bh] 

(It  is  understood  that  the  sides  of  the  section  are  hori- 
zontal and  vertical  respectively  and  that  the  material  if 
homogeneous.) 

246.  Comparative  Stiffness  of  Rectangular  Beams. — Taking  th*. 
deflection  under  the  same  loading  as  an  inverse  measure 
of  the  stiffness,  and  noting  that  in  §§  233,  235,  and  236, 
this  deflection  is  inversely  proportional  to  I=^bhz  = 
the  "moment  of  inertia"  of  the  section  about  its  neutral 
axis,  we  may  state  that : 

For  rectangular  prismatic  beams  of  the  same  length, 
same  material,  same  mode  of  support,  and  same  loading  : 

(1)  The  stiffness  is  proportional  to  the  ^idth  for  beams 
of  the  same  depth. 

(2)  The  stiffness   is  proportional   to  the  cube  of  the 
height  for  beams  of  the  same  width  (b). 

(3)  The  stiffness  is  proportional  to  the  square  of  the 
depth  for  beams  of  equal  volume  (bhl). 

(4)  If  the  length  alone  vary,  the  stiffness  is  inversely 
proportional  to  the  cube  of  the  length. 


247.  Table  of  Moments  of  Inertia. — These  are  here  recapitn 
lated  for  the  simpler  cases,  and  also  the  values  of  <5.  the 
distance  of  the  outermost  fibre  from  the  axis. 

Since  the  stiffness  varies  as  /(other  things  being  equal), 


itu.  T 


274 


MECHANICS   OF   ENGINEERING. 


while  the  strength  varies  as  7-i-e,  it  is  evident  that  a 
square  beam  has  the  same  stiffness  in  any  position  (§89), 
while  its  strength  is  greatest  with  one  side  horizontal,  for 
then  e  is  smallest,  being  =}4b. 

Since  for  any  cross-section  I—  CdF  a2,  in  which  «=the 

distance  of  any  element,  dF,  of  area  from  the  neutral  axis, 
a  beam  is  made  both  stiffer  and  stronger  by  throwing 
most  of  its  material  into  two  flanges  united  by  a  vertical 
web,  thus  forming  a  so-called  "  I-beam  "  of  an  I  shape.  But 
not  without  limit,  for  the  web  must  be  thick  enough  to 
cause  the  flanges  to  act  together  as  a  solid  of  continuous 
substance,  and,  if  too  high,  is  liable  to  buckle  sideways, 
thus  requiring  lateral  stiffening.  These  points  will  be 
Created  later. 


SECTION. 

I 

0 

Rectangle,  width  =  ft,  depth  =  h  (vertical) 

Via  bh* 

%h 

Hollow  Rectangle,  symmet.  about  neutral  axis.     See  1 
Fig.  238  (a)                                                                         f 

Via[&,^»-&8A8a] 

%ht 

Triangle,  width  =&,  height  =  A,  neutral  axis  parallel! 
to  base  (horizontal).                                                           f 

V,e  MS 

%b 

Circle  of  radius  r 

y4nr* 

r 

Ring  of  concentric  circles.   Pig.  288  (b) 

KT(r4,-r*8) 

rt 

Rhombus;  Fig.  238  (c)  h  =  diagonal  which  Is  vertical. 

1/48  MS 

%h 

Square  with  side  b  vertical. 

Vu  &4 

%b 

"      "     *at45°  withhoriz. 

Vl8*4 

^&V2 

248.    Moment  of  Inertia  of  I-beams,   Box-girders,    Etc, — In 
common  with  other  large  companies,  the  N.  J.  Steel  and 


FLEXURE.      SAFE   LOADS. 


275 


Iron  Co.  of  Trenton,  N".  J.  (Cooper,  Hewitt  &  Co.)  manu- 
facture prismatic  rolled  beams  of  wrought-iron  *  variously 
called  /-beams,  deck -beams,  rails,  and  "  shape  iron,"  (in- 
cluding channels,  angles,  tees,  etc.,  according  to  the  form 
of  section.)  See  fig.  239  for  these  forms.  The  company 


|:BEAM. 


CHANNEL.     DECK-BEAM.  RAIL. 

FIG.  230. 


TEE. 


publishes  a  pocket-book  giving  tables  of  quantities  rela- 
ting to  the  strength  and  stiffness  of  beams,  such  as  the 
safe  loads  for  various  spans,  moments  of  inertia  of  their 
sections  in  various  positions,  etc.,  etc.  The  moments  of 
inertia  of  /-beams  and  deck-beams  are  computed  accord- 
ing to  §§  92  and  93,  with  the  inch  as  linear  unit.  The 
/-beams  range  from  4  in.  to  20  inches  deep,  the  deck- 
beams  being  about  7  and  8  in.  deep. 

For  beams  of  still  greater  stiffness  and  strength  com- 
binations of  plates,  channels,  angles,  etc.,  are  riveted  to- 
gether, forming  "  built-beams,"  or  "  plate  girders."  The 
proper  design  for  the  riveting  of  such  beams  will  be  ex- 
amined later.  For  the  present  the  parts  are  assumed  to 
act  together  as  a  continuous  mass.  For  example,  Fig.  240 
shows  a  "  box-girder,"  formed  of  two  "  channels "  and 
two  plates  riveted  together.  If  the  axis  of  symmetry,  N, 
is  to  be  horizontal  it  becomes  the  neu- 
tral axis.  Let  C=  the  moment  of  iner- 
tia, of  one  channel  (as  given  in  the 
pocket-book  mentioned)  about  the  axis 
N  perpendicular  to  the  web  of  the  chan- 
nel. Then  the  total  moment  of  inertia  of 
the  combination  is  (nearly) 


FIG.  940. 


/N  - 


(1) 


*  These  forms  are  now  (1900)  rolled  almost  exclusively  of  steel  ("  struc- 
tnnl  Ptoel  "). 


276 


MECHANICS   OF   ENGINEERING. 


In  (1),  b,  tt  and  d  are  the  distances  given  in  Fig.  240  (d  ex* 
tends  to  the  middle  of  plate)  while  d'  and  t'  are  the  length 
and  width  of  a  rivet,  the  former  from  head  to  head 
(i.e.,  d'  and  t'  are  the  dimensions  of  a  rivet-hole). 

For  example,  a  box-girder  of  wrought-iron  is  formed  of 
two  15-inch  channels  and  two  plates  10  inches  wide  and  1 
inch  thick,  the  rivet  holes  ^  in.  wide  and  1^  in.  long. 
That  is,  b=10;  t=I;  d=8;  *'  =  #;  and  d'  =1%  inches. 
Also  from  the  pocket-book  we  find  that  for  the  channel  in 
question,  (7=376  biquadratic  inches.  Hence,  eq.  (1) 

7N  =  752+2xlOxlx64—  4x1x^(8—  ^)2=1737biquadr.in. 


Also,  since  in  this  instance  e  =  S}£  inches,  and  12000 
Ibs.  per  sq.  inch  (or  6  tons  per  sq.  in.)  is  the  value  for  E' 
(=greatest  safe  normal  stress  en  the  outer  element  of  any 
cross-section)  used  by  the  Trenton  Co.  (for  wrought  iron), 


we  have  — -= 
e 


R'l    12000x1737 


8.5 


=2451700  inch-lbs. 


That  is,  the  box-girder  can  safely  bear  a  maximum  mo- 
ment, Mm,  =  2451700  inch-lbs.  =  1225.8  inch-tons,  as  far 
as  the  normal  stresses  in  any  section  are  concerned. 
(Proper  provision  for  the  shearing  stresses  in  the  section, 
and  in  the  rivets,  will  be  considered  later). 


249.  Strength  of  Cantilevers.- 


In  Fig.  241  with  a  single 
concentrated  load  P  at  the 
projecting  extremity,  we 

°  easily  find  the  moment  at 
n  to  be  M  =Px,  and  the 
max.  moment  to  occur  at 
the  section  next  the  wall, 

p  its  value  being  Mm—Pl. 
The  shear,  J",  is  constant, 

FIG.  241.  FIG.  242.  an(J     =     P    at     all    Sections. 

The  moment  and  shear  diagrams  are  drawn  in  accordance 
with  these  results. 


FLEXURE.       SAFE   LOADS. 


277 


If  the  load  W  =  wl  is  uniformly  distributed  on  the  can- 
tilever, as  in  fig.  242,  by  making  nO  free  we  have,  putting 
.  about  n)  =  0, 


pi  x 

— =wx  .  <)  .'. 


y2  wi. 


Hence  the  moment  curve  is  a  parabola,  whose  vertex  is  at 
(/  and  axis  vertical.  Putting  -T  (vert,  compons.)  =  0  we 
obtain  J  =  wx.  Hence  the  shear  diagram  is  a  triangle, 
and  the  max.  J=  wl  =s  W. 

250.  Resume"  of  the  Four  Simple  Cases. — The  following  table 
shows  the  values  of  the  deflections  under  an  arbitrary 
load  P9  or  W,  (within  elastic  limit),  and  of  the  safe  load ; 


Cantilevers. 

Beams  with  two  end  supports. 

With  one  end 

With  unif.  load 

Load  Pin 

Unif.  load 

loadP 

W  —  wl 

middle 

Fig.  241 

Fig.  242 

Fig.  234 

Fig7235 

Deflection 

U  .PJ3 

y     Wl9 

1    PI* 

5     Wt* 

El 

~~jpf 

48,'EI 

384*  El 

(  Safe  load  (from  ?1I 

J                                                                Q 

R'l 

R'l 

£ 

*  le 

IT 

IT 

Relative  strength 

1 

a 

4 

8 
128 

j  Relative  stiffness 
j  under  same  load 

1 

*/9 

16 

5 

16 

j  Relative  stiffness 

Vl 

4 

5 

i  under  safe  load 

j  Max.  shear  =  Jin,  (and 
j  location, 

P,  (at  wall) 

Wi  (at  wall) 

^P,  (at  supp). 

*  *<«.«, 

also  the  relative  strength,  the  relative  stiffness  (under  the 
same  load),  and  the  relative  stiffness  under  the  safe  load, 
for  the  same  beam. 

The  max.  shear  will  be  used  to  determine  the  proper 
web-thickness  for  I-beams  and  "  built-girders."  The  stu- 
dent should  carefully  study  the  foregoing  table,  noting 
especially  the  relative  strength,  stiffness,  and  stiffness 
under  safe  load,  of  the  same  beam. 

Thus,  a  beam  with  two  end  supports  will  bear  a  double 


278  MECHANICS   OP   ENGIN  EEKING. 

load,  if  uniformly  distributed  instead  of  concentrated  in 
the  middle,  but  will  deflect  ^  more  ;  whereas  with  a  given 
load  uniformly  distributed  the  deflection  would  be  only 
5/%  of  that  caused  by  the  same  load  in  the  middle,  provided 
*he  elastic  limit  is  not  surpassed  i#  either  case. 

251.  B/,  etc.  For  Various  Materials.— The  formula^/=  Mm, 

e 

from  which  in  any  given  case  of  flexure  we  can  compute 
the  value  of  pm,  the  greatest  normal  stress  in  any  outer 
element,  provided  all  the  other  quantities  are  known, 
holds  good  theoretically  within  the  elastic  limit  only. 
Still,  some  experimenters  have  used  this  formula  for  the 
rupture  of  beams  by  flexure,  calling  the  value  of  pm  thus 
obtained  the  Modulus  of  Rupture,  R.  R  may  be  found  to 
differ  considerably  from  both  the  T  or  C  of  §  203  with 
some  materials  and  forms,  being  frequently  much  larger. 
This  might  be  expected,  since  even  supposing  the  relative 
extension  or  compression  (i.e.,  strain)  of  the  fibres  to  be 
proportional  to  their  distances  from  the  neutral  axis  as 
the  load  increases  toward  rupture,  the  corresponding 
stresses,  not  being  proportional  to  these  strains  beyond  the 
elastic  limit,  no  longer  vary  directly  as  the  distances  from  the 
neutral  axis ;  and  the  neutral  axis  does  not  pass  through  the 
centre  of  gravity  of  the  section,  necessarily. 

The  following  table  gives  average  values  for  R,  R',  R ', 
and  E  for  the  ordinary  materials  of  construction.*     E,  the 
modulus  of  elasticity  for  use  in  the  formulae  for  deflection, 
is  given  as  computed  from  experiments  in  flexure,  and  is- 
nearly  the  same  as  Et  and  E& 

In  any  example  involving  R',  e  is  usually  written  equal 
to  the  distance  of  the  outer  fibre  from  the  neutral  axis, 
whether  that  fibre  is  to  be  in  tension  or  compression ; 
since  in  most  materials  not  only  is  the  tensile  equal  to  the 
compressive  stress  for  a  given  strain  (relative  extension 
or  contraction)  but  the  elastic  limit  is  reached  at  about 
the  same  strain  both  in  tension  and  compression. 

*  Wet,  or  unseasoned,  timber  is  very  considerably  weaker  than  that  (such  as 
ordinary  "  dry"  timber)  containing  only  12  per  cent,  of  moisture.  Large  pieces 
of  timber  take  a  much  longer  time  to  season  than  small  ones.  (Johnson.) 


FLEXUKE.     SAFE  LOADS.  279 

TABLE  FOB  USE  IN  EXAMPLES  IN  FLEXURE. 


Timber. 

Cast  Iron. 

Wro't  Iron. 

Steel. 

Max.  safe  stress  in  outer  fi-  ) 

6,000  in  tens. 

15,000 

V 

1,000 

12,000 

to 

bre  —  7?'(lbs.  per  sq.  inch).  } 

12,000  in  comp. 

40,000 

Stress  in  outer  fibre  at  Elas.  J 

17,000* 

to 

30,000 

limit  ^JP'dbs.  per  sq.  in.)  f 

35,000 

and  upward. 

14  Modul.  of  Rupture  "           | 

4,000 
to 

40,000 

50,000 

120,000 

=7?-=lbs.  per  sq.  inch.        ) 

20,000 

Hard  Steel. 

#=Mod.  of  Elasticity,           | 

1,000,000 
to 

17,000,000 

25,000,000 

30,000,000 

=lbs.  per  sq.  inch.              f 

3,000,000 

In  the  case  of  cast  iron,  however,  (see  §  203)  the  elastic 
limit  is  reached  in  tension  with  a  stress  =9,000  Ibs.  per 
sq.  inch  and  a  relative  extension  of  ^  of  one  per  cent., 
while  in  compression  the  stress  must  be  about  double  to 
reach  the  elastic  limit,  the  relative  change  of  form  (strain) 
being  also  double.  Hence  with  cast  iron  beams,  once 
largely  used  but  now  almost  entirely  displaced  by  rolled 
wrought  iron  beams,  an  economy  of  material  was  effected 
by  making  the  outer  fibre  on  the  compressed  side  twice 
as  far  from  the  neutral  axis  as  that  on  the  stretched  side. 
Thus,  Fig.  243,  cross-sections  with  unequal  flanges  were 
used,  so  proportioned  that  the  centre  of 
gravity  was  twice  as  near  to  the  outer 
fibre  in  tension  as  to  that  in  compression, 
i.e.,  e2=%eL;  in  other  words  more  material 
is  placed  in  tension  than  in  compression. 
The  fibre  A  being  in  tension  (within  elas- 
tic limit),  that  at  B,  since  it  is  twice  as  far  from  the  neu- 
tral axis  and  on  the  other  side,  is  contracted  twice  as  much 
as  A  is  extended  ;  i.e.,  is  rjider  a  compressive  strain 
double  the  tensile  strain  at  A,  but  in  accordance  with  the 
above  figures  its  state  of  stress  is  proportionally  as  much 
within  the  elastic  limit  as  that  of  A. 

•  Steel  beams  are  gradually  coming  into  use,  and  may  ul- 
timately replace f  those  of  wrought  iron. 

*  In  the  tests  by  U.  S.  Gov.  in  1879  with  I-beams,  R"  ranged  from  25,000 
to  38,000,  and  the  elastic  limit  was  reached  with  less  stress  in  the  large 
than  in  the  smaller  beams.     Also,  for  the  same  beam,  R'  decreased  with 
larger  spans. 

f  Later  ;    Feb.  1897.     Structural   steel    is  very  extensively  used  now  in 
preference  to  wrought  iron. 


FIG.  243. 


280  MECHANICS   OF   ENGINEERING. 

The  great  range  of  values  of  R  for  timber  is  due  not 
only  to  the  fact  that  the  various  kinds  of  wood  differ 
widely  in  strength,  while  the  behavior  of  specimens  of 
any  one  kind  depends  somewhat  on  age,  seasoning,  etc., 
but  also  to  the  circumstance  that  the  size  of  the  beam  un- 
der experiment  has  much  to  do  with  the  result.  The  ex- 
periments of  Prof.  Lanza  at  the  Mass.  Institute  of  Tech- 
nology in  1881  were  made  on  full  size  lumber  (spruce),  of 
dimensions  such  as  are  usually  taken  for  floor  beams  in 
buildings,  and  gave  much  smaller  values  of  It  (from  3,200 
to  8,700  Ibs.  per  sq.  inch)  than  had  previously  been  ob- 
tained. The  loading  employed  was  in  most  cases  a  con- 
centrated load  midway  between  4he  two  supports. 

These  low  values  are  probably  due  to  the  fact  that  in 
large  specimens  of  ordinary  lumber  the  continuity  of  it& 
substance  is  more  or  less  broken  by  cracks,  knots,  etc., 
the  higher  values  of  most  other  experimenters  having 
been  obtained  with  small,  straight -grained,  selected  pieces, 
from  one  foot  to  six  feet  in  length.  See  footnote  p.  278. 

The  value  .#'=12,000  Ibs.  per  sq.  inch  is  employed  by 
the  N.  J.  Iron  and  Steel  Co.  in  computing  the  safe  loads 
for  their  rolled  wrought  ironf  beams,  with  the  stipulation 
that  the  beams  (which  are  high  and  of  narrow  width)  must 
be  secure  against  yielding  sideways.  If  such  is  not  the 
case  the  ratio  of  the  actual  safe  load  to  that  computed  with 
R= 12,000  is  taken  less  and  less  as  the  span  increases. 
The  lateral  security  referred  to  may  be  furnished  by  the 
brick  arch-filling  of  a  fire-proof  floor,  or  by  light  lateral 
bracing  with  the  other  beams. 

252.  Numerical  Examples. — EXAMPLE  1. — A  square  bar  of 
wrought  iron,  1^  in.  in  thickness  is  bent  into  a  circular  J 
arc  whose  radius  is  200  ft.,  the  plane  of  bending  being  par- 
allel to  the  side  of  the  square.  Required  the  greatest  nor- 
mal stress  pm  in  any  outer  fibre. 

Solution.     From  §§  230  and  231  we  may  write 

ET      wT 

—  =/—  .•.  p=eE-t-p,  i.e.,  is  constant. 

. P         e 

f  For  their  steel  beams,  channels,  etc.,  this  company  uses  R'  =  16,000  ibs. 

per  sq.  inch. 
\  See  portion  OC,  Fig.  231,  p.  261,  for  example. 


FLEXURE.   SAFE  LOADS.  281 

For  the  units  inch  and  pound  (viz.  those  of  the  table  in  § 
251)  we  have  e=^  in.,  p  =2,400  in.,  and  E-  25,000,000  Ibs. 
per  sq.  inch,  anl  .*. 


p  =  pra=^x25,000,000-r2,400  =  7,812  Ibs.  per  sq.  in., 

which  is  quite  safe.  At  a  distance  of  J^  inch  from  the 
neutral  axis,  the  normal  stress  is  =[y2-^-^.]pm  =  /^pm== 
5,208  Ibs.  per  sq.  in.  (If  the  force-plane  (i.e.,  plane  of 
bending)  were  parallel  to  the  diagonal  of  the  square,  e 
would  =y2X  1.5V2  inches,  giving  ^  =  [7,812X^2  ]  Iks. 
per  sq.  in.)  §  238  shows  an  instance  where  a  portion,  0(7, 
Fig.  231,  is  bent  in  a  circular  arc. 

EXAMPLE  2.  —  A  hollow  cylindrical  cast-iron  pipe  of  radii 
3  :/2  and  4  inches*  is  supported  at  its  ends  and  loaded  in 
middle  (see  Fig.  234).  Eequired  the  safe  load,  neglecting 
the  weight  of  the  pipe.  From  the  table  in  §  250  we  have 
for  safety 


le 

From  §  251  we  put  R'=  6,000  Ibs.  per  sq.  in.;  and  from  § 
247  J=^-(r14  —  r24);  and  with  these  values,  r2  being  =-1,  rl  = 
4,  e=r!=4,  7r=-y-  and  Z=144  inches  (the  inch  must  be  the 
unit  of  length  since  72'  =6,000  lb#.  per  sq.  inch)  we  have 


P=4X6,OOOX^«  -f  (256-150)4-  [144x4]  .-.  P=  3,470  Ibs. 
The  weight  of  the  beam  itself  is  G—  Vf9  (§  7),  Le., 

0=^'-^=  f  (16-12^)144x^=443  Ibs. 

(Notice  that  7-,  here,  must  be  Ibs.,  per  cubic  inch).  This 
weight  being  a  uniformly  distributed  load  is  equivalent  to 
half  as  much,  221  Ibs.,  applied  in  the  middle,  as  far  as  the 
strength  of  the  beam  is  concerned  (see  §  250),  .*.  P  must  be 
taken  =3,249  Ibs.  when  the  weight  of  the  beam  is  consid- 
ered. 

*  And  length  of  12  feet,  should  be  added. 


282  MECHANICS   OF   ENGINEERING. 

EXAMPLE  3.  —  A  wrought-iron  rolled  I-beam  supported 
at  the  ends  is  to  be  loaded  uniformly  Fig.  235,  the  span 
being  equal  to  20  feet.  Its  cross-section,  Fig.  244,  has  a 

depth  *  parallel  to  the  web  of  15 
inches,  a  flange  width  of  5  inches. 
In  the  pocket  book  of  the  Trenton 
Co.  it  is  called  a  15-inch  light  I- 
beam,  weighing  150  Ibs.  per  yard, 
FIG.  244.  with  a  moment  of  inertia  =523.  bi- 

quad.  inches  about  a  gravity  axis  perpendicular  to  the 
web  (i.e.,  when  the  web  is  vertical,  the  strongest  position) 
and  =  15  biq.  in.  about  a  gravity  axis  parallel  to  the  web 
(i.e.,  when  the  web  is  placed,  horizontally). 
First  placing  the  web  vertically,  we  have  from  §  250, 

W,=  Safe   load,  distributed,    =8^5.    With  #=12,000, 

le, 


/!=523,  Z=240  inches,  el=7^  inches,  this  gives 
Wl  =  [8  x  12,000  x  523]  +  [240  x  -f  ]  =27,893  Ibs. 

But  this  includes  the  weight  of  the  beam,  6r=20  ft. 
=  1,000  Ibs.;  hence  a  distributed  load  of  26,893  Ibs.,  or  13.45 
tons  may  be  placed  on  the  beam  (secured  against  lateral 
yielding).  (The  pocket-book  referred  to  gives  13.27  tons 
as  the  safe  load.) 

Secondly,  placing  the  web  horizontal, 


of  Wl 


or  only  about  Vi2  of  W±. 

EXAMPLE  4.  —  Ke  quired  the  deflection  in  the  first  case  of 
Ex.  3.     From  §  250  the  deflection  at  middle  is 


73  K 

'2E7~S  " 

Strictly  15T36  inches. 


FLEXUKE.      SAFE   LOADS.  28,5 


.-.  ^=0.384  in. 

EXAMPLE  5.  —  A  rectangular  beam  of  yellow  pine,  of  width 
b=4  inches,  is  20  ft.  long,  rests  on  two  end  supports,  and  is 
to  carry  a  load  of  1,200  Ibs.  at  the  middle  ;  required  the 
proper  depth  h.  From  §  250 

P=4^=4  #  &#     JL 

le  ;     '  T  12  '  JA 

.«.  hz=QPl-:r4:Ii'b.  For  variety,  use  the  inch  and  ton.  For 
this  system  of  units  P=0.60  tons,  .#'=0.50  tons  per  sq.  in., 
Z=240  inches  and  b=  4  inches. 

...  tf=(6x0.6x240)-r-(4x0.5x4)=108  sq.  in.  .-.  ^=10.4  in, 

EXAMPLE  6.  —  Suppose  the  depth  in  Ex.  5  to  be  deter- 
mined by  the  condition  that  the  deflection  shall  be  =  l/sn 
if  the  span  or  length.  We  should  then  have  from  §  250 

d=  I    Z=I   *? 

500        48  El      . 

Using  the  inch  and  ton,  with  ^7=1,200,000  Ibs.  per  sq.  in*, 
which  =  600  tons  per  sq.  inch,  and  I=l/d>h\  we  have 


y=  500x0.60x240x240x12  =  h  ^ 

48x600x4 

As  this  is  >  10.4  the  load  would  be  safe,  as  well. 

EXAMPLE  7.  —  Required  the  length  of  a  wro't  iron  pipe 
supported  at  its  extremities,  its  internal  radius  being  2j{ 
in.,  the  external  2.50  in.,  that  the  deflection  under  its  own 
weight  may  equal  l/m  of  the  length.  579.6  in.  Ans. 

EXAMPLE  8.  —  Fig.  245.  The  wall  is  6  feet  high  and  one 
foot  thick,  of  common  brick  work 
(see  §  7)  and  is  to  be  borne  by  an 
/-beam  in  whose  outer  fibres  no 
greater  normal  stress  than  8,000 


"6  Ibs.  per  sq.  inch  is  allowable.     If 
FIG.  245.  a  number  of  I-beams  is  available, 


284  MECHANICS   OF   ENGINEERING. 

ranging  in  height  from  6  in.  to  15  in.  (by  whole  inches), 
which  one  shall  be  chosen  in  the  present  instance,  if  their 
cross-sections  are  Similar  Figures,  the  moment  of  inertia  of 
the  15-inch  beam  being  800  biquad.  inches  ? 

The  12-inch  beam.  Ana. 


SHEARING  STRESSES  IN  FLEXURE. 

253.  Shearing  Stresses  in  Surfaces  Parallel  to  the  Neutral 
Surface. — If  a  pile  of  boards  (see  Fig.  246)  is  used  to  sup- 
port a  load,  the  boards  being  free  to  slip  on  each  other,  it 
is  noticeable  that  the  end»  overlap,  although  the  boards 


UNLOADED 


FIG.  246.  FIG.  247. 


are  of  equal  length  (now  see  Fig.  247) ;  i.e.,  slipping  has 
occurred  along  the  surfaces  of  contact,  the  combina- 
tion being  no  stronger  than  the  same  boards  side  by 
side.  If,  however,  they  are  glued  together,  piled  as  in  the 
former  figure,  the  slipping  is  prevented  and  the  deflection 
is  much  less  under  the  same  load  P.  That  is,  the  com- 
pound beam  is  both  stronger  and  stiffer  than  the  pile  of 
loose  boards,  but  the  tendency  to  slip  still  exists  and  is 
known  as  the  "  shearing  stress  in  surfaces  parallel  to  the 
neutral  surface."  Its  intensity  per  unit  of  area  will  now 
be  determined  by  the  usual  "  free-body  "  method.  In  Fig. 
248  let  AN'  be  a  portion,  considered  free,  on  the  left  of  any 


--<ftJ   ^ 


N      N 

FIG.  248. 


SHEAR  IN   FLEXURE. 


285 


section  N:,  of  a  prismatic  beam  slightly  bent  under  forces 
in  one  plane  and  perpendicular  to  the  beam.  The  moment 
equation,  about  the  neutral  axis  at  N't  gives 


= M' ;  whence  p'= 


M'e 


e  I 

Similarly,  with  AN  as  a  free  body,  NN'  being  =dx, 


— ==  M ;  whence  p=  — 
e  I 


(1) 


(2) 


p  and  p'  are  the  respective  normal  stresses  in  the  outer 
fibre  in  the  transverse  sections  N  and  N'  respectively. 

Now  separate  the  block  NN't  lying  between  these  two 
consecutive  sections,  as  a  free  body  (in  Fig.  249).     And 


pa 


PART  OF  J 


tWTOF 


FIG.  250. 


furthermore  remove  a  portion  of  the  top  of  the  latter  block, 
the  portion  lying  above  a  plane  passed  parallel  to  the  neu- 
tral surface  and  at  any  distance  z"  from  that  surface.  This 
latter  free  body  is  shown  in  Fig.  250,  with  the  system  of 
forces  representing  the  actions  upon  it  of  the  portions  taken 
away.  The  under  surface,  just  laid  bare,  is  a  portion  of  a  sur- 
face (parallel  to  the  neutral  surface)  in  which  the  above  men- 
tioned slipping,  or  shearing,  tendency  exists.  The  lower  por- 
tion (of  the  block  NN')  which  is  now  removed  exerted  this 


286  MECHANICS    OF   ENGINEERING. 

rubbing,  or  sliding,  force  on  the  remainder  along  the  under 
surface  of  the  latter.  Let  the  unknown  intensity  of  this 
shearing  force  be  JT(per  unit  of  area)  ;  then  the  shearing 
force  on  this  under  surface  is  =  Xy"dx,  (y",=  oa  in  figure, 
being  the  horizontal  width  of  the  beam  at  this  distance  z" 
from  the  neutral  axis  of  N*)  and  takes  its  place  with  the 
other  forces  of  the  system,  which  are  the  normal  stresses 

between  |  ,    and  portions  of  J  and  J',  the  respective 

\_z=z" 

total  vertical  shears.  (The  manner  of  distribution  of  J 
over  the  vertical  section  is  as  yet  unknown  ;  see  next  arti- 
ole.) 

Putting  I  (horiz.  compons.)  =  0  in  Fig.  250,  we  have 

e  ^p'dF—  Ce  Z-pdF—Xy"dx=0 
"  .  e  «/g»     e 

=£=£     fzdF 


But  from  eqs.  (1)  and  (2),  ff—p  =  (M—M)*,*-^  dMy 
while  from  §  240  dM  =  Jdx  ; 

,.Xy"dx=<^  fzdF.'.X  =*-  fldF  (3) 

/  t/3//  ly"  J%n 

as  the  required  intensity  per  unit  of  area  of  the  shearing 
force  in  a  surface  parallel  to  the  neutral  surface  and  at  a 
distance  z"  from  it.  It  is  seen  to  depend  on  the  "  shear  "  J 
and  the  moment  of  inertia  /  of  the  whole  vertical  section; 
upon  the  horizontal  thickness*  y"  of  the  beam  at  the  sur- 

face   in    question  ;    and    upon     the     integral      /      zdFt 

z" 

which  (from  §  23)  is  the  product  of  the  area  of  that  part  of 
the  vertical  section  extending  from  the  surface  in  question  to 
the  outer  fibre,  by  the  distance  of  the  centre  of  gravity  of  that 
part  from  the  neutral  surface. 

*  Thickness  of  actual  substance. 


SHEAR   IN   FLEXURE. 


287 


It  now  follows,  from  §  209,  that  the  intensity  (per  unit 
area)  of  the  shear  on  an  elementary  area  of  the  vertical 
cross  section  of  a  bent  beam,  and  this  intensity  we  may  call 
Z,  is  equal  to  that  X,  just  found,  in  the  horizontal  section 
which  is  at  the  same  distance  (z")  from  the  neutral  axis. 

254.  Mode  of  Distribution  of  J,  the  Total  Shear,  over  the  Verti- 
cal Cross  Section. — The  intensity  of  this  shear,  Z  (Ibs.  per 
sq.  inch,  for  instance)  has  just  been  proved  to  be 


=x=-,   C 

III       «/   ,' 


(4) 


To  illustrate  this,  required  the 
value  of  Z  two  inches  above  the  neu- 
tral axis,  in  a  cross  section  close  to 
the  abutment,  in  Ex.  5,  §  252.  Fig. 
251  shows  this  section.  From  it  we  AS 
have  for  the  shaded  portion,  lying 
above  the  locality  in  question,  y"  = 

4  inches,  and    Ce  ~  '    zdF  =  (area 
•/  3"=  2 

of   shaded    portion)  X  (distance    of 

its  centre  of  gravity  from   NA)  = 

(12.8  sq.  in.)  x  (3.6  in.)  =  46.08  cubic  inches. 

the  total  shear  J  =  the  abutment  reaction  =  600  Ibs., 
while  /  =  L.  W  =  1-  x  4  x  (10.4)3  =  375  biquad.  inches. 
Both  Jand  /refer  to  the  whole  section. 


FIG.  251. 


~ 


600x46.08 


1Q,<oii, 
=18.42  Ibs.  per  sq.  m., 


insignificant.     In  the  neighborhood  of  the  neutral 
axis,  where  z"  =  0,  we  have  y1'  —  4  and 

™'SX  2.6=54.8, 

wh..e  J  and  /  of  course  are  the  same  as  before.     Hence 
for  z"  =0 


288 


MECHANICS   OF   ENGINEERING. 
^=^=21.62  Ibs.  per  sq.  in. 


At  the  outer  fibre  since      6  zdF=0,  z"  being  =  e,  Z  is  — 


FIG.  252 


for  a  beam  of  any  shape. 

For  a  solid  rectangular  section  like  the 
above,  Z  and  z"  bear  the  same  relation  to 
each  other  as  the  co-ordinates  of  the  para- 
bola in  Fig.  252  (axis  horizontal). 

Since  in  equation  (4)  the  horizontal 
thickness,  y",  from  side  to  side  ef  the  sec- 
tion of  the  locality  where  Z  is  desired, 

occurs  in  the  denominator,  and  since   /  ezdF 

z" 

increases  as  z"  grows  numerically  smaller,  the  following 
may  be  stated,  as  to  the  distribution  of  J,  the  shear,  in 
any  vertical  section,  viz.: 

The  intensity  (Ibs.  per  sq.  in.)  of  the  shear  is  zero  at 
the  outer  elements  of  the  section,  and  for  beams  of  ordi- 
nary shapes  is  greatest  where  the  section  crosses  the  neu- 
tral surface.  For  forms  of  cross  section  having  thin  webs 
its  value  may  be  so  great  as  to  require  special  investiga- 
tion for  safe  design. 

Denoting  by  ZQ  the  value  of  Z&i  the  neutral  axis,  (which 
=X0  in  the  neutral  surface  where  it  crosses  the  vertica 
section  in  question)  and  putting  the  thickness  of  the  sub- 
stance of  the  beam  =  bQ  at  the  neutral  axis,  we  have, 

j       (  area   above  )      j  the  dist.  of  ite  cent.  ),„ 
*=JiBS-15  X  I  5?5£g  f  X  {  grav.from  that  axis  Jfo 

255,  Values  of  Zo  for  Special  Forms  of  Cross  Section, — From 
the  last  equation  it  is  plain  that  for  a  prismatic  beam  the 
value  of  ZQ  is  proportional  to  J9  the  total  shear,  and  hence 
to  the  ordinate  of  the  shear  diagram  for  any  particular 
case  of  loading.  The  utility  of  such  a  diagram,  as  obtain- 


SHEAR   IX   FLEXURE. 


289 


ed  in  Figs.  234-237  inclusive,  is  therefore  evident,  for  by 
locating  the  greatest  shearing  stress  in  the  beam  it 
enables  us  to  provide  proper  relations  between  the  load- 
ing and  the  form  and  material  of  the  beam  to  secure  safety 
against  rupture  by  shearing. 

The  table  in  §  210  gives  safe  values  which  the 
maximum  ZQ  in  any  case  should  not  exceed.  It  is 
only  in  the  case  of  beams  with  thin  webs  (see  Figs. 
238  and  240)  however,  that  ZQ  is  likely  to  need  at- 
tention. 

For  a  Rectangle  we  have,  Fig.  253,  (see  eq.  5,  § 


FIG.  263. 


254)  b0=b,  I= 


.:Z0=X0=^-  Z  i.e.,  =  A  (total  shear)-f-  (whole  area) 

Hence  the  greatest  intensity  of  shear  in  the  cross-section 
is  A  as  great  per  unit  of  area  as  if  the  total  shear  were 
uniformly  distributed  over  the  section. 


FIG.  254.  FIG.  255.  FIG.  256. 

For  a  Solid  Circular  section  Fig.  254 
J 


FIG.  257. 


. 

2      3;r     3 


[See  §  26  Prob.  3]. 

For  a  Hollow  Circular  section  (concentric    circles)    Fig; 
255,  we  have  similarly, 


290 


MECHANICS   OF   ENGINEERING. 


ntrf-rWfa-rj     2       3;r 


Applying  this  formula  to  Example  2  §  252,  we  first  have 
as  the  max.  shear  Jm  —Y2P  =1,735  Ibs.,  this  being  the  abut- 
ment reaction,  and  hence  (putting  it  =  (22  -f-  7)) 


- 


which  cast  iron  is  abundantly  able  to  withstand  in  shear- 
ing. 

For  a  Hollow  Rectangular  Beam,  symmetrical  about    its 
neutral  surface,  Fig.  256  (box  girder) 


The  same  equation  holds  good  for  Fig.  257  (I-beam  with 
square  corners)  but  then  b2  denotes  th£  sum  of  the  widths 
of  the  hollow  spaces. 


256.  Shearing  Stress  in  the  Web  of  an  I-Beam.  —  It  is  usual  to 
consider  that,  with  I-beams  (and  box- 
beams)  with  the  web  vertical  the  shear  J, 
in  any  vertical  section,  is  borne  exclusively 
by  the  web  and  is  uniformly  distributed 
over  its  section.  That  this  is  nearly  true 
may  be  proved  as  follows,  the  flange  area 
being  comparatively  large.  Fig.  258.  Let 
Fl  be  the  area  of  one  flange,  and  I*0  that  of 
the  half  web.  Then  since 


Fia.  268. 


/-4 


SHEAH   IX   FLEXUBE.  291 

,(the  last  term  approximate,  y2  li§  being  taken  for  the  radi- 
us of  gyration  of  FI,)  while 


CdF=F+F         (the  first  term  approx.)  we  have 
Jo  2          4 


J 

z     _  wc__, 

A         -V>o6^i+2^o)  '  "Wo* 


if  we  write  (2^+^)  +  (6^+2^)  =  1A  -  But  b^  is  the 
area  of  the  whole  web,  .*.  the  shear  per  unit  area  at  the 
neutral  axis  is  nearly  the  same  as  if  J  were  uniformly  dis- 
tributed over  the  web.  E.  g.,  with  Fl  =  2  sq.  in.,  and  FQ 
=  1  sq.  in.  we  obtain  ZQ  =  1.07  (J-rtyty. 

Similarly,  the   shearing  stress  per   unit   area  at  n,  the 
upper  edge  of  the  web,  is  also  nearly  equal  to  J+ 


eq.,  4  (£254)  for  then   I  f°  (zdF)']  =     F^\       nearly, 

\_y  z"=l/2hQ          J 

while  /  remains  as  before. 

The  shear  per  unit  area,  then,  in  an  ordinary  I-beam  ia 
obtained  by  dividing  the  total  shear  J  by  the  area  of  the 
web  section. 

EXAMPLE.  —  It  is  required  to  determine  the  proper 
thickness  to  be  given  to  the  web  of  the  15-inch  wrought- 
iron  rolled  beam  of  Example  3  of  §252,  the  height  of  web 
being  13  inches,  with  a  safe  shearing  stress  as  low*  as  4000 
Ibs.  per  sq.  in.  (the  practice  of  the  N.  J.  Steel  and  Iron 
Co.,  for  webs),  the  web  being  vertical. 

The  greatest  total  shear,  Jm,  occurring  at  either  support 
and  being  equal  to  half  the  load  (see  table  §250)  we  have 
with  60  =  width  of  web, 


Z0  max.=      -;  i.e.  4000  =  .-.  ^  «  0.26  inches. 

60Xl3 


*  This  low  value  of  4000  in  the  vertical  edge  of  an  element  of  the  web 
may  carry  with  it  a  much  higher  intensity  of  shearing  stress  on  some  internal 
oblique  plane,  it  the  element  is  near  the  junction  with  the  flange  (see  §  270a 
on  p.  319);  hence  the  choice  of  4000. 


292 


MECHANICS   OF   ENGINEERING. 


(Units,  inch  and  pound).  The  15-inch  light  beam  of  the 
N.  J.  Co.  has  a  web  ]/2  inch  thick,  so  as  to  provide  for  a 
shear  double  the  value  of  that  in  the  foregoing  example. 
In  the  middle  of  the  span  Z$  =  0,  since  J  =  0. 

257.  Designing  of  Riveting  for  Built  Beams. — The  latter  are 
generally  of  the  I-beam  and  box  forms,  made  by  riveting 
together  a  number  of  continuous  shapes,  most  of  the  ma- 
terial being  thrown  into  the  flange  members.  E.  g.,  in  fig. 
259,  an  I-beam  is  formed  by  riveting  together,  in  the 
manner  shown  in  the  figure,  a  "  vertical  stem  plate  "  or 
web,  four  "  angle-irons,"  and  two  "  flange-plates,"  each  of 


FIG.  259. 


FIG.  260. 


these  seven  pieces  being  continuous  through  the  whole 
length  of  the  beam.  Fig  260  shows  a  box-girder.  If  the 
riveting  is  well  done,  the  combination  forms  a  single  rigid 
beam  whose  safe  load  for  a  given  span  may  be  found  by 
foregoing  rules  ;  in  computing  the  moment  of  inertia,  how- 
ever, the  portion  of  cross  section  cut  out  by  the  rivet 
holes  must  not  be  included.  (This  will  be  illustrated  in 
a  subsequent  paragraph.)  The  safe  load  having  been  com- 
puted from  a  consideration  of  normal  stresses  only,  and 
the  web  being  made  thick  enough  to  take  up  the  max. 
total  shear,  «/m,  with  safety,  it  still  remains  to  design  the 
riveting,  through  whose  agency  the  web  and  flanges  are 
caused  to  act  together  as  a  single  continuous  rigid  mass. 
It  will  be  on  the  side  of  safety  to  consider  that  at  a  given 


SHEAR  IN  FLEXUKE.  293 

locality  in  the  beam  the  shear  carried  by  the  rivets  con- 
necting the  angles  and  flanges,  per  unit  of  length  of  beam, 
is  the  same  as  that  carried  by  those  connecting  the  angles 
and  the  web  ("vertical  stem -plate").  The  amount  of  this^ 
shear  may  be  computed  from  the  fact  that  it  is  equal  to 
that  occurring  in  the  surface  (parallel  to  the  neutral  sur- 
face) in  which  the  web  joins  the  flange,  in  case  the  web 
and  flange  were  of  continuous  substance,  as  in  a  solid  I- 
beam.  But  this  shear  must  be  of  the  same  amount  per 
horizontal  unit  of  length  as  it  is  per  vertical  linear  unit  in 
the  web  itself,  where  it  joins  the  flange  ;  (for  from  §  254  Z 
=X.)  But  the  shear  in  the  vertical  section  of  the  web, 
being  uniformly  distributed,  is  the  same  per  vertical  linear 
unit  at  the  junction  with  the  flange  as  at  any  other  part 
of  the  web  section  (§  256,)  and  the  whole  shear  on  the  ver- 
tical section  of  web  =  J,  the  "  total  shear  "  of  that  section 
of  the  beam. 

Hence  we  may  state  the  following : 

The  riveting  connecting  the  angles  with  the  flanges,  (or 
the  web  with  the  angles)  in  any  locality  of  a  built  beam, 
must  safely  sustain  a  shear  equal  to  J  on  a  horizontal  length 
equal  to  the  height  of  web. 

The  strength  of  the  riveting  may  be  limited  by  the  re- 
sistance of  the  rivet  to  being  sheared  (and  this  brings 
into  account  its  cross  section)  or  upon  the  crushing  resist- 
ance of  the  side  of  the  rivet  hole  in  the  plate  (and  this  in- 
volves both  the  diameter  of  the  rivet  and  the  thickness  of 
the  metal  in  the  web,  flange,  or  angle.)  In  its  practice  the 
N.  J.  Steel  and  Iron  Co.  allows  7500  Ibs.  per  sq.  inch  shear- 
ing stress  in  the  rivet  (wrought  iron),  and  12500  Ibs.  per 
sq.  inch  compressive  resistance  in  the  side  of  the  rivet- 
hole,  the  axial  plane  section  of  the  hole  being  the  area  of 
reference. 

In  fig.  259  the  rivets  connecting  the  web  with  the  angles 
are  in  double  shear,  which  should  be  taken  into  account  in 
considering  their  shearing  strength,  which  is  then  double  ; 
those  connecting  the  angles  and  the  flange  plates  are  in 


294  MECHANICS   OF   ENGINEERING. 

single  shear.  In  fig.  260  (box-beam)  where  the  beam  ia 
built  of  two  webs,  four  angles,  and  two  flange  plates,  all 
the  rivets  are  in  single  shear.  If  the  web  plate  is  very 
high  compared  with  its  thickness,  vertical  stiffeners  in  the 
form  of  T  irons  may  need  to  be  riveted  upon  them  lat- 
erally [see  §  314]. 

EXAMPLE. — A  built  I-beam  oi  wrought  iron  (see  fig.  259) 
is  to  support  a  uniformly  distributed  load  of  40  tons,  its 
extremities  resting  on  supports  20  feet  apart,  and  the 
height  and  thickness  of  web  being  20  ins.  and  y2  in.  re- 
spectively. How  shall  the  rivets,  which  are  \  in.  in  di- 
ameter, be  spaced,  between  the  web  and  the  angles  which 
are  also  y2  in.  in  thickness?  Eef erring  to  fig.  235  we  find 
that  J  =  y£  W  —  20  tons  at  each  support  and  diminishes 
regularly  to  zero  at  the  middle,  where  no  riveting  will  there- 
fore be  required.  (Units  inch  and  pound).  Near  a  sup- 
port the  riveting  must  sustain  for  each  inch  of  length  of 
beam  a  shearing  force  of  (J  ~r  height  of  web)  =  40000  -r- 
20  in.  =  2000  Ibs.  Each  rivet,  having  a  sectional  area  of 
i^  TT  (^j)2  =  0.60  sq.  inches,  can  bear  a  safe  shear  of  0.60 
X  7500  =  4500  Ibs.  in  single  shear,  and  .-.  of  9000  Ibs.  in 
double  shear,  which  is  the  present  case.  But  the  safe 
compressive  resistance  of  the  side  of  the  rivet  hole  in 
either  the  web  or  the  angle  is  only  7/&  in.  x  %  in.  X  12500 
=  5470  Ibs.,  and  thus  determines  the  spacing  of  the  rivets 
as  follows  : 

2000  Ibs.  -r-  5470  gives  0.36  as  the  number  of  rivets  per 
inch  of  length  of  beam,  i.e.,  they  must  be  1  -f-  0.36  =  2.7 
inches  apart,  centre  to  centre,  near  the  supports;  5.4  inches 
apart  at  ^  the  span  from  a  support;  none  at  all  in  the 
middle. 

However,  "  the  rivets  should  not  be  spaced  closer  than 
2^  times  their  diameter,  nor  farther  apart  than  16  times 
the  thickness  of  the  plate  they  connect,"  is  the  rule  of  the 
N.  J.  Co. 

As  for  the  rivets  connecting  the  angles  and  flange  plates, 
being  in  two  rows  and  opposite  (in  pairs)  the  safe  shear- 


FLEXURE.  BUILT   BEAM. 


295 


ing  resistance  of  a  pair  (each  in  single  shear)  is  9,000  Ibs., 
while  the  safe  compressive  resistance  of  the  sides  of  the 
two  rivet  holes  in  the  angle  irons  (the  flange  plate  being 
much  thicker)  is  =  10,940  Ibs.  Hence  the  former  figure 
(9,000)  divided  into  2,000  Ibs.,  gives  0.22  as  the  number  of 
pairs  of  rivets  per  inch  of  length  of  the  beam  ;  i.e.,  the 
rivets  in  one  row  should  be  spaced  4.5  inches  apart,  centre 
to  centre,  near  a  support ;  the  interval  to  be  increased  in 
inverse  ratio  to  the  distance  from  the  middle  of  span, 
(bearing  in  mind  the  practical  limitation  just  given). 

If  the  load  is  concentrated  in  the  middle  of  the  span, 
instead  of  uniformly  distributed,  «/is  constant  along  each 
half -span,  (see  fig.  234)  and  the  rivet  spacing  must  accord- 
ingly be  made  the  same  at  all  localities  of  the  beam. 


SPECIAL  PROBLEMS  IN  FLEXURE. 

258.  Designing  Cross  Sections  of  Built  Beams. — The  last  par- 
agraph dealt  with  the  riveting  of  the  various  plates ;  we 
now  consider  the  design  of  the  plates  themselves.  Take 
for  instance  a  built  I-beam,  fig.  261 ;  one  vertical  stem- 


.  261 


296  MECHANICS   OF   ENGINEERING, 

plate,  four  angle  irons,  (each  of  sectional  area  =  A,  re- 
maining after  the  holes  are  punched,  with  a  gravity  axis 
parallel  to,  and  at  a  distance  =  a  from  its  base),  and  two 
flange  plates  of  width  =  b,  and  thickness  =  t.  Let  the 
whole  depth  of  girder  =  h,  and  the  diameter  of  a  rivet 
hole  =f.  To  safely  resist  the  tensile  and  compressive 
forces  induced  in  this  section  by  Mm  inch-lbs.  (Mm  being 
the  greatest  moment  in  the  beam  which  is  prismatic)  we 
have  from  §  239, 

Mm=^I  (1) 

6 

R'  for  wrought  iron  =  12,000  Ibs.  per  sq.  inch,  e  is  =  y2  h 
while  1,  the  moment  of  inertia  of  the  compound  section, 
is  obtained  as  follows,  taking  into  account  the  fact  that 
the  rivet  holes  cut  out  part  of  the  material.  In  dealing 
with  the  sections  of  the  angles  and  flanges,  we  consider 
them  concentrated  at  their  centres  of  gravity  (an  approx^ 
imation,  of  course,)  and  treat  their  moments  of  inertia 

about  N  as  single  terms  in  the  series    CdF  z* 

(see  §  85).  The  sub  tractive  moments  of  inertia  for  the 
rivet  holes  in  the  web  are  similarly  expressed  ;  let  bQ  = 
thickness  of  web. 

f  /N  for  web  =   £fa  (h—Zt)*—ZbQt'  [|— t— a']2 
-|  7N  for  four  angles  =  4tA  [± — t — a]2 
(  7N  for  two  flanges  =  2(6—20  t  (1£)2 

the  sum  of  which  makes  the  7N  of  the  girder.  Eq.  (1)  may 
now  be  written 

-^--4  (2) 

aF 

which  is  available  for  computing  any  one  unknown  quan- 
tity. The  quantities  concerned  in  7N  are  so  numerous  and 
they  are  combined  in  so  complex  a  manner  that  in  any 
numerical  example  it  is  best  to  adjust  the  dimensions  of 
the  section  to  each  other  by  successive  assumptions  and 


FLEXURE   BUILT   BEAM.  297 

trials.  The  size  of  rivets  need  not  vary  much  in  different 
cases,  nor  the  thickness  of  the  web-plate,  which  as  used 
by  the  N.  J.  Co.  is  "  rarely  less  than  ^  or  more  than  ^ 
inch  thick."  The  same  Co.  recommends  the  use  of  a 
single  size  of  angle  irons,  viz.,  3"  X  3''  X  ^",  for  built 
girders  of  heights  ranging  from  12  to  36  inches,  and  also 
y±  in.  rivets,  and  gives  tables  computed  from  eq.  (2)  for 
the  proportionate  strength  of  each  portion  of  the  com- 
pound section. 

EXAMPLE. — (Units,  inch  and  pound).  A  built  I-beam 
with  end  supports,  of  span  =  20  ft.  =  240  inches,  is  to 
support  a  uniformly  distributed  load  of  36  tons  —  72,000  Ibs. 
If  y±  inch  rivets  are  used,  angle  irons  3''  X  3"  X  ^",  ver- 
tical web  y^"  in  thickness,  and  plates  1  inch  thick  for 
flanges,  how  wide  (b  =  ?)  must  these  flange-plates  be  ? 
taking  h  =  22  inches  =  total  height  of  girder. 

Solution. — From  the  table  in  §  250  we  find  that  the  max. 
M  for  this  case  is  y&  Wl,  where  W  =  the  total  distributed 
load  (including  the  weight  of  the  girder)  and  I  =  span. 
Hence  the  left  hand  meniber  of  eq.  (2)  reduces  to 

Wl     7i^        72000  x  240  x  22 
16  '  W  ~  16  x  12000 

That  is,  the  total  moment  of  inertia  of  the  section  must 
be  =  1,980  biquad.  inches,  of  which  the  web  and  angles 
supply  a  known  amount,  since  bQ  =  ^!/,  t  =  1",  f=  ^", 
a'  =  1%",  A==  2.0  sq.  in.,  a  =  0.9'',  and  h  =  22",  are 
known,  while  the  remainder  must  be  furnished  by  the 
flanges,  thus  determining  their  width  &,  the  unknown 
quantity. 

The  effective  area,  A,  of  an  angle  iron  is  found  thus : 
The  full  sectional  area  for  the  size  given,  =  3  X  $4  + 
2>^  X  J/2  =2.75  sq.  inches,  from  which  deducting  for  two 
rivet  holes  we  have 

A=  2.75—2  x  %  X  */2==  2.0  sq.  in. 

The  value  a  =  0.90"  is  found  by  cutting  out  the  shape 


298  MECHANICS   OF   ENGINEERING. 


of  two  angles  from  sheet  iron,  thus  : 
and  balancing  it  on  a  knife  edge.  (The 
gaps  left  by  the  rivet  holes  may  be  ignored, 
without  great  error,  in  finding  a).  Hence, 
substituting  we  have 


IN  for  web  =-1-  .  ^x203—  2x^  .  %  [8^]2=282.3 
INfor  four  angles  =  4x2x  [9.10]2=662.5 
IN  for  two  nanges=2(^T)xlX(10^)2=220.4(&—  1.5) 

.-.  1980=  282.3+662.5+(6—  1.5)220.4 
whence  b  =  4.6  +  1.5  =  6.1  inches 

the  required  total  width  of  pach  of  the  1  in.  flange  plates. 
This  might  be  increased  to  6.5  in.  so  as  to  equal  the 
ttnited  width  of  the  two  angles  and  web. 

The  rivet  spacing  can  now  be  designed  by  §  257,  and 
the  assumed  thickness  of  web,  y2  in.,  tested  for  the  max. 
total  shear  by  §  256.  The  latter  test  results  as  follows  : 
The  max.  shear  Jm  occurs  near  either  support  and  = 
y2  W=  36,000  Ibs.  .-.,  calling  &'0  the  least  allowable  thickness 
of  web  in  order  to  keep  the  shearing  stress  as  low  as  4,000 
Ibs.  per  sq.  inch, 

b'  x  20"  x  4000  =36000  .-.  b'  =0.45" 


showing  that  the  assumed  width  of  y2  in.  is  safe. 

This  girder  will  need  vertical  stiffeners  near  the  ends, 
as  explained  subsequently,  and  is  understood  to  be  sup- 
ported laterally.  Built  beams  of  double  web,  or  box- 
form,  (see  Fig.  260)  do  not  need  this  lateral  support. 

259.  Set  of  Moving  Loads. — When  a  locomotive  passes  over 
a  number  of  parallel  prismatic  girders,  each  one  of  which 
experiences  certain  detached  pressures  corresponding  to 
the  different  wheels,  by  selecting  any  definite  position  of 
the  wheels  on  the  span,  we  may  easily  compute  the  reac- 
tions of  the  supports,  then  form  the  shear  diagram,  and 
finally  as  in  §  243  obtain  the  max.  moment,  Mm9  and  the 


FLEXURE.      MOVING   LOADS. 


1>99 


max.  shear  e/m,  for  this  particular  position  of  the  wheels. 
But  the  values  of  Mm  and  Jin  for  some  other  position  may 
be  greater  than  those  just  found.  We  therefore  inquire 
which  will  be  the  greatest  moment  among  the  infinite 
number  of  (Jf^)'»  (one  for  each  possible  position  of  the 
wheels  on  the  span).  It  is  evident  from  Fig.  236  from  the 
nature  of  the  moment  diagram,  that  when  the  pressures  or 
loads  are  detached,  the  Mm  for  any  position  of  the  loads, 
which  of  course  are  in  this  case  at  fixed  distances  apart, 
must  occur  under  one  of  the  loads  (i.e.  under  a  wheel). 
We  begin  .•.  by  asking :  What  is  the  position  of  the  set  of 
moving  loads  when  the  moment  under  a  given  wheel  is 
greater  than  will  occur  under  that  wheel  in  any  other  po- 
sition? For  example,  in  Fig.  262,  in  what  position  of  the 


0 

<  £  1  —  .. 
ft                          •  — 

'   O  Cp 

£KV 

1 

f             T 

fe 

3 

HTl 

FIG  262. 

loads  PX,  P2,  etc.  on  the  span  will  the  moment  M2,  i.e., 
under  P2,  be  a  maximum  as  compared  with  its  value  under 
P2  in  any  other  position  on  the  span.  Let  R  be  the  resultant 
of  the  loads  which  are  now  on  the  span,  its  variable  distance 
from  0  be  =  x,  and  its  fixed  distance  from  P2  =  #';  while 
a,  b,  c,  etc.,  are  the  fixed  distances  between  the  loads 
(wheels).  For  any  values  of  x  ,  as  the  loading  moves 
through  the  range  of  motion  within  which  no  wheel  of  the 
set  under  consideration  goes  off  the  span,  and  no  new 

wheel  comes  on  it,  we  have  Rl=-.—  R,  and  the  moment 
under  P2 

=M2=R,  [7-(^-a')]_  P3&-P4(&+c) 

T> 

(1) 


300  MECHANICS  OF  ENGINEERING. 

In  (1)  we  have  M2  as  a  function  of  x,  all  the  other  quan- 
tities in  the  rightjhand  member  remaining  constant  as  the 
loading  moves  ;  x  may  vary  from  ~x=a+a'  to 
x=l—  (c+&—  a').     For  a  max.  M2t  we  put  dM2-±-dx=0,  i.  e. 

~(l-2x+a')=Q  .-.  a  (for  Max  M.2}  = 


(For  this,  or  any  other  value  of  x,  d-^M^dx2  is  negative, 
hence  a  maximum  is  indicated).  For  a  max.  M2,  then,  R 
must  be  as  as  far  (^2  a')  ofr  one  side  of  the  middle  of  the 
span  as  P2  is  on  the  other  ;  i.e.,  as  the  loading  moves,  the 
moment  under  a  given  wheel  becomes  a  max.  when  that 
wheel  and  the  centre  of  gravity  of  all  the  loads  (then  on 
the  span)  are  equi-distant  from  the  middle  of  the  span. 

In  this  way  in  any  particular  case  we  may  find  the 
respective  max.  moments  occurring  under  each  of  the 
wheels  during  the  passage,  and  the  greatest  of  these  is  the 
Mm  to  be  used  in  the  equation  Mm  =HfI-~e  for  safe  loading.* 

*  As  to  the  shear  J9  for  a  given  position  of  the  wheels  this 
will  be  the  greatest   at   one   or  the   other   support,  and 
equals  the  reaction  at  that  support.     When  the  load  moves 
toward  either  support  the  shear  at  that  end  of  the  beam 
evidently  increases  so  long  as  no  wheel  rolls  completely 
over  and  beyond  it.      To  find  «/max.,  then,  dealing  with 
each  support  in  turn,  we  compute  the  successive  reactions 
at  the  support  when  the  loading  is  successively  so  placed 
that  consecutive  wheels,  in  turn,  are  on  the  point  of  roll- 
ing off  the  girder  at  that  end  ;  the  greatest  of  these  is  the 
max.  shear,  Jm.     As  the  max.  moment  is  apt  to  come  under 
the  heaviest  load  it  may  not  be  necessary  to  deal  with 
more  than  one  or  two  wheels  in  finding  Mm. 

EXAMPLE.  —  Given  the  following  wheel  pressures, 

A<  .  .  8'  .  .  >B<  .  .  5'  .  .  >C<  .  .  4  .  .  <D 
4  tons.  6  tons.  6  tons.  5  tons. 

on  one  rail  which  is  continuous  over  a  girder  of  20  ft.  span, 
under  a  locomotive. 

*  Since  this  maybe  regarded  as  a  case  of  "sudden  application"  of  a  load,  it  is 
customary  to  make  R'  much  smaller  than  for  a  dead  load;  from  one-third  to  one-half 
smaller. 


FLEXURE.      MOVING   LOADS. 


301 


1.  Required  the  position  of  the  resultant  of  A,  B,  and  C  5 

2.  •       "  "  "  "        A,  B,  (7,  and  D ; 

3.  "  "  "  "  B,  (7,  and  D. 
4  In  what  position  of  the  wheels  on  the  span  will  the 

moment  under  B  be  a  max.  ?    Ditto  for  wheel  C?    Required 
the  value  of  these  moments  and  which  is  Mm  ? 

5.  Required  the  value  of  «7m,  (max.  shear),  its  location  and 
the  position  of  loads. 

Results.— (1.)  7.8'  to  right  of  A.  (2.)  10'  to  right  of  A. 
(3.)  4.4'  to  right  of  B.  (4.)  Max.  M*  =  1,273,000  inch  Ibs. 
with  all  the  wheels  on  ;  Max.  MG  =  1,440,000  inch-lbs.  with 
wheels  B,  C,  and  D  on.  (5.)  Jm  =  13.6  tons  at  right  sup- 
port with  wheel  D  close  to  this  support. 

260.  Single  Eccentric  Load. — In  the  following  special  cases 
of  prismatic  beams,  peculiar  in  the  distribution  of  the 

loads,  or  mode  of  support,  or  both, 
the  main  objects  sought  are  the 
values  of  the  max.  moment  Mm,  for 
use  in  the  equation 

7?/r 
Mm=—  (see  §239); 


and   of    the 


e 
max. 


shear  Jm.  from 


I 

FIG.  263. 


which  to  design  the  web  riveting 
in  the  case  of  an  I  or  box-girder. 
The  modes  of  support  will  be  such 
that  the  reactions  are  independent 
of  the  form  and  material  of  the 
beam  (the  weight  of  beam  being 
neglected).  As  before,  the  flexure  is  to  be  slight,  and  the 
forces  are  all  perpendicular  to  the  beam. 

The  present  problem  is  that  in  fig.  263,  the  beam  being 
prismatic,  supported  at  the  ends,  with  a  single  eccentric 
load,  P.  We  shall  first  disregard  the  weight  of  the  beam 
itself.  Let  the  span  =^+1^  First  considering  the  whole 
beam  free  we  have  the  reactions  Rl  =  PI,  +  I  and  R2  = 
*•  'i  ~^~  ^. 

Making  a  section  at  m  and  having  Om  free,  x  being  <  121 
2  (vert,  compons.)  =  0  gives 


302  MECHANICS   OF   ENGINEEBING. 

R2  —  J=0,  i.e.,  J=B2  ; 
while  from  2  (mom.)m==0  we  have 

^-Z-R2x=  0  .'.M  =  Rx= 


I 

These  values  of  J  and  M  hold  good  between  0  and  C,  J 
being  constant,  while  M  is  proportional  to  x.  Hence  for 
0  C  the  shear  diagram  is  a  rectangle  and  the  moment  dia- 
gram a  triangle.  By  inspection  the  greatest  M  f  or  00  is 
for  x  =  Z2,  and  =  PIJ2  •*-  ^  This  is  the  max.  M  for  the 
beam,  since  between  C  and  B,  M  is  proportional  to  the  dis- 
tance of  the  section  fronp.  B. 


.  .m  --  —         —  ---  —      .     .     . 

is  the  equation  for  safe  loading. 

J  =  Rl  in  any  section  along  OB,  and  is  opposite  in  sign 
to  what  it  is  on  0(7;  i.e.,  practically,  if  a  dove-tail  joint 
existed  anywhere  on  OC  the  portion  of  the  beam  on  the 
right  of  such  section  would  slide  downward  relatively  to 
the  left  hand  portion  ;  but  vice  versa  on  CB. 

Evidently  the  max.  shear  Jm  =  Rl  or  R>,  as  12  or  ^  is  the 
greater  segment. 

It  is  also  evident  that  for  a  given  span  and  given  beam 
the  safe  load  P,  as  computed  from  eq.  (1)  above,  becomes 
very  large  as  its  point  of  application  approaches  a  sup- 
port ;  this  would  naturally  be  expected  but  not  without 
limit,  as  the  shear  for  sections  between  the  load  and  the 
support  is  equal  to  the  reaction  at  the  near  support  and 
may  thus  soon  reach  a  limiting  value,  when  the  safety  of 
the  web  or  the  spacing  of  the  rivets,  if  any,  is  considered. 

Secondly,  considering  the  weight  of  the  beam,  or  any 
uniformly  distributed  loading,  weighing  w  Ibs.  per  unit  of 
length  of  beam,  in  addition  to  P,  Fig.  264,  we  have  the 
reactions  • 


Let  ?2  be  >^;   then  for  a  portion  Om  of  length 
moments  about  m  give 


FLEXURE.      SPECIAL   PROBLEMS. 


303 


x—0 

e  ^'2~ 

Le.,  on  OC,  M=E2x  —  y2  wx2  .        .        .        .         (2) 

Evidently  for  x  =  0  (i.e.  at  0)  M  =  0,  while  for  »  =  4  (i.e. 
ut  (7)  we  have,  putting  w  =  W  -f-  Z 


It  remains  to  be  seen  whether  a  value  of  Jf  may  not  exist 
in  some  section  between  0  and  (7,  (i.e.,  for  a  value  of  cc 
<12  in  eq.  (2)),  still  greater  than  Mc.  Since  (2)  gives  M  as 
a  continuous  function  of  x  between  0  and  6Y,  we  put 
dJf  -r-  dx  =  0,  and  obtain,  substituting  the  value  of  the  con- 
stants RZ  and  w, 


(  max. 

=0  .-.  #n  -j  for  J!/  or 
(  min. 

M  max.,   since 


='+ 


(4) 


for 


FIG.  264. 


-  dx*  is  negative 
when  this  value  of  x  is  sub- 
stituted. If  the  particular 
_j  value  of  x  given  by  (4)  is 
<?2,  the  corresponding  value 
of  M  (call  it  Mn)  from  eq. 
(2)  will  occur  on  0(7  and  will 
be  greater  than  MG  (Dia- 
grams II.  in  fig.  264  show 
this  case) ;  but  if  #n  is  >  12, 
we  are  not  concerned  with 
the  corresponding  value  of 
Mj  and  the  greatest  M  on  OC 
would  be  Mc. 

For  the  short  portion  BC, 
which  has  moment  and  shear 
diagrams  of  its  own  not  con- 
tinuous with  those  for  0(7,  it 
may  easily  be  shown  that 
Mc  is  the  greatest  moment  of 
any  section.  Hence  the  M 


304 


MECHANICS   OF   ENGINEERING. 


max.,  or  Mm,  of  the  whole  beam  is  either  Jfc  or  Mnt 
according  as  xn  is  >  or  <  12.  This  latter  criterion  may  be 
expressed  thus,  [with  12  —  *£  I  denoted  by  /3,  the  distance 
of  P  from  the  middle  of  the  span]  : 


and  since  from  (4)  and  (2) 


The  equation  for  safe  loading  is 


and 


- 
W 


.        .        .        .     (6) 

Seeeqs.  (3)  and  (5) 
for  M. c  and  Afu 

If  either  P,  Wt  Z3,  or  ^  is  the  unknown  quantity  sought,  the 
criterion  of  (6)  cannot  be  applied,  and  we  .*.  use  both  equa- 
tions in  (6)  and  then  discriminate  between  the  two  results. 
The  greatest  shear  is  Jm==  ^ ,  in  Fig.  264,  where  12  is 


261.  Two  Equal  Terminal  Loads,  Two  Symmetrical  Supports 
Fig.  265.  [Same  case  as  in  Fig.  231,  §  238].  Neglect 
weight  of  beam.  The  reaction  at  each  support  being  =P, 
(from  symmetry),  we  have  for  a  free  body  Om  with  x  <  Zj 

f^O  .-.  M=Px  (li 


(2) 


while  where  x  >  ^  and  < 


Px-P  (x—l,)—  ^ 


That  is,  see  (1),  M  varies  directly  with  x  between  0  and  (7, 
while  between  C  and  D  it  is  constant.  Hence  for  safe 
loading 

(3) 


FLEXURE.      SPECIAL   PROBLEMS.  305 


The  construction  of  the 
moment  diagram  is  evident 
from  equations  (1)  and  (2). 

As  for  J",  the  shear,  the 
same  free  bodies  give,  from 
2  (vert,  f orces)=  0. 

On  00    .  J=P    .    .    .    (4) 
On  CD    .  J=P— P=zero(5) 


(4)  and  (5)  might  also  be  ob- 
tained  from  (1)  and  (2)  by 
writing  J=d  M-±dx,  but  the 

former  method  is  to  be  preferred  in  most  cases,  since  the 
latter  requires  M  to  be  expressed  as  a  function  of  x  while 
the  former  is  applicable  for  examining  separate  sections 
without  making  use  of  a  variable. 

If  the  beam  is  an  I-beam,  the  fact  that  J  is  zero  any- 
where on  0  D  would  indicate  that  we  may  dispense  with 
a  web  along  0  D  to  unite  the  two  flanges;  but  the  lower 
flange  being  in  compression  and  forming  a  "  long  column  " 
would  tend  to  buckle  out  of  a  straight  line  if  not  stayed  by 
a  web  connection  with  the  other,  or  some  equivalent  brac~ 
ing. 

262,  Uniform  Load  over  Part  of  the  Span.  Two  End  Supports. 
Fig.  266.  Let  the  load  =  W,  extending  from  one  support 
over  a  portion  =  c,  of  the  span,  (on  the  left,  say,)  so  that 
W=  we,  w  being  the  load  per  unit  of  length.  Neglect 
weight  of  beam.  For  a  free  body  Om  of  any  length 
so  <  0  B  (i.e.  <  c),  I  momsm=0  gives 

pi       WX*         -n           *        T[jr       T>             WX*  /tx 

£-H ^ ^=0  .'.M^fitX _         .       .      .      (1) 

€  A  & 

which  holds  good  for  any  section  on  0  B.  As  for  sections 
on  B  (7  it  is  more  simple  to  deal  with  the  free  body  m'O, 
of  length 

x'  <  C  B  from  which  we  have  M=Ez  x!    .     .   (2) 


306 


MECHANICS   OF   ENGINEERING. 


FIG.  266. 


On  OB 

while  on  B  C 


which  shows  the  moment 
curve  for  B  C  to  be  a  straight 
line  DC t  tangent  at  D  to  the 
parabola  0'  D  representing 
eq.  (1.)  (If  there  were  a  con- 
centrated load  at  By  CD 
would  meet  the  tangent  at 
D  at  an  angle  instead  of  co- 
inciding with  it ;  let  the  stu- 
dent show  why,  from  the 
shear  diagram). 

The  shear  for  any  value  of 
a?  on  0  B  is  : 

.  .  J=Rl — ivx  ...  (3) 
,  .  J=  J?2==  constant  .  (4) 


The  shear  diagram  is  constructed  accordingly. 

To  find  the  position  of  the  max.  ordinate  of  the  para- 
bola, (and  this  from  previous  statements  concerning  the 
tangent  at  the  point  D  must  occur  on  0  J5,  as  will  be  seen 
and  will .-.  be  the  Mm  for  the  whole  beam)  we  put  J"=0  in 
eq  (3)  whence 


(5) 


and  is  less  than  c,  as  expected.  [The  value  of  J?1=^ 

t 

*=(wc  -r-l)  (I — §),  (the  whole  beam  free)  has  been  substi- 
tuted].    This  value  of  x  substituted  in  eq.  (1)  gives 

iw  —n       /  CY     /  R'l 

is  the  equation  for  safe  loading. 

The  max.  shear  Jm  is  found  at  0  and  is  =  B^  which  is 
evidently  >.#2,  at  C. 


FLEXURE.      SPECIAL   PROBLEMS. 


307 


263.  Uniform  Load  Over  Whole  Length  With  Two  Symmetric? 
Supports.  Fig.  267. — With  the  notation  expressed  in  the  fig- 
ure, the  following  results  may  be  obtained,  after  having 
divided  the  length  of  the  beam  into  three  parts  for  sepa- 
rate treatment  as  necessitated  by  the  external  forces,  which 
are  the  distributed  load  W,  and 
and  the  two  reactions,  each  = 
y2  W.  The  moment  curve  is 
made  up  of  parts  of  three  dis- 
tinct parabolas,  each  with  its 
axis  vertical.  The  central  par- 
abola may  sink  below  the  hori- 
zontal axis  of  reference  if  the 
supports  are  far  enough  apart, 
in  which  case  (see  Fig.)  the  elas-  '  FlQ  267 

tic  curve  of  the  beam  itself  becomes  concave  upward  be- 
tween the  points  E  and  F  of  "  contrary  flexure."  At  each 
of  these  points  the  moment  must  be  zero,  since  the  radius 
of  curvature  is  oo  and  M  =  El  -r-  p  (see  §  231)  at  any  sec- 
tion ;  that  is,  at  these  points  the  moment  curve  crosses  its 
horizontal  axis. 

As  to  the  location  and  amount  of  the  max.  moment  Mm9 
inspecting  the  diagram  we  see  that  it  will  be  either  at  H9 
the  middle,  or  at  both  of  the  supports  B  and  C  (which  from 
symmetry  have  equal  moments),  i.e.,  (with  I  =  total  length,) 


M 


r     A      IT* 
\  and/.  —   == 

e  J 


W. 


or 


.....  at  IT 
at  B  and  C 


according  to  which  is  the  greater  in  any  given  case ;  i.e. 
according  as  12  is  >  or  <  ^  ^/§m 

The  shear  close  on  the  left  of  B  =  wllf  while  close  to  the 
right  of  B  it  =  ^  W  —  w\.  (It  will  be  noticed  that  in  thia 
case  since  the  beam  overhangs,  beyond  the  support,  the 
shear  near  the  support  is  not  equal  to  the  reaction  there, 
as  it  was  in  some  preceding  cases.) 


308 


MECHANICS   OF   ENGINEERING. 


Hence  J 


=  j  ^  ^_^  | 


according  as  I, 


> 


264,  Hydrostatic  Pressure  Against  a  Vertical  Plank.  —  From 
elementary  hydrostatics  we  know  that  the  pressure,  per 
unit  area,  of  quiescent  water  against  the  vertical  side  of  a 
tank,  varies  directly  with  the  depth,  x,  below  the  surface, 
and  equals  the  weight  of  a  prism  of  water  whose  altitude 
=  a?,  'and  whose  sectional  area  is  unity.  See  Fig.  268. 


FIG.  268. 


plank  is  of  rectangular  cross  section,  its  constant 
breadth,  —  b,  being  r~  to  the  paper,  and  receives  no  sup- 
port except  at  its  two  extremities,  0  and  B,  0  being  level 
with  the  water  surface.  The  loading,  or  pressure,  per  unit 
of  length  of  the  beam,  is  here  variable  and,  by  above  defini- 
nition,  is  =  w=  fxb,  where  f  =  weight  of  a  cubic  unit 
(i.e.  the  heaviness,  see  §  7)  of  water,  and  x  =  Om  =  depth 
of  any  section  m  below  the  surface.  The  hydrostatic  pres- 
sure on  dx  =  wdx.  These  pressures  for  equal  dx's,  vary 
as  the  ordinates  of  a  triangle  ORiB. 

Consider  Om  free.  Besides  the  elastic  forces  of  the  ex- 
posed section  m,  the  forc*es  acting  are  the  reaction  RQ,  and 
the  triangle  of  pressure  OEm.  The  total  of  the  latter  is 


J 


^  —      wdx  = 


xdx  — 


and  the  sum  of  the  moments  of  these  pressures  about  m  is 
equal  to  that  of  their  resultant  ( =  their  sum,  since  they 

x2    x 

are  parallel)  about  m,  and  .*.  =  ?b  -—.  m  - 

2     3 


FLEXURE.      SPECIAL   PROBLEMS.  309 

[From  (1)  when  x  =  I,  we  have  for  the  total  water  pres- 
sure on  the  beam  W{  =  fb  =-  and  since  one-third  of  this 

will  be  borne  at  0  we  have  RQ  ~/1s  r^2.] 

Now  putting  2(  moms,  about  the  neutral  axis  of  m)=0, 
for  Om  free,  we  have 

T*         *10  / 

o       e 

(which  holds  good  from  x  =  0  to  x  =  I).  From  2  (horiz. 
forces)  =  0  we  have  also  the  shear 

as  might  also  have  been  obtained  by  differentiating  (2), 
since  J  =  dM  -r-  dx.  By  putting  J  =  0  (§  240,  corollary) 
77Q  have  for  a  max.  M^x  —  l-^-  \/3,  which  is  less  than  I 
and  hence  is  applicable  to  the  problem.  Substitute  this 
in  eq.  2,  and  reduce,  and  we  have 

Sfl    ..     .       E'l    1      1         "L  (4) 


as  the  equation  for  safe  loading. 

265.  Example.  —  If  the  thickness  of  the  plank  is  h,  re- 
quired h  =  ?,  if  E'  is  taken  =  1,000  Ibs.  per  sq.  in.  for 
timber  (§  251),  and  I  =  6  feet.  For  the  inch-pound-second 
system  of  units,  we  must  substitute  R'  =  1,000  ;  I  =  72 
inches  /  7*  =  0.036  Ibs.  per  cubic  inch  (heaviness  of  water 
in  this  system  of  units);  while  J  =bh*  -5-  12,  (§  247),  and  e 
=  ^  h.  Hence  from  (4)  we  have 

1000  ta3    0.0366x72*  .  . 

=  —  »  •*•  A—o.16  .•.  A  =  2.27  in. 


It  will  be  noticed  that  since  x  for  Mm  =  I  -f-  v7^,  and  not 
%  I,  Mm  does  not  occur  in  the  section  opposite  the  resul- 
tant of  the  water  pressure  ;  see  Fig.  268.  The  shear  curve 
is  a  parabola  here  ;  eq.  (3). 


310  MECHANICS   OF   ENGINEERING. 

266.  The  Four  x-Derivatives  of  the  Ordinate  of  the  Elastic  Curve 
— If  y  =  func.  (x)  is  the  equation  of  the  elastic  curve  for 
any  portion  of  a  loaded  beam,  on  which  portion  the  load 
per  unit  of  length  of  the  beam  is  w  =  either  zero,  (Fig. 
234)  or  =  constant,  (Fig.  235),  or  =  a  continuous  func.  (x) 
^  (as  in  the  last  §),  we  may  prove,  as  fol- 

lows, that   w  =  the  x-derivative  of  the 
_^  shear.     Fig.  269.     Let  N  and  If  be  two 

-  consecutive  cross-sections   of    a   loaded 
—  beam,  and  let  the   block  between  them, 

•  bearing  its  portion,  wdx,  of  a  distributed 
load,   be   considered   free.      The   elastic 


Ui 


forces  consist  of  the  two  stress-couples 
FIG.  269.  (tensions  and  compressions)  and  the  two 

shears,  J"and  J  +  dJ>  e&7  being  the  shear-increment  conse- 
quent upon  x  receiving  its  increment  dxt  By  putting 
^(vert.  components)  =  0  we  have 


j+dJ—wdx—J=0  .-. 

dx 

Q.  E.  D.    But  J  itself  =  dM  +  dx,  (§  240)  and 

M  =  [d2y  -T-  dx2]  EL    By  substitution,  then,  we  have  the 

following  relations  : 


=  ordinate  at  any  point  of  the  elastic  curve  (1) 
_^=  a  =  slope  at  any  point  of  the  elastic  curve     .     .      (2) 

>-72«  y 

El  -r-^-=  M  =  ordinate  (to  scale)  of  the  moment  curve  (3) 

-r  $"u       ,1      1-          T      (  the  ordinate  (to  scale)  )          ,A\ 
EI^  =  ^  '          I  of  the  shear  diagram    }  '  '  & 

(  the  load  per  unit  of  length   } 

El  —  ^  =  w  =  •<  of  beam  =  ordinate  (to  scale)  y      .     (5) 
&&  (  of  a  curve  of  loading.  ) 

If,  then,  the  equation  of  the  elastic  curve  (the  neutral  line 
of  the  beam  itself  ;  a  reality,  and  not  artificial  like  the 


FLEXURE.       SPECIAL   PROBLEMS.  311 

other  curves  spoken  of)  is  given  ;  we  may  by  successive 
differentiation,  for  a  prismatic  and  homogeneous  beam  so 
that  both  E  and  /  are  constant,  find  the  other  four  quan- 
tities mentioned. 

As  to  the  converse  process,  (i.e.  having  given  w  as  a 
function  of  x,  to  find  expressions  for  Jt  M  and  y  as  func- 
tions of  x)  this  is  more  difficult,  since  in  taking  the 
^-anti-derivative,  an  unknown  constant  must  be  added  and 
determined.  The  problem  just  treated  in  §  264,  however, 
offers  a  very  simple  case  since  w  is  the  same  function  of 
x,  along  the  ivhole  beam,  and  there  is  therefore  but  one  elas- 
tic curve  to  be  determined. 

We  .*.  begin,  numbering  backward,  with 

WT  d^  -     -rbr\  sinC6  w  =  ?bx>  S6e  1 

~dtf  T     \  last  §  and  Fig.  268  ('        • 

[N.  B.  —  This  derivative  (dJ-^dx)  is  negative  since 
dx  have  contrary  signs.] 

A  (shear=)E 

But  writing  out  this  equation  for  a?=0,  i.e.  for  the  point 
0,  where  the  shear=7?0,  we  have  11$=  0  -{-Const..:  Const.  = 
RQt  and  hence  write 


(Shear) 


Again  taking  the  cc-anti-derivative  of  both  sides 

(Moment=)EI^=—rb^+Il0x+(Const.=0)  .    (3o) 
a'Ju  o 

[At  0,  x=Q  also  M,  .-.   Const.  =0],     Again, 


At  0,  where  x=0  dy-^dx=c^=\hQ  unknown  slope  of  the 
elastic  line  at  0,  and  hence  Gr=EIa^ 

>    .    .    .    (2a) 


312  MECHANICS  OF  ENGINEERING. 

Passing  now  to  y  itself,  and  remembering  that  at  0,  both 
y  and  x  are  zero,  so  that  the  constant,  if  added,  would= 
zero,  we  obtain  (inserting  the  value  of  £0  from  last  §) 

-r*~+rW~ 

the  equation  of  the  elastic  curve.  This,  however,  contains 
the  unknown  constant  «0=the  slope  at  0.  To  determine 
«0  write  out  eq.  (la)  for  the  point  B,  Fig.  268,  where  x  is 
known  to  be  equal  to  lt  and  y  to  be  =  zero,  solve  for  «0, 
and  insert  its  value  both  in  (la)  and  (2a).  To  find  the 
point  of  max.  y  (i.e.,  of  greatest  deflection)  in  the  elastic 
curve,  write  the  slope,  i.e.  dy  -f-  dx,  =  zero  [see  eq.  2a]  and 
solve  for  x  ;  four  values  will  be  obtained,  of  which  the  one 
lying  between  0  and  I  is  obviously  the  one  to  be  taken. 
This  value  of  x  substituted  in  (la)  will  give  the  maximum 
deflection.  The  location  of  this  maximum  deflection  is 

(2 
X*=     q 


nor  at  the  section  of  max.  moment  (x  =^f- 

The  qualities  of  the  left  hand  members  of  equations  (1) 
to  (5)  should  be  carefully  noted.  E.  g.,  in  the  inch-pound- 
second  system  of  units  we  should  have  : 

1.  y  (B,  linear  quantity)  =  (so  many)  inches. 

2.  dy+dx  (an  abstract  number)  =   (so  many)  abstract 
units. 

3.  M  (a  moment)  =  (so  many)  inch-pounds. 
4  J(a,  shear,  i.e.,  force)  =  (so  many)  pounds. 

6.  w  (force  per  linear  unit)  =  (so  many)  pounds  per  run- 
ning inch  of  beam's  length. 

As  to  the  quantities  E,  and  /,  individually,  E  is  pounds 
per  sq.  in.,  and  /has  four  linear  dimensions,  i.e.  (so  many) 
bi-quadratic  inches. 


FLEXURE       SPECIAL    PROBLEMS.  313 

267.  Resilience  of  Beam  With  End  Supports.—  Fig.  270.  If  a 
mass  whose  weight  is  G  (G  large  com- 
pared with  that  of  beam)  be  allowed  to 
fall  freely  through  a  height  =  h  upon 
the  centre  of  a  beam  supported  at  its 
extremities,  the  pressure  P  felt  by  the 
beam  increases  from  zero  at  the  first 
instant  of  contact  up  to  a  maximum  Pm,  as  already  stated 
in  §233a,  in  which  the  equation  was  derived,  dm  being 
small  compared  with  h, 


The  elastic  limit  is  supposed  not  passed.  In  order  that 
the  maximum  normal  stress  in  any  outer  fibre  shall  at  most 
be=.#/,  a  safe  value,  (see  table  §251)  we  must  put 

jy  T       ~p     7 

—  =  —  ~-   [according  to  eq.  (2)  §241,]  i.e.  in  equation  (a) 

€  4: 

above,  substitute  Pm=  [4  R'l]  +le,  which  gives 


having  put  I=F7i?  (Is  being  the  radius  of  gyration  §85) 
and  Fl=  V  the  volume  of  the  (prismatic)  beam.  From 
equation  (b)  we  have  the  energy,  Gh,  (in  ft.-lbs.,  or  inch- 
Ibs.)  of  the  vertical  blow  at  the  middle  which  the  beam  of 
Fig.  270  will  safely  bear,  and  any  one  unknown  quantity 
can  be  computed  from  it,  (but  the  mass  of  G  should  not 
be  small  compared  with  that  of  the  beam.) 

The  energy  of  this  safe  impact,  for  two  beams  of  the 
same  material  and  similar  cross-sections  (similarly  placed), 
is  seen  to  be  proportional  to  their  volumes;  while  if  further- 
more their  cross  -sections  are  the  same  and  similarly 
placed,  the  safe  Gh  is  proportional  to  their  lengths.  (These 
same  relations  hold  good,  approximately,  beyond  the  elas' 
tic  limit.) 

It  will  be  noticed  that  the  last  statement  is  just  the  re- 


314  MECHANICS   OF   ENGINEERING. 

verse  of  what  was  found  in  §245  for  static  loads,  (the 
pressure  at  the  centre  of  the  beam  being  then  equal  to 
the  weight  of  the  safe  load) ;  for  there  the  longer  the  beam 
(and  .'.  the  span)  the  less  the  safe  load,  in  inverse  ratio. 
As  appropriate  in  this  connection,  a  quotation  will  be 
given  from  p.  186  of  "  The  Strength  of  Materials  and 
Structures,"  by  Sir  John  Anderson,  London,  1884,  viz.: 

"  It  appears  from  the  published  experiments  and  state- 
ments of  the  Railway  Commissioners,  that  a  beam  12  feet 
long  will  only  support  ^  of  the  load  that  a  beam  6  feet 
long  of  the  same  breadth  and  depth  will  support,  but  that 
it  will  bear  double  the  weight  suddenly  applied,  as  in  the 
case  of  a  weight  falling  upon  it,"  (from  the  same  height, 
should  be  added) ;  "  or  if  the  same  weights  are  used,  the 
longer  beam  will  not  break  by  the  weight  falling  upon  it 
unless  it  falls  through  twice  the  distance  required  to  frac- 
ture the  shorter  beam." 

268.  Combined  Flexure  and  Torsion.    Crankshafts.     Fig.  271. 
Let    OiB  be  the  crank,  and  NOi   the  portion  projecting 
o  beyond  the  nearest  bearing 

N.  P  is  the  pressure  of  the 
connecting-rod  against  the 
crank -pin  at  a  definite  in- 
stant, the  rotary  motion  be- 
ing uniform.  Let  a=  the 
perpendicular  dropped  from 
the  axis  OOi  of  the  shaft 
upon  P,  and  1=  the  distance 
of  P,  along  the  axis  0  Ol  from 
the  cross-section  NmN'  of  the 
shaft,  close  to  the  beajing.  Let  NN'^e  a  diameter  of  this 
section,  and  parallel  to  a.  In  considering  the  portion 
NOiB  free,  and  thus  exposing  the  circular  section  Nm  N\ 
we  may  assume  that  the  stresses  to  be  put  in  on  the  ele- 
ments of  this  surface  are  the  tensions  (above  NN')  and 
the  compressions  (below  NN')  and  shears  ~|  to  NN',  due 
to  the  bending  action  of  P ;  and  the  shearing  stress  tan- 


FLEXURE.   SPECIAL  PROBLEMS. 


315 


gent  to  the  circles  which  have  0  as  a  common  centre,  and 
pass  through  the  respective  dF's  or  elementary  areas, 
these  latter  stresses  being  due  to  the  twisting  action  of  P. 
In  the  former  set  of  elastic  forces  let  p  =  the  tensile 
stress  per  unit  of  area  in  the  small  parallelopipedical  ele- 
ment m  of  the  helix  which  is  furthest  from  NN'  (the  neu- 
tral axis)  and  /=  the  moment  of  inertia  of  the  circle  about 
NN';  then  taking  moments  about  'NN'  for  the  free  body, 
(disregarding  the  motion)  we  have  as  in  cases  of  flexure 
(see  §239) 

Pi       PI-  {(,     v-Plr  (a} 

_-  ,e.,  p-  — 

[None  of  the  shears  has  a  moment  about  NN'.]  Next 
taking  moments  about  OOi,  (the  flexure  elastic  forces,  both 
normal  and  shearing,  having  no  moments  about  OOi)  we 
have  as  in  torsion  (§216) 


pj, 

r 


Par 


in  which  p*  is  the  shearing  stress  per  unit  of  area,  in  the 
torsional  elastic  forces,  on  any  outermost  dF,  as  at  m ; 
and  /p  the  polar  moment  of  inertia  of  the  circle  about  its 
centre  0. 

Next  consider  free,  in  Fig.  272,  a  small  parallelepiped 
taken  from  the  helix  at  m  (of  Fig.  271.)  The  stresses  [see 
§209]  acting  on  the  four  faces  |—  to  the  paper  in  Fig.  272 
are  there  represented,  the  dimensions  (infinitesimal)  being 
n  II  to  NN',  b  ||  to  OOi,  and  d  ~\  to  the  paper  in  Fig.  272. 


Fig.  273. 


316  MECHANICS   OF   ENGINEERING. 

By  altering  the  ratio  of  b  to  n  we  may  make  the  angle  0 
what  we  please.  It  is  now  proposed  to  consider  free  the 
triangular  prism,  GHT,  to  find  the  intensity  of  normal 
stress  q,  per  unit  of  area,  on  the  diagonal  plane  GH,  (of 
length  =  c,)  which  is  a  bounding  face  of  that  triangular 
prism.  See  Fig,  273.  By  writing  21  (compons.  in  direc- 
tion of  normal  to  6r-£T)=0,  we  shall  have,  transposing, 

qcd=pnd  sin  0-\-pJ)d  sin  d+psnd  cos  6  ;    and  solving  for  q 
q=p  ~   sin  0+p8  hrsin  0+j.  cos  d  J  ;       .     (I) 

but  n  :  c=sin  d  and  b  :  c=  cos  0    .•. 

q=p  sin20-f-p82  sin  d  cos  0        .        .        (2) 

This  may  be  written  (see  eqs.  63  and  60,  O.  W.  J.  Trigo- 
nometry) 

m20        .        .        (3) 


As  the  diagonal  plane  GH  is  taken  in  different  positions 
(i.e.,  as  6  varies),  this  tensile  stress  q  (Ibs.  per  sq.  in.  for 
instance)  also  varies,  being  a  function  of  0,  and  its  max. 
value  may  be  >p.  To  find  d  for  q  max.  we  put 


^    ,  -p  sin  20+2ps  cos  20,          .        .       (4) 
-=•0,  and  obtain:    tan  [2(0  for  q  max)]  ---  ^  .    .        .     (5) 

Call  this  value  of  0,  0'.     Since  tan  20'  is  negative,  20'  lies 
either  in  the  second  or  fourth  quadrant,  and  hence 

Bin20;-±    y2^8          and  cos  20'-  T     y  ^  (6) 

vy+4$  vy+4$ 

[See  equations  28  and  29  Trigonometry,  O.  W.  J.]     The 


FLEXURE.   CRANK  SHAFT.  317 

tipper  signs  refer  to  the  second  quadrant,  the  lower  to  the 
fourth.     If  we  now  differentiate  (4),  obtaining 


(7) 


we  note  that  if  the  sine  and  cosine  of  the  [20']  of  the  2nd 
quadrant  [upper  signs  in  (6)]  are  substituted  in  (7)  the  re- 
sult is  negative,  indicating  a  maximum  ;  that  is,  q  is  a  max- 
imum for  6=  the  6'  of  eq.  (6)  when  the  upper  signs  are  taken 
(2nd  quadrant).  To  find  q  max.,  then,  put  0'  for  0  in  (3) 
substituting  from  (6)  (upper  signs).  We  thus  find 

q  max  =^[p+Vp2+4ps2.]         .         .        (8) 

A  similar  process,  taking  components  parallel  to  GH9 
Fig.  273,  will  yield  ^s  max.,  i.e.,  the  max.  shear  per  unit  of 
area,  which  for  a  given  p  and  ps  exists  on  the  diagonal 
plane  GH  in  any  of  its  possible  positions,  as  0  varies. 
This  max.  shearing  stress  is 

?8max  =  Xvy-i-4pg2        .        .          (9) 

In  the  element  diametrically  opposite  to  m  in  Fig.  271,  p 
is  compression  instead  of  tension  ;  q  maximum  will  also 
be  compression  but  is  numerically  the  same  as  the  q  max. 
of  eq.  8. 

269.  Example.—  In  Fig.  271  suppose  P=2  tons  =  4,000 
Ibs.,  a=6  in.,  1=5  in.,  and  that  the  shaft  is  of  wrought 
iron.  Kequired  its  radius  that  the  max.  tension  or  com- 
pression may  not  exceed  72'  =12,000  Ibs.  per  sq.  in.;  nor  the 
max.  shear  exceed  /S"=7,000  Ibs.  per  sq.  in.  That  is,  we' 
put  g=12,000  in  eq.  (8)  and  solve  for  r:  also  ^=7,000  in 
(9)  and  solve  for  r.  The  greater  value  of  r  should  be 
taken.  From  equations  (a)  and  (b)  we  have  (see  §§  219  and 
247  for  L  and  2) 


318 


MECHANICS  OF  ENGINBBEIN0. 


which  in  (8)  and  (9)  give 
,.P 


max. 


(8a) 


and 


max.  q, = %  — .  V(4Z)2-|-4(2a)2 


.(9a) 


With  max.  g= 12,000,  and  the  values  of  P,  a,  and  Z,  already 
given,  (units,  inch  and  pound)  we  have  from  (8a),  ^=2.72 
cubic  inches  .*.  r=1.39  inches. 

Next,  with  max.  ^=7,000 ;  P,  a,  and  I  as  before ;  from 
(9a),  r3=2.&4  cubic  inches  .•.  r=1.41  inches. 

The  latter  value  of  r,  1.41  inches,  should  be  adopted.  It 
is  here  supposed  that  the  crank-pin  is  in  such  a  position 
(when  P=4,000  Ibs.,  and  a=6  in.)  that  q  max.  (and  qs 
max.)  are  greater  than  for  any  other  position  ;  a  number 
of  trials  may  be  necessary  to  decide  this,  since  P  and  a  are 
different  with  each  new  position  of  the  connecting  rod.  If 
the  shaft  and  its  connections  are  exposed  to  shocks,  H  and 
#'  should  be  taken  much  smaller. 


P.' 


270.  Another  Example  of  combined  torsion  and  flexure  is 

shown  in  Fig.   274.      The 
PB    work  of  the  working  force 
P!( vertical  cog-pressure)  is 
B  expended  in  overcoming  the 
resistance  (another  vertical 
cog-pressure)  Q^ 

That  is,  the  rigid  body 
consisting  of  the  two  wheels  and  shaft  is  employed  to 
transmit  power,  at  a  uniform  angular  velocity,  and  since 
it  is  symmetrical  about  its  axis  of  rotation  the  forces  act- 
ing on  it,  considered  free,  form  a  balanced  system.  (See 
§  114).  Hence  given  PI  and  the  various  geometrical  quan- 


FIG.  274. 


FLEXURE.       CRANK   SHAFT.  319 

titles  alt  bit  e^c.,  we  may  obtain  QLt  and  the  reactions  P0  an(l 
PB,  in  terms  of  Plt  The  greatest  moment  of  flexure  in  the 
shaft  will  be  either  P0li9  at  (7;  or  PE13,  at  D.  The  portion 
CD  is  under  torsion,  of  a  moment  of  torsion  =  P1a1=  Qjb^ 
Hence  we  proceed  as  in  the  example  of  §  269,  simply  put- 
ting PQ?!  (or  PBZ3,  whichever  is  the  greater)  in  place  of  PI, 
and  P^  in  place  of  Pa.  We  have  here  neglected  the 
weight  of  the  shaft  and  wheels.  If  Qi  were  an  upward  ver- 
tical force  and  hence  on  the  same  side  of  the  sh J!t  as  P19 
the  reactions  P0  and  PB  would  be  less  than  before,  and  one 
or  both  of  them  might  be  reversed  in  direction. 

270a.  Web  of  I-Beam.  Maximum  Stresses  on  an  Oblique 
Plane. — The  analysis  of  pp.  315,  316,  etc.,  also  covers  the 
case  of  an  element  of  the  web  of  a  horizontal  I-beam  under 
stress,  when  this  element  is  taken  near  the  point  of  junction 
with  the  flange.  Supposing  that  the  thickness  of  web  has 
already  been  designed  such  that  the  shearing  stress  on  the 
vertical  (and  therefore  also  on  the  horizontal)  edges  of  such 
an  element  is  at  rate  of  4000  Ibs.  per  sq.  inch  ;  and  that  the 
horizontal  tension  at  each  end  of  this  element  (since  it  is 
not  far  from  the  outer  fibre  of  the  whole  section)  is  at  rate 
of  10,000  Ibs.  per  sq.  in.;  we  note  that  Fig.  272  gives  us  a 
side  view  of  this  element,  with  p3  =  4000,  and  p  —  10,000, 
Ibs.  per  sq.  inch.  GTis  the  lower  edge  of  the  upper  flange, 
corresponding  (in  an  end  view)  to  the  point  n  in  Fig.  258  on 
p.  290.  (We  here  suppose  the  upper  flange  to  be  in  tension  ; 
of  course,  an  illustration  taken  from  the  compression  side 
would  do  as  well.) 

Substitution  in  equations  (8)  and  (9)  of  p.  317  results  in 
giving  as  maximum  stresses  on  internal  oblique  planes : 

q  max.  =  11,400  Ibs.  per  sq.  in.,  tension ; 
and          q8  max.  =    6,460    "       "     "      "     shearing. 

This  shows  why  the  low  value  of  4000  Ibs.  per  sq.  in.  of 
vertical  section  is  adopted  as  a  maximum,  in  designing  the 
thickness  of  webs  of  I-beams,  by  the  N.  J.  Steel  and  Iron 
Co.;  since  the  maximum  shearing  stress  on  some  oblique 
plane  may  be  much  greater  than  that  ,on  a  vertical  plane. 
See  p.  291. 


320  MECHANICS   OF   ENGINEERING. 


CHAPTEK    IV. 


FLEXURE,    CONTINUED, 


CONTINUOUS  GIRDERS. 


271.  Definition. — A  continuous  girder,  for  present 
poses,  may  be  denned  to  be  a  loaded  straight  beam  sup- 
ported in  more  than  two  points,  in  which  case  we  can  no 
longer,  as  heretofore,  determine  the  reactions  at  the  sup- 
ports from  simple  Statics  alone,  but  must  have  recourse 
k>  the  equations  of  the  several  elastic  curves  formed  by  its 
neutral  line,  which  equations  involve  directly  or  indirect- 
ly the  reactions  sought ;  the  latter  may  then  be  found  as 
if  they  were  constants  of  integration.  Practically  this 
amounts  to  saying  that  the  reactions  depend  on  the  man- 
ner in  which  the  beam  bends  ;  whereas  in  previous  cases, 
with  only  two  supports,  the  reactions  were  independent  of 
bhe  forms  of  the  elastic  curves  (the  flexure  being  slight, 
however). 

As  an  Illustration,  if  the  straight  beam  of  Fig.  275  is  placed 
on  three  supports  0,  B,  and  (?,  at  the  same  level,  the 
reactions  of  these  supports  seem  at  first  sight  indeterm- 
inate ;  for  on  considering  the  p i 4 

whole  beam  free,  we  have  three     £37  D        J? 

unknown  quantities  and  only  B 
two   equations,  viz :    S    (vert.  FIG.  275. 

compons.)  —  0  and  S  (moms,  about  some  point)  =  0.     If 
now  0  be  gradually  lowered,  it  receives  less  and  less  pres- 


FLBXUEB.      CONTINUOUS   GIRDERS.  321 

sure,  until  it  finally  reaches  a  position  where  the  beam 
barely  touches  it ;  and  the-n  O's  reaction  is  zero,  and  B  and 
0  support  the  beam  as  if  0  were  not  there.  As  to  how 
low  0  must  sink  to  obtain  this  position,  depends  on  the 
stiffness  and  load  of  the  beam.  Again,  if  0  be  raised 
above  the  level  of  B  and  C  it  receives  greater  and  greater 
pressure,  until  the  beam  fails  to  touch  one  of  the  other 
supports.  Still  another  consideration  is  that  if  the  beam 
were  tapering  in  form,  being  stiffest  at  0,  and  pointed  at 
B  and  C,  the  three  reactions  would  be  different  from  their 
values  for  a  prismatic  beam.  It  is  therefore  evident  that 
for  more  than  two  supports  the  values  of  the  reactions  de- 
pend on  the  relative  heights  of  the  supports  and  upon  the 
form  and  elasticity  of  the  beam,  as  well  as  upon  the  load. 
The  circumstance  that  the  beam  is  made  continuous  over 
the  support  0,  instead  of  being  cut  apart  at  0  into  two 
independent  beams,  each  covering  its  own  span  and  hav- 
ing its  own  two  supports,  shows  the  significance  of  the 
term  "  continuous  girder." 

All  the  cases  here  considered  will  be  comparatively 
simple,  from  the  symmetry  of  their  conditions.  The 
beams  will  all  be  prismatic,  and  all  external  forces  (i.e. 
loads  and  reactions)  perpendicular  to  the  beam  and  in  the 
same  plane.  All  supports  at  the  same  level, 

272.  Two  Equal  Spans;  Two  Concentrated  Loads,  One  in  the  Mid- 
dle of  Each.  Span.  Prismatic  Beam. — Fig.  275.  Let  each  half- 
span  =  y2  I.  Neglect  the  weight  of  the  beam.  Required 
the  reactions  of  the  three  supports.  Call  them  fB9  P0  and 
Pc.  From  symmetry  PB  =  Pc,  and  the  tangent  to  the 
elastic  curve  at  0  is  horizontal ;  and  since  the  supports 
are  on  a  level  the  deflection  of  C  (and  B)  below  O's  tangent 
is  zero.  The  separate  elastic  curves  OD  and  DC  have  a 
common  slope  and  a  common  ordinate  at  D. 

For  the  equation  of  OD,  make  a  section  n  anywhere  be- 
tween 0  and  D,  considering  nC  a  free  body.  Fig.  276  (a) 


322 


MECHANICS   OF   ENGINEERING. 


FIG.  276. 


with  origin  and  axis  as  there  indicated.      From  I  (moms 
about  neutral  axis  of  ri)  =  0  we  have  (see  §  281) 


(1 


=0)    .   (2) 


The  constant  =  0,  for  at  0  both  x,  and  dy  -f-  dxt  =  0. 
Taking  the  x-anti-derivative  of  (2)  we  have 


Here  again  the  constant  is  zero  since  at  O,x  and  y  both  =0. 
(3)  is  the  equation  of  OD,  and  allows  no  value  of  x  <0 
or>y.  It  contains  the  unknown  force  Pc. 

For  the  equation  of  DC,  let  the  variable  section  n  be  made 
anywhere  between  D  and  (7,  and  we  have  (Fig.  276  (£A;  x 
may  now  range  between  ^l  and  I) 

(4) 


(5)' 


To  determine  (7',  put  x  =  y2l  both  in  (5)'  and  (2),  and 
equate  the  results  (for  the  two  curves  have  a  common 
tangent  line  at  D)  whence  C'  =  l/%  PI1 


(6) 


FLEXUBE.      CONTINUOUS  GIHDEB8.  323 

Hence    Ely    =  %  PPx-P,  fJtl+C"        .        .      (6)' 


At  D  the  curves  have  the  same  y,  hence  put  x  =  L  in  the 
right  hand  member  both  of  (3)  and  (6)',  equating  results, 


and  we  derive  C"=—  ± 


(6) 


which  is  the  equation  of  DC,  but  contains  the  unknown 
reaction  Pc.  To  determine  Pc  we  employ  the  fact  that  O's 
tangent  passes  through  Cy  (supports  on  same  level)  and 
hence  when  x  =  I  in  (6),  y  is  known  to  be  zero.  Making 
these  substitutions  in  (6)  we  have 


From  symmetry  PB  also  =  -£P,   while  P0  must   =  gP, 

since  PB  +  P0  -f-  Pc  =  2  P  (whole  beam  free).  [NoTE.  — 
If  the  supports  were  not  on  a  level,  but  if,  (for  instance) 
the  middle  support  0  were  a  small  distance  =  h0  below 
the  level  line  joining  the  others,  we  should  put  x  =  I  and 
y  =  —  AQ  in  eq.  (6),  and  thus  obtain  PB  =  Pc=  J£  P  -f- 

3El!~,  which  depends  on  the  material   and  form   of  the 
i> 

prismatic  beam  and  upon  the  length  of  one  span,  (whereas 
with  supports  all  on  a  levd,  PB  =  PG  =  A  P  is  independent 
of  the  material  and  form  of  the  beam  so  long  as  it  is  ho- 
mogeneous and  prismatic.)  If  P0,  which  would  then  = 
?  P  —  6  El  (kQ-7-J?\  is  found  to  be  negative,  it  shows  that 
0  requires  a  support  from  above,  instead  of  below,  to 
cause  it  to  occupy  a  position  h0  below  the  other  supports, 
i.e.  the  beam  must  be  "  latched  down  "  at  0.] 

The  moment  diagram  of  this  case  can  now  be  easily  con- 
structed ;  Fig.  277.  For  any  free  body  nC,  n  lying  in  DCt 
we  have 


324 


MECHANICS   OF  ENGINEERING. 


i.e.,  varies  directly  as  x,  un- 
til x  passes  D  when,  for  any 
point  on  DO, 


Fie.  277. 

figure)  and  at  0,  where x=l,  becomes  — ^ 


which  is  =0,  (point  of  in- 
flection of  elastic  curve) 
for  x=*/n  I  (note  that  x  is 
measured  from  C  in  this 


32 


Hence,  since  M  max.  *=~P19  the  equation  for  safe  loading 
is 


B'I=6 
e       32 


PI 


.    (7) 


t  while  on 


The  shear  at  C  and  anywhere  on 
it  *=  J^P  in  the  opposite  direction 

.     r  _np  /o\ 

The  moment  and  shear  diagrams  are  easily  constructed, 
as  shown  in  Fig.  277,  the  former  being  svmmetrical  about 
a  vertical  line  through  0,  the  latter  about  the  point  0" 
Both  are  bounded  by  right  lines. 

273.    Two  Equal   Spans.    Uniformly   Distributed   Load   Over 
w   wl  Whole  Length.  Prismatic  Beam. 

J?  W"w*        c   —Fig.  278.    Supports  B,  0, 

BM   J  UJJ_LLn  J   J  1    1     @>   on  a  ^^-     Total  load 

f^~       "^"     ui    ^T^  ~~—-\  =  2  W=  %wl  and  may  include 

|PO  j     w(i:z)      I  that  of  the  beam  ;  w  is  con- 


l  I J  J  J  |  I  stant.  As  bef pre,  from  sym- 
i  ^£-~~~^^^%  metry  PB=PC,  the  unknown 
pc|  reactions   at    the   extremi- 
278.  ties. 


FLEXURE.      CONTINUOUS   GIRDERS.  325 

On=x  ;  then  with  n  G  free,  2  moms,  about  n=  0  gives 

w 


=0]  (2) 

ax      z  o 

[Const.  =0  for  at  0  both  dy+dx  the  slope,  and  x,  are  =0] 

•'•  EIy=  - 


[Const.  =0  for  at  0  both  x  and  y  are  =0].  Equations  (1), 
(2),  and  (3)  admit  of  any  value  of  x  from  0  to  I,  i.e.,  hold 
good  for  any  point  of  the  elastic  curve  0(7,  the  loading  on 
which  follows  a  continuous  law  (viz. :  w—  constant).  But 
when  x=l,  i.e.,  at  (7,  y  is  known  to  be  equal  to  zero,  since 
0,  B  and  C  are  on  the  axis  of  X,  (tangent  at  0).  With 
these  values  of  x  and  y  in  eq.  (3)  we  have 

0=  ^-  .  *-  —  jAP<P  .'.  Pc=3£wl=$£W 
2        4 

.'.  PB=^  JPand  P0=2W—  2PC=£  W 

The  Moment  and  Shear  Diagrams  can  now  be  formed  since 

all  the  external  forces  are 
known.  In  Fig.  279  meas- 
ure x  from  (7.  Then  in  any 
section  n  the  moment  of  the 


"  stress -couple  "  is 


wx* 
2~ 


(1) 


which    holds   good  for   any 
value  of  x  on   (70,  i.e.,  from 
x=0  up  to  x=l.     By  inspec- 
PIO.  279.  tion  it  is  seen  that  for  x=Q, 

M=0  ;  and  also  for  x=ffl,  M=0,  at  the  inflection  point  G, 
beyond  which,  toward  0,  the  upper  fibres  are  in  tension 


326  MECHANICS   OF   ENGINEERING. 

the  lower  in  compression,  whereas  between  C  and  G  they 
are  vice  versa.  As  to  the  greatest  moment  to  be  found  on 
CG,  put  dM-^-dx—^  and  solve  for  x.  This  gives 


$6  W—wx=Q  .-.  [x  for  M  max.]  =}$        .       (2) 
which  in  eq.  (1)  gives 

+AJF7        .        .     (2) 


But  this  is  numerically  less  than  M0(=  —  }fa  Wl)  hence  the 
stress  in  the  outer  fibre  at  0  being 

P,=X~,      ...       (3) 
ike  equation  for  safe  loading  is 

#m    .   \vf   r.-    .  (4) 

the  same  as  if  the  beam  were  cut  through  at  0,  each  half, 
of  length  I,  retaining  the  same  load  as  before  [see  §  242  eq. 
(2)].  Hence  making  the  girder  continuous  over  the  mid- 
dle support  does  not  make  it  any  stronger  under  a  uni- 
formly distributed  load  ;  but  it  does  make  it  considerably 
stiver. 

As  for  the  shear,  «7,  we  obtain  it  for  any  section  by  tak- 
ing the  cc-derivative  of  M  in  eq.  (1),  or  by  putting  ^"(ver- 
tical forces)  =0  for  the  free  body  nC,  and  thus  have  for 
any  section  on  00 

J=#W—wx        .        .        .          (5) 

e/is  zero  for  x  =  ^l  (where  M  reaches  its  calculus  maxi- 
mum My  ;  see  above)  and  for  x=l  it  =  —  %  W  which  is  nu 
merically  greater  than  #  W,  its  value  at  C.  Hence 


FLEXURE.       CONTINUOUS   GIRDERS. 


327 


The  moment  curve  is  a  parabola  (a  separate  one  for  each 
span),  the  shear  curve  a  straight  line,  inclined  to  the  hor- 
izontal, for  each  span. 

Problem. — How  would  the  reactions  in  Fig.  278  be 
changed  if  the  support  0  were  lowered  a  (small)  distance 
A0  below  the  level  of  the  other  two  ? 

274.  Prismatic  Beam  Fixed  Horizontally  at  Both  Ends  (at 
Same  Level).  Single  Load  at  Middle.— Fig.  280.  [As  usual 

the  beam  is  understood  to 
be  homogeneous  so  that  E 
is  the  same  at  all  sections]. 
The  building  in,  or  fixing, 
of  the  two  ends  is  supposed 
to  be  of  such  a  nature  as  to 
cause  no  horizontal  con- 
FIG.  280.  straint ;  i.e.,  the  beam  does 

not  act  as  a  cord  or  chain,  in  any  manner,  and  hence  the 
sum  of  the  horizontal  components  of  the  stresses  in  any 
section  is  zero,  as  in  all  preceding  cases  of  flexure.  In 
other  words  the  neutral  axis  still  contains  the  centre  of 
gravity  of  the  section  and -the  tensions  and  compressions 
are  equivalent  to  a  couple  (the  stress-couple)  whose  mo- 
ment is  the  "  moment  of  flexure." 

If  the  beam  is  conceived  cut  through  close  to  both  wall 
faces,  and  this  portion  of  length=Z,  considered  free,  the 
forces  holding  it  in  equilibrium  consist  of  the  downward 
force  P  (the  load) ;  two  upward  shears  J0  and  Jc  (one  at 
each  section) ;  and  two  "  stress-couples  "  one  in  each  sec- 
tion, whose  moments  are  M0  and  Mc.  From  symmetry  we 
know  that  J0=JC,  and  that  M0=MC.  From  I  7=0  for  the 
free  body  just  mentioned,  (but  not  shown  in  the  figure), 
and  from  symmetry,  we  have  J0=  ^  P  and  Jc=  *4  P  ',  but 
to  determine  Mn  and  Mf,  the  form  of  the  elastic  curves 
0  B  and  B  G  must  be  taken  into  account  as  follows  : 
Equation  of  OB,  Fig.  280.  I  [mom.  about  neutral  axis 
of  any  section  n  on  0  B]  =  0  (for  the  free  body  nC  which 


328 


MECHANICS    OF   ENGINEERING. 


has  a  section  exposed  at  each  end,  n  being  the  variable 
section)  will  give 


[Note,  In  forming  this  moment  equation,  notice  that 
Mc  is  the  sum  of  the  moments  of  the  tensions  and  com- 
pressions at  G  about  the  neutral  axis  at  n,  just  as  much  as 
about  the  neutral  axis  of  G;  for  those  tensions  and  com- 
pressions are  equivalent  to  a  couple,  and  hence  the  sum  of 
their  moments  is  the  same  taken  about  any  axis  whatever 
"|  to  the  plane  of  the  couple  (§32).] 

Taking  the  #-anti-derivative  of  each  member  of  (1), 

Ei^-=p(y2 1 x — y£  #*)+  MG x — y2  P(i x — y2  #2)   .   (2) 

(The  constant  is  not  expressed,  as  it  is  zero).  Now  from 
symmetry  we  know  that  the  tangent-line  to  the  curve  0  B 
at  B  is  horizontal,  i.e.,  for  x=^l,  dy-r-dx=Q,  and  these 
values  in  eq.  (2)  give  us 


0= 


whence  M=  M=      PI 


(3) 


Safe  Loading.  Fig.  281.  Having  now  all  the  forces  which 
act  as  external  forces  in  straining  the  beam  OG,  we  are 
ready  to  draw  the  moment  diagram  and  find  Mm.  For  con- 
venience measure  x  from  (7.  For  the  free  body  nC,  we 
have  [see  eq.  (3)] 

e 

Eq.  (4)  holds  good  for  any 
section  on  CB.  By  put- 
7*  ting  x=0  we  have  M=MC= 
ys  PI-,  l&yo8HC'=Mc  to 
scale  (so  many  inch-pounds 
moment  to  the  inch  of  pa- 
per). At  B,  for  x=y&  I, 
JfB= — yi  PI  ;  hence  lay 
off  B'D=%  PI  on  the  op- 
FIO.  281.  posite  side  of  the  axis  O'G' 


FLEXURE.      CONTINUOUS   GIRDERS. 


329 


from  HC',  and  join  DH.  DK,  symmetrical  with  DH  about 
BfDt  completes  the  moment  curves,  viz.:  two  right  lines. 
The  max.  M  is  evidently  =y&  PI  and  the  equation  of  safe 
loading 


=#  PI 

6 


(5) 


Hence  the  beam  is  twice  as  strong  as  if  simply  supported 
at  the  ends,  under  this  load  ;  it  may  also  be  proved  to  be 
four  times  as  stiff. 

The  points  of  inflection  of  the  elastic  curve  are  in  the 
middles  of  the  half-spans,  while  the  max.  shear  is 


(6) 


275.    Prismatic  Beam  Fixed  Horizontally  at  Both  Ends  [at  Same 
Level],     Uniformly  Distributed  Load  Over  the    Whole    Length. 

Fig.  282.  As  in  the  preceding  problem,  we  know  from 
symmetry  that  JQ=JQ=^W=%  ivl,  and  that  M0=MC)  and 
determine  the  latter  quantities  by  the  equation  of  the 
curve  0(7,  there  being  but  one  curve  in  the  present  in- 
stance, instead  of  two,  as  there  is  no  change  in  the  law  of 
loading  between  0  and  O.  With  nO  free,  I  (momn)=0 
gives 


Ei^y=- 


wot? 


(1) 


and.% 


(2) 


330 


MECHANICS   OF   ENGINEERING. 


The  tangent  line  at  0  being  horizontal  we  have  for  a?=0,— $  = 

doc 

0,  .'.  (7=0.     But  since  the  tangent  line  at  C  is  also  hori- 
zontal, we  may  for  x=l  put  dy-r-dx=0,  and  obtain 

0=—*X  JFP+J^H-^titf;  whence  MQ=L  Wl     .     (3) 

as  the  moment  of  the  stress-couple  close  to  the  wall  at  0 
and  at  C. 

Hence,  Fig.  283,  the  equation  of  the  moment  curve  (a 
single  continuous  curve  in  this  case)  is  found  by  putting 
2  (momn)=0  for  the  free  body  nO,  of  length  x,  thus 
obtaining 


j^^^ 

FIG.  283. 


(4) 


an  equation  of  the  second  degree,  indicating  a  conic.  At  0, 
M=M0  of  course,=  A.  Wl ;  at  ^by  putting  x=y2  I  in  (4),  we 
have  ME= — ^  Wl,  which  is  less  than  Jf0,  although  MB  is  the 
calculus  max.  (negative)  for  M,  as  may  be  shown  by  writ- 
ing the  expression  for  the  shear  (J=}£  W- — wx)  equal  to 
zero,  etc. 


FLEXURE.       CONTINUOUS    GIRDERS. 

Hence  Mm=^Wl,  and  the  equation  for  safe  loading  is 

~^k  Wl (5) 

Since  (with  this  form  of  loading)  if  the  beam  were  not 
built  in  but  simply  rested  on  two  end  supports,  the  equa- 
tion for  safe  loading  would  be  [R'I+e']  =  }iWl,  (see  §242), 
it  is  evident  that  with  the  present  mode  of  support  it  is  50 
per  cent,  stronger  as  compared  with  the  other  ;  i.e.,  as  re- 
gards normal  stresses  in  the  outer  elements.  As  regards 
shearing  stresses  in  the  web  if  it  has  one,  it  is  no  stronger* 
since  Jm=^  Win  both  cases. 

As  to  stiffness  under  the  uniform  load,  the  max.  deflec- 
tion in  the  present  case  may  be  shown  to  be  only  |-  of  that 
in  the  case  of  the  simple  end  supports.  EK 

It  is  noteworthy  that  the  shear  diagrani  in  Fig.  283  is 
identical  with  that  for  simple  end  supports  §242,  under 
uniform  load  ;  while  the  moment  diagrams  differ  as  fol- 
lows :  The  parabola  KB' A,  Fig.  283,  is  identical  with  tha* 
in  Fig.  235,,  but  the  horizontal  axis  from  which  the  ordi- 
nates  of  the  former  are  measured,  instead  of  joining  the 
extremities  of  the  curve,  cuts  it  in  such  a  way  as  to  have 
equal  areas  between  it  and  the  curve,  on  opposite  sides 

i.e.,  areas  [JT<7'#'-M£'0']=area  H'G'B' 

In  other  words,  the  effect  of  fixing  the  ends  horizontally 
is  to  shift  the  moment  parabola  upward  a  distance  =  Mc 
(to  scale),  =  i  Wl,  with  regard  to  the  axis  of  reference, 
O'B',  in  Fig.  235. 

276.  Remarks. — The  foregoing  very  simple  cases  of  con- 
tinuous girders  illustrate  the  means  employed  for  deter- 
mining the  reactions  of  supports  and  eventually  the  max. 
moment  and  the  equations. for  safe  loading  and  for  deflec- 
tions When  there  are  more  than  three  supports,  with 
spans  of  unequal  length,  and  loading  of  any  description, 
the  analysis  leading  to  the  above  results  is  much  more 
complicated  and  tedious,  but  is  considerably  simplified 


332  MECHANICS   OF   ENGINEERING. 

and  systematized  by  the  use  of  the  remarkable  theorem  of 
three  moments,  the  discovery  of  Clapeyron,  in  1857.  By 
this  theorem,  given  the  spans,  the  loading,  and  the  vertical 
heights  of  the  supports,  we  are  enabled  to  write  out  a  rela- 
tion between  the  moments  of  each  three  consecutive  sup- 
ports, and  thus  obtain  a  sufficient  number  of  equations  to 
determine  the  moments  at  all  the  supports  [p.  641  Rankine's 
Applied  Mechanics.]  From  these  moments  the  shears 
close  to  each  side  of  each  support  are  found,  then  the 
reactions,  and  from  these  and  the  given  loads  the  moment 
at  any  section  can  be  determined ;  and  hence  finally  the 
max.  moment  M.tl)t  and  the  max.  shear  Jm. 

The  treatment  of  the  general  case  of  continuous  girders 
by  graphic  methods,  however,  is  comparatively  simple, 
and  its  presentation  is  therefore  deferred,  §  391. 


THE    DANGEROUS    SECTION    OF   NON-PRIS- 
MATIC   BEAMS. 

277.  Remarks.  By  "  dangerous  section  "  is  meant  that  sec- 
tion (in  a  given  beam  under  given  loading  with  given  mode 
of  support)  where  p,  the  normal  stress  in  the  outer  fibre, 
at  distance  e  from  its  neutral  axis,  is  greater  than  in  the 
outer  fibre  of  any  other  section.  Hence  the  elasticity  of 
the  material  will  be  first  impaired  in  the  outer  fibre  of 
this  section,  if  the  load  is  gradually  increased  in  amount 
(but  not  altered  in  distribution). 

In  all  preceding  problems,  the  beam  being  prismatic,  /, 
the  moment  of  inertia,  and  e  were  the  same  in  all  sections, 

hence  when  the  equation    £—=M   [§239]  was  solved  for  p, 

a 

.   .  Me  /1N 

giving  P=~T         '         *         '         '        '  ' 

we  found  that  p  was  a  max.,  =  pm,  for  that  section  whose 
M  was  a  maximum,  since  p  varied  as  M,  or    the  moment 


FLEXURE       NON-PRISMATIC    BEAMS.  333 

of  the  stress-couple,  as  successive  sections  along  the  beam 
were  examined. 

But  for  a  non-prismatic  beam  I  and  e  change,  from  sec- 
tion to  section,  as  well  as  M,  and  the  ordinate  of  the 
moment  diagram  no  longer  shows  the  variation  of  pt  nor 
is  p  a  max.  where  M  is  a  max.  To  find  the  dangerous 
section,  then,  for  a  non-prismatic  beam,  we  express  the  Mt 
the  /,  and  the  e  of  any  section  in  terms  of  xt  thus  obtain- 
ing j9=func.  (cc),  then  writing  dp-r-dx=Q,  and  solving  for  x. 

278.  Dangerous  Section  in  a  Double  Truncated  Wedge.  Two 
End  Supports.  Single  Load  in  Middle.  —  The  form  is  shown  in 
Fig.  284.  Neglect  weight  of  beam  ;  measure  x  from  one  sup- 

port 0.     The 
—  PA 

reaction     at 


| 


each  support 
is  y2  P.  The 
width  of  the 

FIG.  284.  beam  —   &   at 

all  sections,  while  its  height,  v,  varies,  being  =  h  at  0. 
To  express  the  e  =  y2  v,  and  the  /  =  1  fo3  (§247)  of  any 
section  on  0(7,  in  terms  of  x,  conceive  the  sloping  faces 
of  the  truncated  wedge  to  be  prolonged  to  their  intersec- 
tion Ay  at  a  known  distance  =  c  from  the  face  at  0.  We 
then  have  from  similar  triangles 

v  :  x   +  c  :  :  h  :  c,  .-.  v  =  -  (x  +  c)      .     .     (I) 

c 

and  .\e  =  %-  (x+c)  and  /=  i  b  -*|>4-c]8     .     (2) 
c  c 

For  the  free  body  nO,  I  (moms.n)  =  0  gives 

...    (3) 
[That  is,  the  M  =  tf  Px.]     But  from  (2),  (3)  becomes 


x 


dp      OT5   c2     (x+cf  —  2x(x+c) 

=  3P     •  —  - 


By  putting  dp  -=-  dx  =  0  we  obtain  both  x  =  —  e,  and 


334 


MECHANICS    OF   ENGINEERING. 


#  =  -f  c,  of  which  the  latter,  x  —  +  c,  corresponds  to  a 
maximum  for  p  (since  it  will  be  found  to  give  a  negative 
result  on  substitution  in  d2p  -r-  dx*). 

Hence  the  dangerous  section  is  as  far  from  the  support 
0,  as  the  imaginary  edge,  A,  of  the  completed  wedge,  but 
of  course  on  the  opposite  side.  This  supposes  that  the 
half -span,  ^?,  is  >  c\  if  not,  the  dangerous  section  will 
be  at  the  middle  of  the  beam,  as  if  the  beam  were 
prismatic. 

the  equation  for  safe  (  ft'iiv 

-g-=^PJ        (5) 

( 


=        o 


i,-i        -4.1,  )  the  equation  for  safe 
while  with  I  load£    is.  (      ta.=c 

j  and  p=R'  in  [3]) 
(aee  §239.) 


279.    Double  Truncated  Pyramid  and  Cone.     Fig.  285.     For 


FIG.  285. 


the  truncated  pyramid  both  width  =  u,  and  height  =  v, 
are  variable,  and  if  b  and  h  are  the  dimensions  at  0,  and 
c  =  Qj£  =  distance  from  0  to  the  imaginary  vertex  A,  we 


shall  have  from  similar  triangles  u=~  (#+c)and  v=  -(a? 4- c). 

,  in  the  moment 


Hence,  substituting  e=*4v  and  7=1 
equation 


x 


dp  _ 

= 


— 

ne 


(x+cf 


(7) 


.     .    (8) 


FLEXUEE.      NON-PKISMATIC   BEAMS.  335 

Putting  this  =  0,  we  have  x  =  —  c,  x  =  —  c,  and  x  = 
+  y2  c,  hence  the  dangerous  section  is  at  a  distance  x  =  y2  c 
from  0,  and  the  equation  for  safe  loading  is 

either  ^^1=  ft  PI  .  .  .  .  if  ft  Hs  <  %  c  ....  (9) 
(in  which  &'  and  hf  are  the  dimensions  at  mid-span) 

or  B£  4&)j%_fc)2  =  I/4  Pc  if  ^  ,.g  >  l/2  c    _  _  _  (10) 

For  the  truncated  cone  (see  Fig.  285  also,  on  right)  where 
e  =  the   variable  radius   r,  and  7  =  ^  it  r*,  we  also  have 

f  =  [Const.]    .  -JL        ....     (11) 


and  hence  p  is  a  max.  for  x  =  ^  c,  and  the  equation  for 
safe  loading 


either  =  #  PZ,  for  #  Z  <  #  c     .      .     .     .     .     (12) 

(where  r'  =  radius  of  mid-span  section)  ; 


or 


(where  r0  =  radius  of  extremity.) 


>  f  or  ft  I  >  y2  c     .....     (13) 


NON-PRISMATIC    BEAMS    OF    "UNIFORM 
STRENGTH." 

280.  Remarks,  A  beam  is  said  to  be  of "  uniform 
strength  "  when  its  form,  its  mode  of  support,  and  the  dis- 
tribution of  loading,  are  such  that  the  normal  stress  p  has 
the  same  value  in  all  the  outer  fibres,  and  thus  one  ele- 
ment of  economy  is  secured,  viz. :  that  all  the  outer  fibres 
may  be  made  to  do  full  duty,  since  under  the  safe  loading, 
p  will  be  =  to  R'  in  all  of  them.  [Of  course,  in  all  cases 
of  flexure,  the  elements  between  the  neutral  surface  and 


336  MECHANICS   OF   ENGINEERING. 

the  outer  fibres  being  under  tensions  and  compressions 
less  than  R'  per  sq.  inch,  are  not  doing  full  duty,  as 
regards  economy  of  material,  unless  perhaps  with  respect 
to  shearing  stresses.]  In  Fig.  265,  §261,  we  have  already 
had  an  instance  of  a  body  of  uniform  strength  in  flexure, 
viz. :  the  middle  segment,  CD,  of  that  figure ;  for  the 
moment  is  the  same  for  all  sections  of  CD  [eq.  (2)  of  that 
§],  and  hence  the  normal  stress  p  in  the  outer  fibres  (the 
beam  being  prismatic  in  that  instance). 

In  the  following  problems  the  weight  of  the  beam  itself 
is  neglected.  The  general  method  pursued  will  be  to  find 
an  expression  for  the  outer -fibre-stress  p,  at  a  definite  sec- 
tion of  the  beam,  where  the  dimensions  of  the  section  are 
known  or  assumed,  then  an  expression  for  p  in  the  varia- 
ble section,  and  equate  the  two.  For  clearness  the  figures 
are  exaggerated,  vertically. 

281.    Parabolic  Working  Beam.    UnsymmetricaL      Fig.   286 

,p 


t; 


FIG.  286. 


CBO  is  a  working  beam  or  lever,  B  being  the  fixed  fulcrum 
or  bearing.  The  force  P0  being  given  we  may  compute  P0 
from  the  mom.  equation  P^Q  =  Pcllt  while  the  fulcrum 
reaction  is  PB=P0+PG.  All  the  forces  are  "|  to  the  beam. 
The  beam  is  to  have  the  same  width  b  at  all  points,  and  ia 
to  be  rectangular  in  section. 

Kequired  first,  the  proper  height  J^,  at  B,  for  safety. 
From  the  free  body  BO,  of  length  =  lot  we  have  I  (momsB) 
=  0;  Le., 


FLEXURE.      NON-PRISMATIC   BEAMS.  337 

Hence,  putting  pB  =  R',  ^  becomes  known  from  (1). 

Ee  quired,  secondly,  the  relation  between  the  variable 
height  v  (at  any  section  n)  and  the  distance  x  of  n  from  0. 
For  the  free  body  nO,  we  have  (<T  momsn  =  0) 


Q  = 

y2V  bv2 

But  for  "  uniform  strength  "  pn  must  =  pR  ;  hence 
equate  their  values  from  (1)  and  (2)  and  we  have 

5  =    ^    which  may  be  written  (#  v)2  = 

V2  A! 

so  as  to  make  the  relation  between  the  abscissa  x  and  the 
ordinate  j£  v  more  marked  ;  it  is  the  equation  of  a  para- 
bola, whose  vertex  is  at  0. 

The  parabolic  outline  for  the  portion  BC  is  found  simi- 
larly. The  local  stresses  at  (7,  JSt  and  0  must  be  proper- 
ly provided  for  by  evident  means.  The  shear  J  =  P0,  at 
0,  also  requires  special  attention. 

This  shape  of  beam  is  often  adopted  in  practice  for  the 
working  beams  of  engines,  etc. 

The  parabolic  outlines  just  found  may  be  replaced  by 
trapezoidal  forms,  Fig.  287,  without  using  much  more  ma- 
terial, and  by  making  the  slop- 
ing plane  faces  tangent  to  the 
parabolic  outline  at  points  T0 
and  TU  half-way  between  0  and 
J5,  and  C  and  B,  respectively.  It  Fie.  237. 

can  be  proved  that  they  contain  minimum  volumes,  among 
all  trapezoidal  forms  capable  of  circumscribing  the  given 
parabolic  bodies.  The  dangerous  sections  of  these  trape- 
zoidal bodies  are  at  the  tangent  points  T0  and  Tlt  This  is 
as  it  should  be,  (see  §  278),  remembering  that  the  subtan- 
gent  of  a  parabola  is  bisected  by  the  vertex. 

282.  I-Beam  of  Uniform  Strength.  —  Support  and  load  same 
as  in  the  preceding  §.  Fig.  288.  Let  the  area  of  the 


338 


MECHANICS   OF   ENGINEERING. 


flange -sections  be  =  F  and  let  it  be  the  same  for  all  values 
of  x.  Considering  all  points  of  F  at  any  one  section  as  at 
the  same  distance  z  from  the  neutral  axis,  we  may  write 
I  =  s?F,  and  assuming  that  the  flanges  take  all  the  tension 
and  compression  while  the  (thin)  web  carries  the  shear,  the 
free  body  of  length  x  in  Fig.  288  gives  (moms,  about  n) 

¥-=Pcx  ;  i.e.  ±—  ——P..X  :  or,  since  p  is  to  be  constant, 

6  Z 

z  =  [Const.],  x        ;-      .        .      ;,"          (1) 

i.  e.  z  must  be  made  proportional  to  x. 

Hence  the  flanges  should  be  made  straight.     Practically, 
if  they  unite  at  (7,  the  web  takes  but  little  shear. 

283.  Rectang.  Section.    Height  Constant.    Two  Supports  (at  Ex- 
tremities). Single  Eccentric  Load.  .  P 
— Fig.   289.    b  and  h  are  the            |P°  PO 
dimensions  of  the  section  at                ^ — 7^">^ 
B.     With  #0free  we  have           ^~"           b/    "^^} 


(1) 


eB 


FIG.  289. 


At  any  other  section  on  BO,  as  n,  where  the  width  is  u, 
the  variable  whose  relation  to  x  is  required,  we  have  for 
nOfree 


(2) 
(3) 


Equating  pK  and  pn  we  have  u  :  b  ::  x  : 


FLEXURE.       BEAMS   OF    UNIFORM    STRENGTH.         339 


That  is,  BO  must  be  wedge-shaped  with  its  edge  at  0,  ver- 
tical. 

284.  Similar  Rectang.  Sections.  Otherwise  as  Before.  —  Fig.289  a, 
b  and  h  are  the  dimen- 
sions at  £;  at  any  other 
section  n,  on  BO,  the  height 
v  and  width  u,  are  the 
variables  whose  relation  to 
x  is  desired  and  by  hypoth- 
esis are  connected  by  the 
relation 


u  :  v  ::  b  :  h 


(1) 


(since  the  section  at  n  is  a  rectangle  similar  to  that  at  B). 


For  the  free  body  BO  ____  pB  = 


For  the  free  body  nO  ____  p«  = 


uv 


Writing  pa  =  pB  we  obtain  IQ  -4-  bh*  =  x 
put  u  —  bv  -f-  h,  from  (1)  ;  whence 


..     (2) 

.        (3) 
*,  in  which 


which  is  the  equation  of  the  curve  (a  cubic  parabola) 
whose  abscissa  is  x  and  ordinate  %  v  ;  i.e.,  of  the  upper 
curve  of  the  outline  of  the  central  longitudinal  vertical 
plane  section  of  the  body  (dotted  line  BO)  which  is  sup- 
posed symmetrical  about  such  a  plane.  Similarly  the 
central  horizontal  plane  section  will  cut  out  a  curve  a 
quarter  of  which  (dotted  line  B?  0)  has  an  equation 


(5) 


That  is,  the  height  and  width  must  vary  as  the  cube  root 


340  MECHANICS   OF   ENGINEERING. 

of  the  distance  from  the  support.  The  portion  CB  will 
give  corresponding  results,  referred  to  the  support  G. 

[If  the  beam  in  this  problem  is  to  have  circular  cross- 
sections,  let  the  student  treat  it  in  the  same  manner.] 

286.    Uniform  Load.    Two    End  Supports.     Rectangular  Cr. 
Sections.    Width   Constant.  —  j_wl  ;  Lwl 

"Weight  of  beam  neglected. 
How  should  the  height  vary, 
the  height  and  width  at  the 
middle  being  h  and  b  ?  Fig. 
289  b.  From  symmetry  each 
reaction  =  y2  W  =  y2  wl.  FIG.  239  &. 

At  any  cross  section  n,  the  width  is  =  ft,  (same  as  that  at 
the  middle)  and  the  height  =  v,  variable.  I  (moms.n)  =0, 
for  the  free  body  n  0,  gives 

.    .    .     (i) 


But  for  a  beam  of  uniform  strength,  pn  is  to  be  =  pB  as 
computed  from  I  (moms.B)  =  0  for  the  free  body  .  .  BO, 
i.e  from 


=  _ 

22 
Hence  solve  (1)  for  pn  and  (2)  for  pB  and  equate  the  results, 

whence  v*=      L[lx-J]  ;  or  (i^)2=[^^]     •     (3) 


This  relation  between  the  abscissa  x  and  the  ordinate 
of  the  curve  CEO,  shows  it  to  be  an  ellipse  since  eq.  (3) 
is  that  of  an  ellipse  referred  to  its  principal  diameter  and 
the  tangent  at  its  vertex  as  co-ordinate  axes. 

In  this  case  eq.  (3)  covers  the  whole  extent  of  both 
upper  and  lower  curves,  i.e.  the  complete  outline,  of  the 
curve  CBOB',  whereas  in  Figs.  286,  289,  and  289  a,  such  is 
not  the  case. 


FLEXURE.       BODIES   OF   UNIFORM   STRENGTH.      341 


287.  Cantilevers  of  Uniform  Strength. — Beams  built  in  at 
one  end,  horizontally,  and  projecting  from  the  wall  with- 
out support  at  the  other,  should  have  the  forms  given  be- 
low, for  the  given  cases  of  loading,  if  all  cross-sections  are 
to  be  Rectangular  and  the  weight  of  beam  neglected.  Sides 
of  sections  horizontal  and  vertical.  Also,  the  sections  are 
symmetrical  about  the  axis  of  the  piece,  b  and  h  are  the 
dimensions  at  the  wall.  1=  length.  No  proofs  given. 


FIG.  290. 


Width    Constant. 


ig.  290,  (6). 


end  load. 

Height  constant.  -^ 
Single    end  load. 
Horizontal   outline  f 
triangular.  J 


Constant  ratio  of  ^ 
height  v  to  width  u. 
Both    outlines    cu-  \™*  290'  (c>- 
bic  parabolas.  J 


(2) 

(3) 


342  MECHANICS   OF   ENGINEERING. 

Uniform    L  o  a  d.  T 

Width  constant.!  ..# 

Vertical  outline  tri-  \*>*  291>  <«>        <*0=<#A)T     •  (4) 
angular. 

Uniform  Load. 
Height  constant. 
Horiz.  outline  is 
two  parabolas  meet- 
ing at  0  (vertex) 
with  geomet.  axes 
II  to  wall. 

Uniform  Load.- 
Both  outlines  semi- 
cubic     parabolas. 
Sections    similar 
rectangles. 


Fig.  291,  (6).        ^=(>  •    .(5) 


7      (6) 


^  Fig.  291,  (c). 


289. — Beams  and  cantilevers  of  circular  cross-sections 
may  be  dealt  with  similarly,  and  the  proper  longitudinal 
outline  given,  to  constitute  them  "  bodies  of  uniform 
strength."  As  a  consequence  of  the  possession  of  this 
property,  with  loading  and  mode  of  support  of  specified 
character,  the  following  may  be  stated ;  that  to  find  the 
equation  of  safe  loading  any  cross-section  ivhatever  may  be 
employed.  This  refers  to  tension  and  compression.  As 
regards  the  shearing  stresses  in  different  parts  of  the  beam 
the  condition  of  "  uniform  strength  "  is  not  necessarily  ob- 
tained at  the  same  time  with  that  for  normal  stress  in  the 
outer  fibres. 


DEFLECTION     OF      BEAMS     OF      UNIFORM 
STRENGTH. 

$90.  Case  of  §  283,  the  double  wedge,  but  symmetrical, 
i.e.,  II=IQ=  y2l,  Fig.  292,     Here  we  shall  find  the  use  of  the 


BEAMS   OF   UNIFORM   STRENGTH.  343 


FIG.  292. 


El 

form  (of  the  three  forms  for  the  moment  of  the  stress 

P 
couple,  see  eqs.  (5),  (6)  and  (7),  §§  229  and  231)  of  the  most 

direct  service  in  determining  the  form  of  the  elastic  curve 
OB,  which  is  symmetrical,  and  has  a  common  tangent  at 
B,  with  the  curve  BC.  First  to  find  the  radius  of  curva- 
ture, />,  at  any  section  n,  we  have  for  the  free  body  n09 
JT(moms.u=0),  whence 

—  ~  +  y2Px=0  ;  but  I  (3)  g1  ^3   |  *=  \V*1  and  *-Vu«P 

we  have  yi2  -   iM-ftP  |  and  .-.  p-%  ^  .  T        (1) 

from  which  all  variables  have  disappeared  in  the  right 
hand  member  ;  i.e.,  p  is  constant,  the  same  at  all  points  of 
the  elastic  curve,  hence  the  latter  is  the  arc  of  a  circle, 
having  a  horizontal  tangent  at  B, 

To  find  the  deflection,  d,  at  B,  consider  Fig.  292,  (b\ 
where  d=KB,  and  the  full  circle  of  radius  BH=p  is 
drawn. 

The  tnangle_ZrO.ZHs  similar  to  YOB, 


But  OB=y2l,  KB=d  and  Z5=2/> 

•'•  d==  (Jfr>  and  -  •  from  e<l-  W»  *~  #OT     « 

From  eq.  (4)  §233  we  note  that  for  a  beam-  of  the  same 
material  but  prismatic  (parallelopipedical  in  this  case,) 
having  the  same  dimensions,  b  and  h,  at  all  sections  as  at 

1    P$          PI? 
the  middle,  deflects  an  amount  =  jg-  jjjf=  J  under  a 


344  MECHANICS   OF   ENGINEERING. 

load  P  in  the  middle  of  the  span.  Hence  the  tapering 
beam  of  the  present  §  has  only  ^  the  stiffness  of  the  pris* 
matic  beam,  for  the  same  6,  h,  I,  Et  and  P. 

291.  Case  of  §  281  (Parabolic  Body),  With  ^=4),  i.e.,  Symmet- 
rical— Fig.  293,(a).     Kequired  the  equation  of  the  neutral 


Jl_l 


V 

(a)  (*) 

FIG.  293. 

line  OB.    For  the  free  body  nO,  J(moms.n)=0  gives  us 


Fig.  293,  (6),  shows  simply  the  geometrical  relations  of  the 
problem,  position  of  origin,  axes,  etc.  OnB  is  the  neutral 
line  or  elastic  curv^  whose  equation,  and  greatest  ordinate 
d,  are  required.  (The  right  hand  member  of  eq.  (1)"  is  made 
negative  because  d2y-±-dx2  is  negative,  the  curve  being  con- 
cave to  the  axis  Xin  this,  the  first  quadrant.) 

Now  if  the  beam  were  prismatic,  I,  the  "  moment  of  in- 
ertia" of  the  cross-section  would  be  constant,  i.e.,  the  same 
for  all  values  of  x,  and  we  might  proceed  by  taking  the  x- 
anti-derivative  of  each  member  of  (1)"  and  add  a  constant  ; 
but  it  is  variable  and  is 


x,    (fromeq.  3,  §281,  putting  k= 
hence  (1)"  becomes 


To  put  this  into  the  form  Const.  X  ^=func.  of  (x\  we  need 


BEAMS   OF    UNIFORM   STRENGTH.  345 

3 

only  divide  through  by  xz  ,.(and  for  brevity  denote 
i-4-  (}4tyT  by  A)  and  obtain 

•*:..-..   (i) 


We  can  now  take  the  a?-anti-derivative  of  each  member,  and 
have 

Ad£-=—y2P(2x+y2)+  C    ....      (2)' 
cLx 

To  determine  the  constant  (7,  we  utilize  the  fact  that  at  Bt 
where  #=j^Z,  the  slope  dy-r-dx  is  zero,  since  the  tangent 
line  is  there  horizontal,  whence  from  (2)' 


.-.  (2)'  becomes  A         =P^~%l-x     ]    .....     (2) 
dx 


Ay=P  [l.x-%  ^+IC'=0]    ...    (3) 


(C"=0  since  for  x=Q,  2/=0).  We  may  now  find  the  deflec- 
tion d  (Fig.  293(6))  by  writing  x=  y2l  and  y=d,  whence,  after 
restoring  the  value  of  the  constant  At 


JOTS 

and  is  twice  as  great  [being=2.  .]^  as  if  the  girder 

i     3 

*  See  §  233,  putting  I  =^  bh  in  eq.  (4). 

were  parallelopipedical.     In  other  words,  the  present  girdei 
is  only  half  as  stiff  as  the  prismatic  one. 

292.    Special  Problem.    (I.)     The  symmetrical  beam  ^n  Fig. 
294  is  of  rectangular  cross-section  and  constant  width  ~=  5, 


346 


MECHANICS   OF   ENGINEERING. 


but  the  height  is  constant  only  over  the  extreme  quarter 
spans,  being  =^=^ A,  i.  e.,  half  the  height  h  at  mid-span. 
The  convergence  of  the  two  truncated  wedges  forming  the 
middle  quarters  of  the  beam  is  such  that  the  prolongations 


FIG.  294. 

of  the  upper  and  lower  surfaces  woM  meet  over  the  supports 
(as  should  be  the  case  to  make  h=2hl).  Neglecting  the 
weight  of  the  beam,  and  placing  a  single  load  in  middle,  it 
is  required  to  find  the  equation  for  safe  loading ;  also  the 
equations  of  the  four  elastic  curves ;  and  finally  the  deflec- 
tion. 

The  solutions  of  this  and  the  following  problem  are  left 
to  the  student,  as  exercises.  Of  course  the  beam  here 
given  is  not  one  of  uniform  strength. 

293,  Special  Problem.  (II).  Fig.  295.  Kequired  the  man- 
ner in  which  the  width  of  the  beam  must  vary,  the  height 
being  constant,  cross-sections  rectangular,  weight  of  beam 


FIG.  295. 


neglected,  to  be  a  beam  of  uniform  strength,  if  the  load  19 
uniformly  distributed  ? 


OBLIQUE  FORCES.  34V 


CHAPTEE  V. 

FLEXURE  OF  PRISMATIC  BEAMS  UNDER 
OBLIQUE  FORCES. 


294  Eemarks,  By  "  oblique  forces  "  will  be  understood 
external  forces  not  perpendicular  to  tho  beam,  but  theso 
external  forces  will  be  confined  to  one  plane,  called  the 
force-plane,  which  contains  the  axis  of  the  beam  and  also 
cuts  the  beam  symmetrically.  The  curvature  induced  by 
these  external  forces  will  as  before  be  considered  very 
slight,  so  that  distances  measured  along  the  beam  will  be 
treated  as  unchanged  by  the  flexure. 

It  will  be  remembered  that  in  previous  problems  the 
proof  that  the  neutral  axis  of  each  cross  section  passes 
through  its  centre  of  gravity,  rested  on  the  fact  that  when 
a  portion  of  the  beam  having  a  given  section  as  one  of  its 
bounding  surfaces  is  considered  free,  the  condition  of 
equilibrium  Jf  (compons.  ||  to  beam)=0  does  not  introduce 
any  of  the  external  forces,  since  these  in  the  problems  re- 
ferred to,  were  "|  to  the  beam ;  but  in  the  problems  of  the 
present  chapter  such  is  not  the  case,  and  hence  the  neutral 
axis  does  not  necessarily  pass  through  the  centre  of  gravity 
of  any  section,  and  in  fact  may  have  only  an  ideal,  geomet- 
rical existence,  being  sometimes  entirely  outside  of  the 
section ;  in  other  words,  the  fibres  whose  ends  are  exposed 
in  a  given  section  may  all  be  in  tension,  (or  all  in  compres- 
sion,) of  intensities  varying  with  the  distance  of  each  from 
the  neutral  axis.  It  is  much  more  convenient,  however,  to 
take  for  an  axis  of  moments  the  gravity  axis  parallel  to  the 


348 


MECHANICS   OF   EXGIXEEBING. 


neutral   axis  instead  of  the  neutral  axis  itself,  since  this 
gravity  axis  has  always  a  known  position. 


295.  Classification  of  the  Elastic  Forces.  Shear,  Thrust,  and 
Stress-Couple.  Fig.  296.  Let  AKMbe  one  extremity  of  a 
portion,  considered  free,  of  a  prismatic  beam,  under  oblique 
forces.  C  is  the  centre  of  gravity  of  the  section  ex- 
posed, and  GO  the  gravity  axis  "|  to  the  force  plane  CAK. 
The  stresses  acting  on  the  elements  of  area  (each  =dF )  of 
the  section  consist  of  shears  (whose  sum=e/,  the  "total 
shear")  in  the  plane  of  the  section  and  parallel  to  the  force 
plane,  and  of  normal  stress  parallel  to  AK  and  proportion- 
al per  unit  of  area  to  the  distances  of  the  dF's  on  which 
they  act  from  the  neutral  axis  NC"9  real  or  ideal  (ideal  in 
this  figure).  Imagine  the  outermost  fibre  KA,  whose  dis- 
tance from  the  gravity  axis  is=e  and  from  the  neutral  axis 


FIG.  296. 


=**€  -fa,  to  be  prolonged  an  amount  AAf,  whose  length  by 
some  arbitrary  scale  represents  the  normal  stress  (tension 
01  compression)  to  which  the  dF  at  A  is  subjected.  Then, 
ii  a  plane  be  passed  through  A'  and  the  neutral  axis  NC"> 
tne  lengths,  such  as  mr,  parallel  to  A  A',  intercepted  between 
this  plane  and  the  section  itself,  represent  the  stress-inten- 


FLEXURE.       OBLIQUE     FORCES.  349 

eities  (i.  e.,  per  unit  area)  on  the  respective  dF's.  (In  this 
particular  figure  these  stresses  are  all  of  one  kind,  all  ten- 
sion or  all  compression  ;  but  if  the  neutral  axis  occurs 
within  the  limits  of  the  section,  they  will  be  of  opposite 
kinds  on  the  two  sides  of  NC."}  Through  C",  the  point 
determined  in  A'NC"  by  the  intercept  CC'  of  the  centre  of 
gravity,  pass  a  plane  A"M"T"  parallel  to  the  section  it- 
self ;  it  will  divide  the  stress -intensity  AA'  into  two  parts 
p{  and  p2,  and  will  enable  us  to  express  the  stress-intensity 
mr,  on  any  dF  at  a  distance  a  from  the  gravity -axis  GC,  in 
two  parts  ;  one  part  the  same  for  all  dF's,  the  other  depen- 
dent on  z,  thus  : 

[Stress-intensity  on  any  dF]  =  PI+ ^pz    •     •  (1) 

€ 

and  hence  the 

[actual  normal  stress  on  any  dF~\  =  pvd  F  +  — ^  p2  dF    (2) 

For  example,  the  stress-intensity  on  the  fibre  at  T,  where 
z  =  —  61,  will  be  Pi  —  —  p2t  and  it  is  now  seen  Low  we  may 

C 

find  the  stress  at  any  dF  when  pl  and  p2  have  been  found. 
If  the  distance  a,  between  the  neutral  and  gravity  axes  is 
desired,  we  have,  by  similar  triangles 

Pz  :  e  ::  C'C :  a  whence  a  =  ^  .  e  (3) 

ft 

It  is  now  readily  seen,  graphically,  that  the  stresses  or  elas- 
tic forces  represented  by  the  equal  intercepts  between  the 
parallel  planes  A M T  and  A"  M"  T",  constitute  a  uniform- 
ly distributed  normal  stress,  which  will  be  called  the  "  uni- 
form thrust,"  or  simply  the  thrust  (or  pull,  as  the  case  may 

be)  of  an  intensity  =  plt  and  .•.  of  .an  amount  =  Cp^dF  = 

'pifdF  =  pf. 

It  is  also  evident  that  the  positive  intercepts  forming  the 


350 


MECHANICS   OF    ENGINEERING. 


wedge  A'A'Gr'C'  and  the  negative  intercepts  forming  the 
wedge  M"MQr'C'  form  a  system  of  "graded  stresses" 
whose  combination  (algebraic)  with  those  of  the  "thrust " 
shows  the  two  sets  of  normal  stresses  to  be  equivalent  to 
the  actual  system  of  normal  stresses  represented  by  the 
small  prisms  forming  the  imaginary  solid  AMT  . .  A'M*  T'. 
It  will  be  shown  that  these  graded  stresses  constitute  a 
"  stress-couple," 

Analytically,  the  object  of  this  classification  of  the  nor- 
mal stresses  into  a  thrust  and  a  stress-couple,  may  be  made 
apparent  as  follows : 

In  dealing  with  the  free  body  KA M  Fig.  296,  we  shall 
have  occasion  to  sum  the  components,  parallel  to  the  beam, 
of  all  forces  acting  (external 
and  elastic),  also  those  ~|  to 
the  beam;  and  also  sum  their 
moments  about  some  axis  "| 
to  the  force  plane.  Let  this 
axis  of  moments  be  GC  the 
gravity -axis  of  the  section 
(and  not  the  neutral  axis) ; 
also  take  the  axis  X  ||  to  the 
beam  and  F"|  to  it  (and  in 
force-plane).  Let  us  see 
what  part  the  elastic  forces 
will  play  in  these  three  summations. 


See  Fig.  297,  which 


gives  merely  a  side  view.     Referring  to  eq.  (2)  we  see  that 


[see  eq.  (4)  §  23].     But  as  the  s's  are  measured  from  G  a 
gravity  axis,  z  must  be  zero.     Hence 


[The  IX  of  the  Elastic  forces]  =#fe=  j  ^  ^^  | 


FLEXUltE.      OBLIQUE   FORCES.  351 

Also, 

[The  2  Tot  the  Elastic  forces]  =  </:==  the  shear,      .      (5) 
while  for  moments  about  G  [see  eq.  (1)] 
[The  2  (moms.G)  of  the  elastic  forces]  = 


And  hence  finally 


where  /G,  —  C  z*dF,  is  the  "moment  of  inertia  "  oi 

the  section  about  the  gravity  axis  G,  (not  the  neutral  axis). 
The  expression  in  (6)  may  be  called  the  moment  of  the 
stress-couple,  understanding  by  stress-couple  a  couple  to 
which  the  graded  stresses  of  Fig.  297  are  equivalent.  That 
these  graded  stresses  are  equivalent  to  a  couple  is  shown 
by  the  fact  that  although  they  are  X  forces  they  do  not 
appear  in  eq.  4,  for  IX\  hence  the  sum  of  the  tensions 

i  C\dF  1  equals  that  of  the  compressions  pi  r°zdF~\ 


in  that  set  of  normal  stresses. 

We  have  therefore  gained  these  advantages,  that,  of  the 
three  quantities  J"(lbs.),  p1  (Ibs.  per  sq.  inch),  and  p2  (Ibs. 
per  sq.  inch)  a  knowledge  of  which,  with  the  form  of  the 
section,  completely  determines  the  stresses  in  the  section, 
equations  (4),  (5),  and  (6)  contain  only  one  each,  and  hence 
algebraic  elimination  is  unnecessary  for  finding  any  one 
of  them  ;  and  that  the  axis  of  reference  of  the  moment  of 
inertia  /is  the  same  axis  of  the  section  as  was  used  in 
former  problems  in  flexure. 

Another  mode  of  -stating  eqs.  (4),  (5)  and  (6)  is  this  :  The 
sum  of  the  components,  parallel  to  the  beam,  of  the  exter- 
nal forces  is  balanced  bf  the  thrust  or  puU;  those  perpen- 


352  MECHANICS   OF   ENGINEERING. 

dicular  to  the  beam  are  balanced  by  the  shear;  while  the 
sum  of  the  moments  of  the  external  forces  about  the 
gravity  axis  of  the  section  is  balanced  by  that  of  the  stress- 
couple.  Notice  that  the  thrust  can  have  no^moment  about 
the  gravity  axis  referred  to. 

The  Equation  for  Safe  Loading,  then,  will  be  this : 
(a)      .     (pi^p-z)  max.     "]  whichever 

is  j-     -IT    .        .     (7) 


•       2>i-2  max. 

For  R't  see  table  in  §  251.  The  double  sign  provides  for 
the  cases  where  pL  and  p2  are  of  opposite  kinds,  one  tension 
the  other  compression.  Of  course  (pi+p*)  max  is  not  the 
same  thing  as  [^  max.  +p-2  max.].  Inmost  cases  in  prac- 
tice el=  et  and  then  the  part  (6)  of  eq.  (7)  is  unnecessary. 

295a.  Elastic  Curve  with  Oblique  Forces. — (By  elastic  curve 
is  now  meant  the  locus  of  the  centres  of  gravity  of  the  sec- 
tions.) Since  the  normal  stresses  in  a  section  differ  from 
those  occuring  under  perpendicular  forces  only  in  the  ad- 
dition of  a  uniform  thrust  (or  pull),  whose  effect  on  the 
short  lengths  (=dx)  of  fibres  between  two  consecutive  sec- 
tions U'  V  and  Z70F0,  Fig.  297,  is  felt  equally  by  all,  the  loca- 
tion of  the  centre  of  curvature  R,  is  not  appreciably  differ- 
ent from  what  it  would  be  as  determined  by  the  stress- 
couple  alone. 

Thus  (within  the  elastic  limit),  strains  being  proportional 
to  the  stresses  producing  them,  if  the  forces  of  the  stress- 
couple  acted  alone,  the  length  dx=  GQG'  of  a  small  portion 
of  a  fibre  at  the  gravity  axis  would  remain  unchanged,  and 
the  lengthening  and  shortening  of  the  other  fibre-lengths 
between  the  two  sections  Z70  V0  and  IT  V,  originallj  parallel, 
would  occasion  the  turning  of  U'  V  through  a  small  angle 
(relatively  to  UQ  VQ)  about  6r',  into  the  position  which  it  oc- 
cupies in  the  figure  (297),  and  GQRl  would  be  the  radius  of 
curvature.  But  the  effect  of  the  uniform  pull  (added  to 
that  of  the  couple)  is  to  shift  V  V  parallel  to  itself  into 
the  position  UV,  and  hence  the  radius  of  curvature  of  the 


FLEXURE       OBLIQUE   FORCES. 


353 


elastic  curve,  of  which  GQG  is  an  element,  is  GQR  instead 
of  6r072',  But  the  difference  between  GQE  and  GQR'  is  very 
small,  being  the  same,  relatively,  as  the  difference  between, 
GQ G  and  G0G' ;  for  instance,  with  wrought-iron,  even  it  pl9 
the  intensity  of  the  uniform  pull,  were  as  high  as  22,000 
Ibs.  per  sq.  in.  [see  §  203]  GQG  would  exceed  GQG  by  only 
yi2  of  one  per  cent.  (=0.0008) ;  hence  by  using  GE'  instead 
of  GR  as  the  radius  of  curvature  p,  an  error  is  introduced 
of  so  small  an  amount  as  to  be  neglected. 

But  from  §  231,  eqs.  (6)  and  (7),  0*  =  EI^  =  M,    the 

f)  dVu 

the  sum  of  the  moments  of  the  external  forces ;  hence  for 
prismatic  beams  under  oblique  forces  we  may  still  use 


as  one  form  for  the  .T(moms.)  of  the  elastic  forces  of  the 
section  about  the  gravity-axis  ;  remembering  that  the  axis 
X  must  be  taken  parallel  to  beam. 

296.  Oblique  Cantilever  with  Terminal  Load. — Fig.  298.  Let 
i—  length.  The  "  fixing  "  of  the  lower  end  of  the  beam  is 
its  only  support.  Measure  x  along  the  beam  from  0.  Let 


FIG.  298. 


Fio.  299. 


ii  be  the  gravity  axis  of  any  section  and  nT,  =x  sin  a,  the 
length  of  the  perpendicular  let  fall  from  n  on  the  line  of 
action  of  the  force  P  (load).  The  flexure  is  so  slight  that 
nT  is  considered  to  be  the  same  as  before  the  load  is  al- 


354  MECHANICS   OF   ENGINEERING. 

lowed  to  act.  [If  «  were  very  small,  however,  it  is  evident 
that  this  assumption  would  be  inadmissible,  since  then  a 
large  proportion  of  nT  would  be  due  to  the  flexure  caused 
by  the  load.] 

Consider  nO  free,  Fig.  299.  In  accordance  with  the  pre- 
ceding paragraph  (see  eqs.  (4),  (5),  and  (6))  the  elastic 
forces  of  the  section  consist  of  a  shear  J,  whose  value  may 
be  obtained  by  writing  2T=0 

whence  J—P  sin  a  ;        .        .        .        .     (1) 

of  a  uniform  thrust  =p\Ft  obtained  from  2X=Q,  viz  : 

P  cos  a—plF=0  .'.  p,F=P  cos  a  ;  .  (2) 
and  of  a  stress-couple  whose  moment  [which  we  may  wrify 
either  &_,  or  El  —  ^  ]  is  determined  from  J(moms.n)=0  or 

6  aX? 

P*l—Px  sin  a=0,  or  ^=Px  sin  a     .        .      (3) 
e  e 

As  to  the  strength  of  the  beam,  we  note  that  the  stress-in- 
tensity, p^  of  the  thrust  is  the  same  in  all  sections,  from  0 
to  L  (Fig.  298),  and  that  p.2,  the  stress  -intensity  in  the  outer 
fibre,  (and  this  is  compression  if  e—no'  of  Fig.  299)  due  to 
the  stress  -couple  is  proportional  to  x  ;  hence  the  max.  of 
[Pi+#j]  will  be  in  the  lower  outer  fibre  at  L,  Fig.  298, 
where  x  is  as  great  as  possible,  =1  ;  and  will  be  a  compres- 
sion, viz.  : 

max.=P 


.%  the  equation  for  Safe  Loading  is 

"      .        .        .    (5) 


since  with  e^=  e,  as  will  be  assumed  here,[jo  —  —  pi\    max. 


FLEXURE.       OBLIQUE   FORCES. 


355 


can  not  exceed,  numerically,  [^1+^2]  max.  The  stress- 
intensity  in  the  outer  fibres  along  the  upper  edge  of  the 
beam,  being  —pl — p2  (supposing  el=e)  will  be  compressive 
at  the  upper  end  near  0,  since  there  p2  is  small,  x  being 
small ;  but  lower  down  as  x  grows  larger,  p2  increasing,  a 
section  may  be  found  (before  reaching  the  point  L)  where 
p2=pl  and  where  consequently  the  stress  in  the  outer  fibre 
is  zero,  or  in  other  words  the  neutral  axis  of  that  section 
passes  through  the  outer  fibre.  In  any  section  above  that 
section  the  neutral  axis  is  imaginary,  i.e.,  is  altogether  out- 
side the  section,  while  below  it,  it  is  within  the  section,  but 
cannot  pass  beyond  the  gravity  axis.  Thus  in  Fig.  300,  O'U 


FIG.  300. 


FIG.  301. 


is  the  locus  of  the  positions  of  the  neutral  axis  for  successive 
sections,  while  OL  the  axis  of  the  beam  is  the  locus  of  the 
gravity  axes  (or  rather  of  the  centres  of  gravity)  of  the 
sections,  this  latter  line  forming  the  "  elastic  curve  "  un- 
der flexure.  As  already  stated,  however,  the  flexure  is  to 
be  but  slight,  and  a  must  not  be  very  small.  For  in- 
stance, if  the  deflection  of  0  from  its  position  before  flex- 
ure is  of  such  an  amount  as  to  cause  the  lever-arm  OR  of 
P  about  L  to  be  greater  by  10  per  cent,  than  its  value 
(=1  sin  a)  before  flexure,  the  value  of  p2  as  computed  from 
eq.  (3)  (with  x—T}  will  be  less  than  its  true  value  in  the 
same  proportion. 

The  deflection  of  0  from  the  tangent  at  L,  by  §  237,  Fig. 
229(a)  is  d=(P  8ina)P-^-3EIt  approximately,  putting  P  sin  a 


356 


MECHANICS   OF   ENGINEERING. 


for  the  P  of  Fig.  229  ;  but  this  very  deflection  gives  to  the 
other  component,  P  cos  a,  ||  to  the  tangent  at  L,  a  lever 
arm,  and  consequent  moment,  about  the  gravity  axes  of  all 
the  sections,  whence  for  2  (moms.L)=0  we  have,  (more  ex- 
actly than  from  eq.  (3)  when  x=l) 


.     ,     .     (6) 

(We  have  supposed  P  replaced  by  its  components  ||  and 
"|  to  the  fixed  tangent  at  L,  see  Fig.  301).  But  even  (6) 
will  not  give  an  exact  value  for  p2  at  L  ;  for  the  lever  arm 
of  P  cos  a,  viz.  d,  is  >(P  sina)Z3-^3.E7,  on  account  of  the 
presence  and  leverage  of  P  cos  a  itself.  The  true  value  of 
d  in  this  case  may  be  obtained  by  a  method  similar  to  that 
indicated  in  the  next  paragraph. 

297.    Elastic  Curve  of  Oblique  Cantilever  with  Terminal  Load. 
More  Exact  Solution.     For  variety  place  the  cantilever  as  in 

Fig.  302,  so  that  the  deflection 
OY=d  tends  to  decrease  the 
moment  of  P  about  the  gravity 
axis  of  any  section,  n.  "We 
may  replace  P  by  its  X  and  J 
components,  Fig.  303,  ||  and 
"1  respectively  to  the  fixed 
tangent  line  at  L.  The  origin, 

IP  +'/  0,  is  taken  at  the  free  end  of 

FIG.  302.  FIG.  3o3.       the  beam.     Let  a=  angle  bet- 

ween P  and  X.  For  a  free  body  On,  n  being  any  section, 
we  have  2  (moms.n)=0 


whence 


El  —  ^-==P(cosa)y  —  P(sina)  x 
dzr 


(1) 


[See  eq.  (1)  §  295a],  In  this  equation  the  right  hand 
member  is  evidently  (see  fig.  303)  a  negative  quantity; 
this  is  as  it  should  be,  for  Eldty-^dx*  is  negative,  the  curve 
being  concave  to  the  axis  X  in  the  first  quadrant.  (It 
must  be  noted  that  the  axis  JTis  always  to  be  taken  ||  to 
the  beam,  for  Eld^y-i-dx*  to  represent  the  moment  of  the 
stress-couple.) 


FLEXURE.       OBLIQUE     FORCES.  357 

Eq.  (1)  is  not  in  proper  form  for  taking  the  #-anti-deri- 
vative  of  both  members,  since  one  term  contains  the  vari- 
able ?/,  an  unknown  function  of  x.  Its  integration  is  in- 
cluded in  a  more  general  case  given  in  some  works  on  cal- 
culus, but  a  special  solution  by  Prof.  Robinson,  of  Ohio, 
is  here  subjoined  for  present  needs.* 

We  thus  obtain  as  the  equation  of  the  elastic  curve  in 
Fig.  303, 

/Pcos«  I"~g9;  i  g— ^l~|  rfginaVc (cos#\w"]=  sin  #[~eqaj 6~qx"l    (2) 

V  ~~ET  Ln      n  JL  J          L "       l  J 

In  which  en  denotes  the  NaperianBase=2.71828,  an  abstract 
number,  and  q  for  brevity  stands  for  VPcosa-^jET". 

To  find  the  deflection  d,  we  make  x=l  in  (2),  and  solve 
for  y;  the  result  is  d. 

The  uniform  thrust  at  L  is  pLfJ=Pcosa  ....  (3) 
while  the  stress  intensity  p2  in  the  outer  fibre  at  L,  is  ob- 

*  Denoting  Fcoa+EI\)y  g*  and  Psin  a-i-^r/by  jo2,  eq.  (1)  becomes^- =q*y—p*x  .  .  (Q) 
Differentiate  (6)  then  ~d~^—^~/~  ~P*-  Differentiate  again :  whence  ^-=^a—-y-  .  (7) 

Xetting  ^=w    .     (8)    so  that  u=q'*y-p'*x,  from  (Q)     we  have  0-=-^-and  |^=^, 
See  (7)  —L=q*Ut  which,  mult,  by  2  du,  gives  -^-5  2  C?M  du=2q*udu 

•'-    [     dx  constant     J.  -3-5       I  2dudu=2q*       I   udu-\-G   .'.  -r-n=q*u*4-C.    whence 
\  /    ax      j  j  dx* 


,/^!    ["«+  V^+C    1  ;or  C"«-Si 
n        ^'    L  2a    J  n 


(9) 


a  2o-a;        o'a;  n  qx     Cen 

Square  each   side  of  (9);  then    C'e    —2C'e    w-fw«=«H-^  '.    .'.  «=}^  C"«    — 


n       2C'q* 
\    (pee  eqs.  6  and  8)  ^y-^x=%  C'e*  -  1™          Consolidating  the 


358  MECHANICS   OF   ENGINEERING. 

tained  from  the  moment  equation  for  the  free  body  0  L 
viz:  -M=P(sin«)Z—  P(cos«)d    .....  (4) 


in  which  e=  distance  of  outer  fibre  from  the  gravity  axis. 
The  equation  for  safe  loading  is  written  out  by  placing 
the  values  of  plt  p2,  and  d,  as  derived  from  equations  (2), 
(3),  and  (4)  in  the  expression 


To  solve  the  resulting  equation  for  P,  in  case  that  is  the 
unknown  quantity,  can  only  be  accomplished  by  successive 
assumptions  and  approximations,  since  it  occurs  trans- 
cendentally. 

In  case  a  horizontal  tension-member  of  a  bridge-truss  is 
subjected  to  a  longitudinal  tension  P'  (due  to  its  position 
in  the  truss  and  the  load  upon  the  latter)  and  at  the  same 
time  receives  a  vertical  pressure=P"  at  the  middle,  each 
half  will  be  bent  in  the  same  manner  as  the  cantilever  in 
I'ig.  302  ;  y^P"  corresponding  to  P  sin  «,  and  P'  to  Pcos  a, 


pi  qx       —qx 

constant  factors  we  now  write  y  =  -^x-}-me    —  ne        .    .    the  equation  required  .   (10) 


To  determine  the  constants  m  and  n(m=  C'-t-'2q*;  n=  C+2C'q4)  we  first  find  dy+dxt  i.e. 
by  differentiating  (10)  ^=^+4™^+$™*  **  •    -    •    •    GD          a=0  fory=0 
.'.  (10)  gives     .    .    .    0=0+m«n—  /K?n  =0    i.e.  m— n=0  .'.  m=n   .     (12)    Also  for  x=l 


«a  Ql     —Ql~] 

.'.  (11)   gives  0^+tf     m<?n -fn<2n         .    .    (13);  /.  with  n=m  we  have  m=»=* 

ql   — ql  I 
~-p»-».0»    «n-Hn  •    •    .(14).    The  equation  of  the  curve,  then,  substituting  (14)  in 

qx    —qx 
(10)  is  y=^  x-p*>  gfl     gn  .  (15)    .-.,  Substituting  for  p  and  qvr* 


,  (as  in  §897)      / —£7-  I  ^ -H«  *         I  xsina — ycosa       =sina 

A/          Ln     l   J   L  J 


FLEXURE.       OBLIQUE   FORCES. 


359 


298.  Inclined  Beam  with  Hinge  at  One  End. — Fig.  304.  Let 
6  =  €i.  Required  the  equation  for  safe  loading  ;  also  the 
n^ximum  shear,  there  being  but  one  load,  P,  and  that  in 
tl»e  middle.  The  vertical  wall  being  smooth,  its  reaction, 


FIG.  305.     0 


Fio.306. 


B,  at  0  is  horizontal,  while  that  of  the  hinge-pin  being  un- 
known, both  in  amount  and  direction,  is  best  replaced  by 
its  horizontal  and  vertical  components  B0  and  PJ,,  unknown 
in  amount  only.  Supposing  the  flexure  slight,  we  find 
these  external  forces  in  the  same  manner  as  in  Prob.  1  § 
37,  by  considering  the  whole  beam  free,  and  obtain 


=r  cota  ;  H0  also  =  ^  cota  ;  VQ  =  P 


(1) 


For  any  section  n  between   0  and  Bt  we  have,  from  the 
free  body  nO,  Fig.  305, 

uniform  thrust  =  p^F  =  H  cos  a        •        .    (2) 
and  from  S  (moms.n)  =  0, 


(8) 


-  =Hx  sin  a 

e 


and  the  shear  =  J  =•  H  sin  «  =  ^  P  cos  a  (4) 

The  max.  (Pi+p2)  to  be  found  on  OB  is  /.  close  above  3t 
where  x  =  ^A  I,  and  is 


360  MECHANICS  OF   ENGINEERING. 

=  Jcos«+^sin«  wMch  =  p  cog    rcot, ;+.  ,  (g) 


In  examining  sections  on  CB  let  the  free  body  be  Cn't 
Fig.  306.     Then  from  I  (longitud.  comps.)  =  0 

(the  thrust  =)  p^F  =  VQ  sin  a  -f-  HQ  cos  a  (6)' 

Le.  j?1^T=P[sin  a  H-  y2  cos  a  cot  a]  (6) 

while,  from  2'  (moms.n')  =  0, 

£^=  Fia;'  cos  a—B<p'  sin  a  (7V 

e 

Le.  ^i=Xpcos«^  (7) 

6 

Hence  (p}  •}-  p*)  for  sections  on  CB  is  greatest  when  xr 
is  greatest,  which  is  when  x'  =  *^ltx'  being  limited  be- 
tween x'  =  0  and  a?'  =  j£  I,  and  is 


=P  cos  «[tan  "+/  Cot  a+^]       (8) 


which  is  evidently  greater  than  the  max. 

see  eq.  (5).     Hence  the  equation  for  safe  loading  is 


^'  =  P  cos  «r*2BJL±?l5!2*ff+j(^]    ....  (9) 
/^  /J 


in  which  R'  is  the  safe  normal  stress,  per  square  unit,  for 
the  material. 

The  shear,  J,  anywhere  on  GB,hom2  (transverse comp.) 

=0  in  Fig.  306,  is 

j 

J=  YQ  cos  «  —  HQ  sin  a  —  ]/2  P  cos  a       .         .  (10) 


FLEXURE.       OBLIQUE   FORCES. 


361 


As  showing  graphically  all  the  results  found,  moment^ 
thrust,  and  shear  diagrams  are  drawn 
in  Fig.  307,  and  also  a  diagram  whose  / !      //9 

ordinates  represent  the  variation  of 
(pi-\-pz)  along  the  beam.  Each  ordi- 
nate  is  placed  vertically  under  the 
gravity  axis  of  the  section  to  which  it 
refers. 

299.  Numerical  Example  of  the  Forego-  INCH.LBS. 
ing. — Fig.  308.  Let  the  beam  be  of 
wrought  iron,  the  load  P  =  1,800  Ibs., 
hanging  from  the  middle.  Cross  sec- 
tion rectangular  2  in.  by  1  in.,  the  2 
in.  being  parallel  to  the  force-plane. 
Required  the  max.  normal  stress  in 
any  outer  fibre  ;  also  the  max.  total 
shear. 

This  max.  stress -intensity  will  be  in 
the  outer  fibres  in  the  section  just  below  B  and  on  the 
upper  side,  according  to  §  298,  and  is  given  by  eq.  (8)  of 
that  article ;  in  which,  see  Fig.  308,  we  must  substitute 
(inch -pound-second-system)  P  =  1,800  Ibs.;  F  =  2  sq.  in.; 
I  =  V1202  +  122  =  120.6  in.;  e  =  1  in,,  I  =  ±  bW  =  n  =  % 
biquad.  inches  ;  cot  a  =  JL  ;  cos  a  =  .0996  ;  and  tan  a  =  10. 


FIG.  307. 


max. 


120.6  x  11  _ 


FIG.  309. 


FIG.  310. 


9000  Ibs.  persq.  inch,  very  nearly,  compression.    This  is  i» 


362  MECHANICS   OF   ENGINEERING. 

the  upper  outer  fibre  close  under  B.  In  the  lower  outer  fibre 
just  under  B  we  have  a  tension  =  p2  —  p^  =  7,200  Ibs.  per 
sq.  in.  (It  is  here  supposed  that  the  beam  is  secure  against 
yielding  sideways.) 

300.  Strength  of  Hooks.  —  An  ordinary  hook,  see  Fig.  309, 
may  be  treated  as  follows  :  The  load  being  =  P,  if  we 
make  a  horizontal  section  at  AB,  whose  gravity  axis  g  is 
the  one,  of  all  sections,  furthest  removed  from  the  line  of 
action  of  P,  and  consider  the  portion  C  free,  we  have  the 
shear  =  J  =  zero  .......  (1) 

the  uniform  pull  =  p{F=P      .        .        .     (2) 

•while  the  moment  of  the  stress-couple,  from  21  (moms.g)  = 
0,is 


Pa  (3) 

For  safe  loading  pl  -f  p2  must  =  R'9  ie. 


Tt  is  here  assumed  that  e  =  elt  and  that  the  maximum 
[P\~\~P2\  occurs  at  AB. 

301  Crane.  —  As  an  exercise  let  the  student  investigate  the 
strength  of  a  crane,  such  as  is  shown  in  Fig.  310. 


FL.EXUKE.      LOXG   COLUMNS. 


CHAPTEK  VL 


FLEXURE   OF   "  LONG  COLUMNS/' 


302.  Definitions. — By  "  long  column  "  is  meant  a  straight 
beam,  usually  prismatic,  which  is  acted  on  by  two  com- 
pressive  forces,  one  at  each  extremity,  and  whose   length 
is  so  great  compared  with  its  diameter  that  it  gives  way 
(or  "  fails  ")  by  buckling  sideways,  i.e.  by  flexure,  instead 
of  by  crushing  or  splitting  like  a  short  block  (see  §  200). 
The  pillars  or  columns  used  in  buildings,  the  compression 
members    of  bridge-trusses  and  roofs,  the  "  bents  "  of  a 
trestle  work,  and  the  piston-rods  and  connecting-rods  of 
steam-engines,  are  the  principal  practical  examples  of  long 
columns.     That  they  should  be  weaker  than  short  blocks 
of  the  same  material  and  cross-section  is  quite  evident,  but 
their  theoretical  treatment  is  much   less   satisfactory  than 
in  other  cases  of  flexure,  experiment  being  very  largely 
relied  on  not  only  to  determine   the  physical    constants 
which  theory  introduces  in  the  formulae  referring  to  them, 
but  even  to  modify  the  algebraic  form  of  those  formulae, 
thus  rendering  them  to  a  certain  extent  empirical. 

303.  End  Conditions. — The  strength  of  a  column  is  largely 
dependent  on   whether  the  ends  are  free  to  turn,    or  are 
fixed  and  thus  incapable  of  turning.     The  former  condi- 
tion is  attained  by  rounding  the  ends,  or  providing  them 
with  hinges  or  ball-and-socket-joints  ;  the  latter  by  facing 
off  each  end  to  an  accurate  plane  surface,  the  bearing  on 
which  it  rests  being  plane  also,  and  incapable  of  turning. 
In  the  former  condition  the  column  is  spoken  of  as  having 


364 


MECHANICS    Or   ENGINEERING. 


round  ends  ;  *  Fig.  311,  (a) ;  in  the  latter  as  having  fixed  ends, 
(or  flat  bases ;  or  square  ends),  Fig.  311,  (b). 


(a) 


FIG.  312. 


FIG.  311. 

Sometimes  a  column  is  fixed  at  one  end  while  the  other 
end  is  not  only  round  but  incapable  of  lateral  deviation  from 
the  tangent  line  of  the  other  extremity ;  this  state  of  end 
conditions  is  often  spoken  of  as  "Pin  and  Square,"  Fig. 
311,  (c). 

If  the  rounding  *  of  the  ends  is  produced  by  a  hinge  or 
"pin  joint,"  Fig.  312,  both  pins  lying  in  the  same  plane 
and  having  immovable  bearings  at  their  extremities,  the 
column  is  to  be  considered  as  round-ended  as  regards  flex- 
ure in  the  plane  ~|  to  the  pins,  but  as  square-ended  as  re- 
gards flexure  in  the  plane  containing  the  axes  of  the  pins. 

The  "  moment  of  inertia  "  of  the  section  of  a  column  will 
be  understood  to  be  referred  to  a  gravity  axis  of  the  sec- 
tion which  is  T  to  the  plane  of  flexure  (and  this  corres- 
ponds to  the  "  force-plane  "  spoken  of  in  previous  chap- 
ters), or  plane  of  the  axis  of  column  when  bent. 

303#.  Euler's  Formula. — Taking  the  case  of  a  round-ended 
column,  Fig.  313  (a),  assume  the  middle  of  the  length  as 
an  origin,  with  the  axis  X  tangent  to  the  elastic  curve  at 
that  point.  The  flexure  being  slight,  we  may  use  the  form 
dot?  for  the  moment  of  the  stress-couple  in  any 


*  With  round  ends,  or  pin  ends,  it  should  be  understood  that  the  force 
at  each  end  must  be  so  applied  as  to  act  through  the  centre  of  gravity  of  the 
base  (plane  figure)  of  the  prismatic  column  at  that  end  ;  and  continue  to  do 
so  as  the  column  bends. 


FLEXURE.       LONG   COLUMNS. 


365 


fa) 


FIG.  31c 


FIG.  314. 


section  n,  remembering  that  with  this  notation  the  axis  Jf 
must  be  ||  to  the  beam,  as  in  the  figure  (313).  Considering 
the  free  body  nC,  Fig,  313  (6),  we  note  that  the  shear  is 
zero,  that  the  uniform  thrust  =P,  and  that  2T(moms.n)=0 
gives  (a  being  the  deflection  at  0) 


Multiplying  each  side  by  dy  we  have 


ci/x 


dyd2y=Pady  —  Py  dy 


(2) 


Since  this  equation  is  true  for  the  y,  dx,  dy,  and  d?y  of  any 
element  of  arc  of  the  elastic  curve,  we  may  suppose  it 
written  out  for  each  element  from  0  where  y=Q,  and  dy=Q, 
up  to  any  element,  (where  dy—dy  and  y=y)  (see  Fig.  314) 
and  then  write  the  sum  of  the  left  hand  members  equal  to 
that  of  the  right  hand  members,  remembering  that,  since 
dx  is  assumed  constant,  1-^cfcc2  is  a  common  factor  on  the 
left.  In  other  words,  integrate  between  0  and  any  point 
of  the  curve,  n.  That  is, 


[dy-]d[dy-]=Pa   C   dy—P  f*  ydy         (3) 

-Q  Jo  Jo 

The  product  dy  d2y  has  been  written  (dy)d(dy\  (for  d?y  is 


366  MECHANICS   OF   ENGINEERING. 

the  differential  or  increment  of  dy)  and  is  of  a  form  like 
xdx,  or  ydy.  Performing  the  integration  we  have 

£•*•*»-*{  •    •    •    •  <*> 

which  is  in  a  form  applicable  to  any  point  of  the  curve, 
and  contains  the  variables  x  and  y  and  their  increments 
tfo  and  dy.  In  order  to  separate  the  variables,  solve  for  dx, 
and  we  have 


ordx=J^. (6) 


ZUVY 

a     \a  I 


r 

j. 


** 


sn 


|)       . 


(6)  is  the  equation  of  the  elastic  curve  J>0<7,  Fig.  313  (a), 
and  contains  the  deflection  a.  If  P  and  a  are  both  given, 
y  can  be  computed  for  a  given  x,  and  vice  versa,  and  thus 
the  curve  traced  out,  but  we  would  naturally  suppose  a  to 
depend  on  P,  i  or  in  eq.  (6)  when  x=y2l,y  should  =a.  Mak- 
ing these  substitutions  we  obtain 

i/2l=  Jljl  (vers.  sin  —  1.00)  ;  i.e.,  ^=  J*jj    \      (7) 

Since  a  has  vanished  from  eq.  (7)  the  value  for  P  ob- 
tained from  this  equation,  viz.: 

Pt=SI$     ....       (8) 

is  independent  of  a,  and 

is  .•.  to  be  regarded  as  that  force  (at  each  end  of  the  round" 

ended  column  in  Fig.  313)  which  will  hold  the  column  at 

any  small  deflection  at  which  it  may  previously  have  been 

set. 


FLEXURE.   LONG  COLUMNS. 


In  other  words,  if  the  force  is  less  than  P0  no  flexure  at 
all  will  be  produced,  and  hence  P0  is  sometimes  called  the 
force  producing  "  incipient  flexure."  [This  is  roughly  ver* 
ified  by  exerting  a  downward  pressure  with  the  hand  oij 
the  upper  end  of  the  flexible  rod  (a  T-squaj  e-blade  for  in^ 
stance)  placed  vertically  on  the  floor  of  a  room  ;  the  pres- 
sure must  reach  a  definite  value  before  a  decided  buckling 
takes  place,  and  then  a  very  slight  increase  of  pressure  oc- 
casions a  large  increase  of  deflection.] 

It  is  also  evident  that  a  force  slightly  greater  than  P0 
would  very  largely  increase  the  deflection,  thus  gaining  for 
itself  so  great  a  lever  arm  about  the  middle  section  as  to 
cause  rupture.  For  this  reason  eq.  (8)  may  be  looked 
upon  as  giving  the  Breaking  Load  of  a  column  with  round 
ends,  and  is  called  Euler's  formula. 

Keferring  now  to  Fig.  311,  it  will  be  seen  that  if  the  three 
parts  into  which  the  flat-ended  column  is  di-  , 

vided  by  its  two  points  of  inflection  A  and  B 
are  considered  free,  individually,  in  Fig.  315, 
the  forces  acting  will  be  as  there  shown,  viz.: 
At  the  points  of  inflection  there  is  no  stress- 
couple,  and  no  shear,  but  only  a  thrust,  =P> 
and  hence  the  portion  AB  is  in  the  condition 
of  a  round -ended  column.  Also,  the  tangents 
to  the  elastic  curves  at  0  and  C  being  pre- 
served vertical  by  the  f  rictionless  guide-blocks 
and  guides  (which  are  introduced  here  simply 
as  a  theoretical  method  of  preventing  the  ends 
from  turning,  but  do  not  interfere  with  verti- 
cal freedom)  OA  is  in  the  same  state  of  flex- 
ure as  half  of  AB  and  under  the  same  forces. 
Hence  the  length  AB  must  =  one  half  the 
total  length  I  of  the  flat-ended  column.  In 
other  words,  the  breaking  load  of  a  round- 
ended  column  of  length  = 


I,  is  the  same  as 

that    of   a   flat-ended  column   of   length   =L  — * 
Hence  for  the  I  of  eq.  (8)  write   y2l  and  we  |  p 

have  as  the  breaking  load  of  a  column  with        FlG-  315« 
flat -ends  and  of  length  =1. 


368  MECHANICS   OF   ENGINEERING. 


(9) 


Similar  reasoning,  applied  to  the  "  pin-and-square  " 
mode  of  support  (in  Fig.  311)  where  the  points  of  inflec- 
tion are  at  B,  approximately  ^  I  from  C,  and  at  the 
extremity  0  itself,  calls  for  the  substitution  of  2/$  lioil  in 
eq.  (8),  and  hence  the  breaking  load  of  a  "pin-and-square  "" 
column,  of  length  =  I,  is 

P2=|  EI*       .        .        .     "(10) 

Comparing  eqs.  (8),  (9),  and  (10),  and  calling  the  value  of 
PI  (flat-ends)  unity,  we  derive  the  following  statement  : 
The  breaking  loads  of  a  given  column  are  as  the  numbers 


I  flat-ends 


9/16 
pin  -and-square 


round-ends 


according  to  the 
mode  of  support. 


These  ratios  are  approximately  verified  in  practice. 

Euler's  Formula  [i.e.,  eq.  (8)  and  those  derived  from  it, 
(9)  and  (10)]  when  considered  as  giving  the  breaking  load 
is  peculiar  in  this  respect,  that  it  contains  no  reference  to 
the  stress  per  unit  of  area  necessary  to  rupture  the  material 
of  the  column,  but  merely  assumes  that  the  load  producing 
"  incipient  flexure  ",  i.e.,  which  produces  any  bending  at 
all,  will  eventually  break  the  beam  because  of  the  greater 
and  greater  lever  arm  thus  gained  for  itself.  In  the  canti- 
lever of  Fig.  241  the  bending  of  the  beam  does  not  sensibly 
affect  the  lever-arm  of  the  load  about  the  wall-section,  but 
with  a  column,  the  lever- arm  of  the  load  about  the  mid- 
section  is  almost  entirely  due  to  the  deflection  produced. 

304.  Example.  Euler's  formula  is  only  approximately 
verified  by  experiment.  As  an  example  of  its  use  when 
considered  as  giving  the  force  producing  "  incipient  flex- 
ure "  it  will  now  be  applied  in  the  case  of  a  steel  T-square- 
blade  whose  ends  are  free  to  turn.  Hence  we  use  the 
round-end  formula  eq.  (8)  of  §303,  with  the  modulus  of 
elasticity  ^=30,000,000  Ibs.  per  sq.  inch.  The  dimensions 


FLEXURE,   LONG  COLUMNS.  369 

are  as  follows  :  the  length  I  =  30  in.,  thickness  =.  ^  of  an 
inch,  and  width  ~  2  inches.  The  moment  of  inertia,  Ir 
about  a  gravity  axis  of  the  section  ||  to  the  width  (the 
plane  of  bending  being  ||  to  the  thickness)  is  (§247) 


.-.  ,  with  TT  =  22  -f-  7, 

«>     30,000,000 


inches' 


-2031bs 
~AUt 


o-       -j-     162,000         72 

Experiment  showed  that  the  force,  a  very  small  addition 
to  which  caused  a  large  increase  of  deflection  or  side-buck- 
ling, was  about  2  Ibs. 

305.  Hodgkinson's  Formulae  for  Columns.—  The  principal 
practical  use  of  Euler's  formula  was  to  furnish  a  general 
form  of  expression  for  breaking  load,  to  Eaton  Hodgkin- 
eon,  who  experimented  in  England  in  1840  upon  columns 
of  iron  and  timber. 

According  to  Euler's  formula  we  have  for  cylindrical 

columns,  /being  =J^  TIT*  =  ~  nd*  (§247), 

U4 

for  flat-ends        *         .          J\  =i  ET?  .  * 

i.fc  ,  proportional  to  the  fourth  power  of  the  diameter,  and 
inversely  as  the  square  of  the  length.  But  Hodgkinson's 
experiments  gave  for  wrought-iron  cylinders 

^3.55  .J3.&1 

Pl  =  (const.)  x  ,2  —  ;  and  for  cast  iron  Pl=  (const.)  x  ^-7— 

Again,  for  a  square  column,  whose  side  =  bt  Euler's  for* 
mula  would  give 


whilo  Hodgkinson  found  for  square  pillars  of  wood 

fe4 
P!=  (const.)  X 


370  MECHANICS   OF   ENGINEERING. 

Hence  in  the  case  of  wood  these  experiments  indicated  the 
same  powers  for  b  and  I  as  Euler's  formula,  but  with  a  dif- 
ferent constant  factor;  while  for  cast  and  wrought  iron 
the  powers  differ  slightly  from  those  of  Euler. 

Hodgkinson's  formulae  are  as  follows,  and  evidently 
not  homogeneous  ;  the  prescribed  units  should  .'.  be  care- 
fully followed,  d  denotes  the  diameter  of  the  cylindrical 
columns,  b  the  side  of  square  columns,  1=  length. 

(  For  solid  cylindrical  cast  iron  columns,  flat-ends  ; 


(  For  solid  cylindrical  wrought  iron  columns,  flat-ends  ; 
\  Breaking  load  in  tons  )      IQ^^/J-    ,-   ,o™\3<55  .    n\     n  \ 
(     of  2,240  Ibs.  each    }  = 

f  For  solid  square  columns  of  dry  oak,  flat-ends  • 

t  J5tSs  iffiS"  }  =1^  ><  (>  -  -w  -  9  !-* 


{For  solid  square  columns  of  dry  fir  ',  flat-ends  ; 
Brroko^1i(^d  ^  ^  I  =7.81  X  (b  in  inches)4  -f-  (Z  in  ft.)* 
of  2,240  Ibs.  each     ) 

Hodgkinson  found  that  when  the  mode  of  support  was 
•'  pin-and-square,"  the  breaking  load  was  about  y2  as 
great  ;  and  when  the  ends  were  rounded,  about  ^  as  great 
as  with  flat  ends.  These  ratios  differ  somewhat  from  the 
theoretical  ones  mentioned  in  §303,  just  after  eq.  (10.) 

Experiment  shows  that,  strictly  speaking,  pin  ends  are 
not  equivalent  to  round  ends,  but  furnish  additional 
strength  ;  for  the  friction  of  the  pins  in  their  bearings 
hinders  the  turning  of  the  ends  somewhat.  As  the  lengths 
become  smaller  the  value  of  the  breaking  load  in  Hodg- 
kinson's formulae  increases  rapidly,  until  it  becomes  larger 
than  would  be  obtained  by  using  the  formula  for  the 
crushing  resistance  of  a  short  block  (§201)  viz.,  FC,  i.e., 
the  sectional  area  X  the  crushing  resistance  per  unit  of 
area. 

In  such  a  case  the  pillar  is  called  a  short  column,  or  "  short 
block,"  and  the  value  FG  is  to  be  taken  as  the  breaking 


FLEXURE.   LONG  COLUMNS.  371 

load.  This  distinction  is  necessary  in  using  Hodgkinson's 
formulae ;  i.e.,  the  breaking  load  is  the  smaller  of  the  two 
values,  FC  and  that  obtained  by  Hodgkinson's  rule. 

In  present  practice  Hodgkinson's  formulae  are  not  often 
used  except  for  hollow  cylindrical  iron  columns,  for  which 
with  d.2  and  d^  as  the  external  and  internal  diameters,  we 
have  for  fiat-ends 

Breaking  load  in  tons  )      n       ,        (d  in  in.)355  —  (c?t  in  in.)3-5! 
of  2,240  Ibs.  each     f=  ~(Z  in  feet)" 

in  which  the  const.  =  44.16  for  cast  iron,  and  134  for 
wrought,  while  n  =  1.7  for  cast-iron  and  =  2  for  wrought. 

306.  Examples  of  Hodgkinson's  Formulae. — Example  1.  Re- 
quired the  breaking  weight  of  a  wrought-iron  pipe  used 
as  a  long  column,  having  a  length  of  12  feet,  an  internal 
diameter  of  3  in.,  and  an  external  diameter  of  3^  inches, 
the  ends  having  well  fitted  flat  bases. 

If  we  had  regard  simply  to  the  sectional  area  of  metal, 
which  is  F  =  1.22  sq.  inches,  and  treated  the  column  as  a 
short  block  (or  short  column)  we  should  have  for  its  com- 
pressive  load  at  the  elastic  limit  (see  table  §203)  P"=FC" 
-=1.22  X  24,000=29,280  Ibs.  and  the  safe  load  Pl  may  be 
taken  at  16,000  Ibs. 

But  by  the  last  formula  of  the  preceding  article  we  have 

Breaking  load   in   )      1<unv  (3.25)355  -  3s-55       1Rn7  . 
ton  of  2,240  Ibs.  each  [  =  ~J^~ 

i.e.=  15.07  X  2240=33,768  Ibs. 
Detail     [log.  3.25]  x  3.55=  0.511883  x  3.55=  1.817184 ; 

[log.  3.00]  x 3.55=0.477,121x3.55=1.693,779  ; 

and  the  corresponding  numbers  are  65.6  and  49.4 ;  theil 
difference  =  16.2,  hence 

Br.  load  in  long  tons  =  134*la2= 15.072   long  tons. 

=33,768  Ibs. 


372  MECHANICS  OF   ENGINEERING. 

With  a  "  factor  of  safety  "  (see  §205)  of  four,  we  have,  as 
the  safe  load,  p'  =  8,442  Ibs.  This  being  less  than  the 
16000  Ibs.  obtained  from  the  "  short  block  "  formula,should 
be  adopted. 

If  the  ends  were  rounded  the  safe  load  would  be  one- 
third  of  this  i.e.,  would  be  2,814  Ibs  ;  while  with  pin-and- 
square  end-conditions,  we  should  use  one-half,  or  4,221  Ibs. 

EXAMPLE  2.  Eequired  the  necessary  diameter  to  be 
given  a  solid  cylindrical  cast-iron  pillar  with  flat  ends,  that 
its  safe  load  may  be  13,440  Ibs.  taking  6  as  a  iactor  of 
safety.  Let  d  =  the  unknown  diameter.  Using  the  proper 
formula  in  §  305,  and  hence  expressing  the  breaking  load, 
which  is  to  be  six  times  the  given  safe  load,  in  long  tons 
vre  have  (the  length  of  column  being  16  ft.) 

13440x6  _  44.16  (d  in  inches)3-55  ,-  , 

2240  1617 


ie.[<2  in  inches]3-5^  (2) 

or  log.d=JLj[log.  36+1.7  x  log.  16-log.  44.16]      .     .       (3) 
A  log.d=3JL_[1.958278]  =0.551627  .-.  d  -  3.56  ins. 

This  result  is  for  flat  ends.  If  the  ends  were  rounded, 
we  should  obtain  d  =  4.85  inches. 

307.  Rankine's  Formula  for  Columns.  —  The  formula  of  this 
name  (some  times  called  Gordon's,  in  some  of  its  forms)  has 
a  somewhat  more  rational  basis  than  Euler's,  in  that  it  in- 
troduces the  maximum  normal  stress  in  the  outer  fibre  and 
is  applicable  to  a  column  or  block  of  any  length,  but  still 
contains  assumptions  not  strictly  borne  out  in  theory,  thus 
introducing  some  co-efficients  requiring  experimental  de- 
termination. It  may  be  developed  as  follows  : 

Since  in  the  flat-ended  column  in  Fig.  315  the  middle 
portion  AB,  between  the  inflection  points  A  and  B,  is 
acted  on  at  each  end  by  a  thrust  ==  P,  not  accompanied  by 
any  shear  or  stress-couple,  it  will  be  simpler  to  treat  thai 


FLEXUKE.   LONG  COLUMNS.  373 

portion  alone  Fig.   316,  (a),  since  the  thrust  and  stress- 
couple  induced  in   the   section   at    _A 
R,  the  middle  oiAB,  will  be  equal   \   \\  K\\  (&) 

to  those  at  the  flat  ends,  0  and  6Y,   I 
in  Fig.  315.     Let  a  denote  the  de-   | 
flection  of  R  from  the  straight  line   ] 
AB.     Now  consider  the   portion  » 
AR  as  a  free  body  in  Fig.  316,  (6),   i 
putting  in  the  elastic  forces  of  the 
section  at  R,  which  may  be  clas-    |    |  /  (a) 
sified    into    a  uniform  thrust   = 
p\F,  and  a  stress  couple  of  moment  FIG.  sie. 

=  ^L.,  (see  §  294).      (The  shear   is  evidently  zero,  from 
e 

I  (hor  comps.)  ==  0).  Here  pl  denotes  the  uniform  pres- 
sure (per  unit  of  area),  due  to  the  uniform  thrust,  and  p2 
the  pressure  or  tension  (per  unit  of  area),  in  the  elastic 
forces  constituting  the  stress-couple,  on  the  outermost 
element  of  area,  at  a  distance  e  from  the  gravity  axis  ("] 
to  plane  of  flexure)  of  the  section.  F  is  the  total  area  of 
the  section.  I  is  the  moment  of  inertia  about  the  said 
gravity  axis,  g 

I  (vert,  comps.)  —  0  gives  P  -=  p±F        .        .      (f) 


—  Ogives  Pa  =  ....    (2) 

€> 

For  any  section,  n,  between  A  and  R,  we  would  evidently 
have  the  same^  as  at  R,  but  a  smaller  p2,  since  Py  <  Pa 
while  e,  I,  and  F9  do  not  change,  the  column  being  pris- 
matic. Hence  the  max.  (pi~\-p»)  is  on  the  concave  edge  at 
R  and  for  safety  should  be  no  more  than  G  -f-  n,  where  G 
is  the  Modulus  of  Crushing  (§  201)  and  n  is  a  "  factor  of 
safety."  Solving  (1)  and  (2)  for  pi  andj^,  and  putting  their 
sum  =  C  -f-  ft;  we  have 

P    Pae    G  .„. 

~ 


We  might  now  solve  for  P  and  call  it  the  safe  load,  bat 


374  MECHANICS   OF   ENGINEERING. 

is  customary  to  present  the  formula  in  a  form  for  giving 
the  breaking  load,  the  factor  of  safety  being  applied  after- 
ward. Hence  we  shall  make  n  =  1,  and  solve  for  P,  call- 
ing it  then  the  breaking  load.  Now  the  deflection  a  is  un- 
known, but  may  be  expressed  approximately,  as  follows, 
in  terms  of  e  and  I. 

Suppose   two   columns  of       lengths   =   Z'  and  Z",  each 

bearing  its  safe  load.     Then  at  the  point  R,  —  —  ._=^-^-t  iae.9 

P         e 
E'e'  =   p'  p2'.     Considering  the  curve  AB  as  a  circular  arc 

We  have  (see  §  290)  a'  =  Z'2  -^-  32  p'  ,  i.e.  a'  =o^->  .  -'  5  and 

r>tf         Z"2 
similarly  for  the  other  column,  a"  =    ^2       .  _  .      If   the 

€ 


columns  are  of  the  same  material  E'  =  E"  ,  and  if  each  is 
bearing  its  safe  load  we  may  assume  p2'  =  p2"  nearly,  in 
which  case  the  term  p%f  ~  E"  =  p2'  -4-  E  ',  and  we  may 
say  that  the  deflection  a,  under  safe  load,  is  proportional 
to  (length)2  -f-  e,  approximately,  i.  e.,  that  ae  =  ftl2,  where 
(3  is  a  constant  (an  abstract  number  also)  dependent  on 
experiment  and  different  for  different  materials,  and  I  the 
full  length.  We  may  also  write,  for  convenience,  1  =  Fk2, 
k  being  the  radius  of  gyration  (see  §  85).  Hence,  finally, 
we  have  from  eq.  (3) 

Breaking    load  )  _p  _     FG  />n 

for  flat  ends        j  =      l~  -  W  *          V? 


This  is  known  as  Rankine's  formula. 

By  the  same  reasoning  as  in  §  303,  for  a  round-ended 
column  we  substitute  2  Z  for  Z ;  for  a  pin-and-square  col- 
umn -t  Z  for  Z ;  and  .'.  obtain 

Breaking  load  l-.pss       ^^  (X\ 

for  a  round-ended  column  j  """    °~  J*      •    •    •    •   W 


Ar2 

)     -7,  ^C 

a  pin-and-square  column 


Breaking  load  for  )  —r>  FC 

(         °=     .    .. 


FLEXURE.      LONG   COLUMNS 


375 


These  formulae,  (4),  (5),  and  (6),  unlike  Hodgkinson's, 
are  of  homogeneous  form.  Any  convenient  system  of  units 
may  therefore  be  used  in  them. 

Rankine  gives  the  following  values  for  C  and  /?,  to  be 
o«ed  in  these  formulae.  These  are  based  on  Hodgkinson's 
experiments.* 


CAST  IRON. 

WB'T  IRON. 

TIMBER. 

C  in  Ibs.  per  sq.  in. 

80,000  f 

36,000 

7,200 

/?  (abstract  number) 

1 

1 

1 
3,000 

6,400 

36,000 

If  these  numerical  values  of  C  are  used  F  must  be  ex- 
pressed in  Sq.  Inches  and  P  in  Pounds.  Rankine  recom- 
mends 4  as  a  factor  of  safety  for  iron  in  quiescent  struct- 
ures, 5  under  moving  loads  ;  10  for  timber.  The  N.  J. 
Iron  &  Steel  Co.  use  Eankine's  formula  for  their  wrought 
iron  rolled  beams,  when  used  as  columns,  with  a  factor  of 
safety  of 


308.  Examples,  Using  Eankine's  Formula.  —  EXAMPLE  1.— 
Take  the  same  data  for  a  wrought  :ron  pipe  used  as  a 
column,  as  in  example  1,  §  306  ;  i.e.,  1=12  ft.  =144  inches, 
F=}£[n(3%)2—7:32]  =1.227  sq.  inches,  while  1$  for  a  nar- 
row circular  ring  like  the  present  section  may  be  put 
=  ^(1^)2  (see  §  98)  sq.  inches.  With  these  values,  and 
(7=36,000  Ibs.  per  sq.  in.,  and  ft=^^  (for  wrought  iron), 
we  have  from  eq.  (4),  for  flat  ends, 


.=   L2fx86n°!L  =3°743-6  »* 

1    |_  •*-  (  JL  ji^t ) 

3pOO'  i  [1.625]2 


(1) 


This  being  the  breaking  load  th*»  safe  load  may  be  taken 
=  J^  or  y5  of  30743.6  Ibs.,  according  as  the  structure  o! 

*  For  "structural  steel"  the  Cambria  Steel  Co.  uses  C—  45,000  to  50,000 
Ibs.  per  sq.  in.,  and  ft  =  ^inr- 

f  In  1897  experiments  in  New  York  (Eng.  News,  Jan.  1898,  p.  27)  indi- 
cated values  of  Cfrom  40,000  to  70,000  ;  80,000  is  therefore  too  high. 


376  MECHANICS   OF   ENGINEERING. 

which  the  column  is  a  member  is  quiescent  or  subject  to 
vibration  from  moving  loads.  By  Hodgkinson's  formula 
33,768  Ibs.  was  obtained  as  a  breaking  load  in  this  case 
(§  306). 

For  rounded  ends  we  should  obtain  (eq.  5) 

P0=16,100.  Ibs.,  as  break,  load  .         (2) 

and  for  pin-and-square,  eq.  (6) 

P2=24,908.  Ibs.  as  break,  load        .        .    (3) 

EXAMPLE  2.—  (Same  as  Example  2,  §  306).     Kequired  by 

Rankine's  formula  the  necessary  diameter,  d,  to  be  given 

a  solid  cylindrical  cast-iron  pillar,  16  ft.  in  length,  with 

rounded  ends,  that  its  safe  load  may  be  six  long  tons  (i.e., 

of  2,240  Ibs.  each)  taking  6  as  a  factor  of  safety.     F=^  , 

while  the  value  of  I?  is  thus  obtained.     From  §  247,  /  for 
a  full  circle  about  its  diameter  =%>r*=;rr2.J^r8  .'.  ^= 
Hence  eq.  (5)  of  §  307  becomes. 


Fo  the  breaking  load  is  to  be  =6x6x2,240  Ibs.,  0  for  cast- 
iron  is  80,000  Ibs.  per  sq.  inch,  while  ft  (abstract  number) 
~6Sob-  Solving  for  d  we  have  the  biquadratic  equation  : 

dt__  28x6x6x2,240^  28x6x6x2,240xl62xl22x4 
22x80,000     —  22x80,000x400 

vhence  e£*=  0.641  (1±  33.92),  and  taking  the  upper  sign, 
finally,  d=  A/22.4  =4.73  inches.  (By  Hodgkinson's  rule 
we  obtained  4.85  inches). 

309.  Radii  of  Gyration.  —  The  following  table,  taken  from 
p.  523  of  Kankine's  Civil  Engineering,  gives  values  of  &*, 
the  square  of  the  least  radius  of  gyration  of  the  given  cross- 
section  about  a  gravity-axis.  By  giving  the  least  value  oi 


FLEXURE.   LONG  COLUMNS. 


377 


#  it  is  implied  that  the  plane  of  flexure  is  not  determined 
by  the  end-conditions  of  the  column ;  (i.e.,  it  is  implied 
that  the  column  has  either  flat  ends  or  round  ends.)  If 
either  end  (or  both)  is  a,pi,n-joint  the  column  may  need  to 
be  treated  as  having  a  flat-end  as  regards  flexure  in  a  plane 
containing  the  axis  of  the  column  and  the  axis  of  the  pin. 
if  the  bearings  of  the  pin  are  firm ;  while  as  regards  flex- 
tire  in  a  plane  perpendicular  to  the  pin  it  is  to  be  consid- 
ered round-ended  at  that  extremity. 

In  the  case  of  a  "  thin  cell "  the  value  of  ^  is  strictly 
true  for  metal  infinitely  thin  and  of  uniform  thickness  ;  stiH 
if  that  thickness  does  not  exceed  l/h  of  the  exterior  diame- 
ter, the  form  given  is  sufficiently  near  for  practical  pur- 
poses ;  similar  statements  apply  to  the  branching  forms. 


FIG.  317. 


Solid  Eectangle. 
h=  least  side. 
Thin  Square  Cell 
=  h. 


h=  least  side. 
Solid  Circular  Section. 
Diameter  =d. 
Thin  Circular  Cell. 
Exterior  diam,  =  d. 
Angle-Iron  of  Equal 
rius 


Fie.  318. 


Fig.  317(o). 

*•*$*• 

Fig.  317  (&> 

^=ih" 

Fig.  317  (c). 

f-  h" 

'  12 

Fig.  317  (d). 

**=-* 

Fig.  317  (e). 

*=»* 

Fig.  317  (f) 

ArJ=-^.0^ 

378 


MECHANICS   OF   ENGINEEEING. 


Angle-Iron  of  unequal 

ribs. 

Cross  of  equal  arms. 
I-Beam  as  a  pillar. 
Let  area  of  web  =_Z?. 

"     "  loth  flanges 

=A. 

Channel 
Iron. 


Fig.  318  (a).      tf= 
Fig.  318  (6).      &=4P 

Fig.  318  (c).      **=g 


Fig.318(,*).      ^ 


Let  area  of  web  =B ;  of  flanges   =A  (both),     h  extends 
from  edge  of  flange  to  middle  of  web. 


FIG.  319. 


PHCENIX   COLUMN. 


FIG.  320. 


310.  Built  Columns. — The  "  compression  members  "  of 
wrought-iron  bridge  trusses  are  generally  composed  of 
several  pieces  riveted  together,  the  most  common  forms 
being  the  Phoenix  column  (ring-shaped,  in  segments,)  and 
combinations  of  channels,  plates,  and  lattice,  some  of  which 
are  shown  in  Figs.  319  and  320. 

Experiments  *  on  full  size  columns  of  these  kinds  were 
made  by  the  U.  S.  Testing  Board  at  the  Watertown  Arse- 
nal about  1880. 

The  Phoenix  columns  ranged  from  8  in.  to  28  feet  in 
length,  and  from  1  to  42  in  the  value  of  the  ratio  of  length 
to  diameter.  The  breaking  loads  were  found  to  be  some- 
what in  excess  of  the  values  computed  from  Bankine'g 
formula  ;  from  10  to  40  per  cent,  excess.  In  the  pocket- 
book  issued  by  the  Phoenix  company  they  give  the  follow* 
ing  formula  for  their  columns,  (wrought-iron.) 

*  See  also  p.  112  of  the  author's  "Notes  and  Examples  in  Mechanics " 
for  the  Pencoyd  experiments  on  wrought  iron  columns. 


FLEXUKE.      LO^G   COLUMNS.  379 


Breaking  load  in  Ibs.   )        50,000  F 
for  flat-ended  columns  J  ""    "  ~ 


where  F  =  area  in  sq.  in.,  I  =  length,  and  h  =  external 
diameter,  both  in  the  same  unit. 

Many  different  formulae  have  been  proposed  by  different 
engineers  to  satisfy  these  and  other  recent  experiments  on 
columns,  but  all  are  of  the  general  form  of  Bankine's. 
For  instance  Mr.  Bouscaren,  of  the  Keystone  Bridge  Co., 
claims  that  the  strength  of  Phoenix  columns  is  best  given 
by  the  formula 

Breaking  load  in  }  _    38,000  F 
Ibs.  for  flat-ends. 


)  = 

j  " 


100,000£a 

(F  must  be  in  square  inches.) 

The  moments  of  inertia,  7,  and  thence  the  value  of  k*  = 
/  -i-  Fy  for  such  sections  as  those  given  in  Figs.  319  and 
320  may  be  found  by  the  rules  of  §§  85-93,  (see  also  §  258.) 

(For  the  "  Straight-line  Formula,"  see  §  314a,  p.  385.) 

311.  Moment  of  Inertia  of  Built  Column.  Example. — It  is  pro- 
posed to  form  a  column  by  joining  two  I-beams  by  lattice- 
work, Fig.  321,  (a).  (While  the  lattice-work  is  relied  upon 
to  cause  the  beams  to  act  together  as  one  piece,  it  is  not 
regarded  in  estimating  the  area  F9  or  the  moment  of  iner- 
tia, of  the  cross  section).  It  is  also  required  to  find  the 
proper  distance  apart  =  x,  Fig.  321,  at  which  these  beams 
must  be  placed,  from  centre  to  centre  of  webs,  that  the 
liability  to  flexure  shall  be  equal  in  all  axial  planes,  i.e. 
that  the  1  of  the  compound  section  shall  be  the  same 
about  all  gravity  axes.  This  condition  will  be  ful- 
filled if  /Y  can  be  made  —7^*  (§89),  0  being  the  centre 
of  gravity  of  the  compound  section,  and  X  perpendicular 
to  the  parallel  webs  of  the  two  equal  I-beams. 

Let  F'  =  the  sectional  area  of  one  of  the  I-beams,  Fr 
(see  Fig.  321(a)  its  moment  of  inertia  about  its  web-axis, 
that  about  an  axis  "]  to  web.     (These  quantities  can  bo 

*  That  is, 'with  flat  ends  or  ball  ends  ;  but  with  pin  ends.  Fig.  312,  if  the 
pin  Is  ||  to  Xt  put  4/Y  =  /x ;  if  II  to  r,  put  4/x  =  /Y  . 


380 


MECHANICS   OF   ENGINEERING. 


found  in  the  hand-book  of  the  iron  company,  for  each  size 
of  rolled  beam). 
Then  the 


total  Jx  =  2PX  ;  and  total  7Y  ==  2v  + 


[ 


(see  §88  eq.  4.)    If  these  are  to  be  equal,  we  write  them  so 
and  solve  for  xt  obtaining 


«=  /iZiEZ 

V        v 


312.  Numerically ;  suppose  each  girder  to  be  a  10^  inch 
light  I-beam,  105  Ibs.  per  yard,  of  the  N.  J.  Steel  and  Iron 
Co.,  in  whose  hand-book  we  find  that  for  this  beam  7'x  = 
185.6  biquad.  inches,  and  I\  =  9.43  biquad.  inches,  while 
F'  =  10.44  sq.  inches.  With  these  values  in  eq.  (1)  we 
have 


8.21  inches 


Cai 


L_    I 

\ 

? 

fT* 

p 

-JS 

^ 

\ 
\ 

j^s 

(&) 

Pie.  881. 


The  square  of  the  radius  of  gyration  will  be 

^=2Px-s-2jP'=  371.2  -J-20.88  =17.7  sq.  in.      .      (2) 

and  is  the  same  for  any  gravity  axis  (see  §  89). 

As  an  additional  example,  suppose  the  two  I-beams  united 
by  plates  instead  of  lattice.  Let  the  thickness  of  the  plate 
«=  t,  Fig.  321,  (b).  Neglect  the  rivet-holes.  The  distance 
a  is  known  from  the  hand-book.  The  student  may  derive 
a  formula  for  x,  imposing  the  condition  that  (total  Jx)^  ^Y* 


TLEXUEE.   LOXG  COLUMNS.  381 

313.    Trussed  Girders. — When  a  horizontal  beam  is  trussed 


FIG.  322. 


in  the  manner  indicated  in  Fig.  322,  with  a  single  post  or 
strut  under  the  middle  and  two  tie-rods,  it  is  subjected  to 
a  longitudinal  compression  due  to  the  tension  of  the  tie- 
rods,  and  hence  to  a  certain  extent  resists  as  a  column,  the 
plane  of  whose  flexure  is  vertical,  (since  we  shall  here  sup- 
pose the  beam  supported  later ally. )Taking  the  case  of  uni- 
form loading,  (total  load  =  W  )and  supposing  the  tie-rods 
screwed  up  (by  sleeve  nuts)  until  the  top  of  the  poet  is  on 
a  level  with  the  piers,  we  know  that  the  pressure  between 
the  post  and  the  beam  is  P'  =  %  JF(see  §  273).  Hence 
by  the  parallelogram  of  forces  (see  Fig.  322)  the  tension 
in  each  tie-rod  is 


Q 


W 


2  cos  a.     16  *    cos  a 
At  each  pier  the  horizontal  component  of  Q  is 

5 


=  Q  sin  a=: 


~ 
16 


•  (i) 


Hence  we  are  to  consider  the  half -beam  B  0  as  a  "  pin-and- 
square  "  column  under  a  compressive  force  P=5/ie  W  tan  at 
as  well  as  a  portion  of  a  continuous  girder  over  three 
equidistant  supports  at  the  same  level  and  bearing  a  uni- 
form load  W.  In  the  outer  fibre  of  the  dangerous  section, 
0,  (see  also  §  273  and  Fig.  278)  the  compression  per  sq. 
inch  due  to  both  these  straining  actions  must  not  exceed 
a  safe  limit,  Rf,  (see  §  251).  In  eq.  (6)  §  307,  where  P2  is 
tJie  breaking  force  for  a  pin-and-square  column,  the  great- 


382  MECHANICS   OF   ENGINEERING. 

est  stress  in  any  outer  fibre  —  G  (  =  the  Modulus  of  Crush- 
ing) per  unit  of  area.  If  then  we  write  pcol.  instead  of  G 
in  that  equation,  and  5/16  JFtan  «  instead  of  Pa  we  have 

j  max.  stress  due  )  =  5      JPtan  an,  ,    16  ^  I2  ~|. 

(  to  column  action  j       ^co1      Jg  *        ~]?  "9"*  P  'T&V 

while  from  eq.  (3),  p.  326,  we  have  (remembering  that  our 
present  W  represents  double  the  Wot  §273). 


max.  stress  due  )  =      =  1    Wle^  1      Wle 
to  girder  action  j      "     i$     2    ~16 


By  writing  pco\.-\-psf=R'=  a  safe  value  of  compression  per 
unit-area,  we  have  the  equation  for  safe  loading 

.  (2) 

Here  I  =  the  half-span  OS,  Fig.  322,  e  =  the  distance  of 
outer  fibre  from  the  horizontal  gravity  axis  of  the  cross 
section,  h*  =  the  radius  of  gyration  of  the  section  referred 
to  the  same  axis,  while  F  =  area  of  section.  y9  should  be 
taken  from  the  end  of  §307. 

EXAMPLE.—  If  the  span  is  30  ft.  =  360  in.,  the  girder  a  15 
inch  heavy  I-beam  of  wrought  iron,  200  Ibs.  to  the  yard,  in 
which  e  =  ^  of  15  =  7^  inches,  -F=20  sq.  in.,  and  F  = 
35.3  sq.  inches  (taken  from  the  Trenton  Co.'s  hand-book), 
required  the  safe  load  Wt  the  strut  being  5  ft.  long. 
From  §307,  p  =  1  :  36,000  ;  tan  a  =  15-^5  =  3.00.  Hence, 
using  the  units  pound  and  inch  throughout,  and  putting 
R'  =  12,000  Ibs.  per  sq.  in.  =  max.  allowable  compression 
stress,  we  have  from  eq.  (2) 

™       16x20x12,000  ' 


0)2i 

.3  _ 


9  ^36,000  '  35.3  _P      35.3 

L  e.,  69,111  Ibs.  besides  the  weight  of  the  beam. 

If  the  middle  support  had  been  a  solid  pier,  the  safe  load 
would  have  been  48  tons  ;  while  if  there  had  been  no 
middle  support  of  any  kind,  the  beam  would  bear  safely 


FLEXURE.   LONG  COLUMNS. 


383 


only  11.5  tons, 
the  strut)]. 


[Let  the  student  design  the  tie-rods  (and 


314-  Buckling  of  Web-Plates  in  Built  Girders. — In  §257  men- 
tion was  made  of  the  fact  that  very  high  web  plates  in 
built  beams,  such  as  /beams  and  box-girders,  might  need 
to  be  stiffened  by  riveting  T-irons  on  the  sides  of  the  web. 
(The  girders  here  spoken  of  are  horizontal  ones,  such  as 
might  be  used  for  carrying  a  railroad  over  a  short  sj^an  of 
20  to  30  feet. 

An  approximate  method  of  determining  whether  such 
stiffening  is  needed  to  prevent  lateral  buckling  of  the  web, 
may  be  based  upon  Rankine's  formula  for  a  long  column 
and  will  now  be  given. 

In  Fig.  323  we  have,  free,  a  portion  of  a  bent  I-beam, 
between  two  vertical  sections  at  a  distance  apart=  ^  = 
the  height  of  the  web.  In  such  a  beam  under  forces  L  to 
its  axis  it  has  been  proved  (§256)  that  we  may  consider 
the  web  to  sustain  all  the  shear,  J,  at  any  section,  and  the 
flanges  to  take  all  the  tension  and  compression,  which 
form  the  "  stress -couple"  of  the  section.  These  couples 
and  the  two  shears  are  shown  in  Fig.  323,  for  the  two 
exposed  sections.  There  is  supposed  to  be  no  load  on  this 
portion  of  the  beam,  hence  the  shears  at  the  two  ends  are 


SLl 

i 

J 

ik, 

j 

<r  —  h-  —  ». 

H 

•f 

1 


FIG.  323. 


equal.  Now  the  shear  acting  between  each  flange  and  the 
horizontal  edge  of  the  web  is  equal  in  intensity  per  square 
inch  to  that  in  the  vertical  edge  of  the  web  ;  hence  if  the 
web  alone,  of  Fig.  323,  is  shown  as  a  free  body  in  Fig.  324, 
we  must  insert  two  horizontal  forces  =  J,  in  opposite 


384  MECHANICS   OF   ENGINEERING. 

directions,  on  its  upper  and  lower  edges.  Each  of  thesa 
f=  J  since  we  have  taken  a  horizontal  length  Aj  =  height 
of  web.  In  this  figure,  324,  we  notice  that  the  effect  of 
the  acting  forces  is  to  lengthen  the  diagonal  BD  ana 
shorten  the  diagonal  AC,  both  of  those  diagonals  making 
an  angle  of  45°  with  the  horizontal. 

Let  us  now  consider  this  buckling  tendency  along  AC, 
by  treating  as  free  the  strip  A  (7,  of  small  width  =  6,.  This 
is  shown  in  Fig.  325.  The  only  forces  acting  in  the  direc- 
tion of  its  length  AC  SLTQ  the  components  along  AC  of  the 
four  forces  J'  at  the  extremities.  We  may  therefore  treat 
the  strip  as  a  long  column  of  a  length  I  —  hi  A/2,  of  a  sec- 
tional area  F  =  bblf  (where  b  is  the  thickness  of  the  web 
plate),  with  a  value  of  k2  =  1/12  62  (see  §  309),  and  with 
fixed  (or  flat)  ends.  Now  the  sum  of  the  longitudinal 
components  of  the  two  JVs  at  A  is  Q  =  2  J'  }4  V2 

=  J'  A/2;  but  «/'  itself  —  ^r.  b   l/2  k  A/2,  since  the  small 

rectangle  on  which  J'  acts  has  an  area  =  b  y2  \  A/2,  and 
/he  shearing  stress  on  it  has  an  intensity  of  (J  -j-  bhj  per 
unit  of  area.  Hence  the  longitudinal  force  at  each  end  of 
this  long  column  is 


(1) 


According  to  eq.  (4)  and  the  table  in  §  307,  the  safe  load 
(factor  of  safety  =  4)  for  a  wi*ought-iron  column  of  this 
form,  with  flat  ends,  would  be  (pound  and  inch) 


^^36,000  9,000  66, 


,ov 


If,   then,   in   any   particular    locality   of    the    girder  (ol 
wrought-irori)  we  find  chat  Q  is  >  Pl9  i.e. 


if       is  >  (pound  and  inch)    .    .    (3) 

*«      i  .      l     hl 
1+  1,500'  fi2 

then  vertical  stiffeners  will  be  required  laterally. 

When  these  are  required,  they  are  generally  placed  at 


FLEXURE.   LONG  COLUMNS  385 

intervals  equal  to  hlt  (the  depth  of  web),  along  that  part 
of  the  girder  where  Q  is  >  P^ 

EXAMPLE  Fig.  326. — Will  stiffening  pieces  be  required 
in  a  built  girder  of  20  feet  span,  bearing  a  uniform  load  of 
40  tons,  and  having  a  web  24  in.  deep  and  ^  in.  thick  ? 

From  §   242  we  know  that  the      w  -40  TONS 

greatest  shear,  e/max.,  is  close  to 
either  pier,  and  hence  we  investi-  |  \ — 
gate  that  part  of  the  girder  first. 

J    max.    =    y2    W  =   20    tons 
=40,000  Ibs. 
.-.  (inch  and  lb.),  see  (3), 

Fie  886. 


.905.0 (6) 


Ai         24 

while,  see  (3),  (inch  and  pound), 
9,000  x^ 

+i>6oo*o$F 

which  is  less  than  1666.66. 

Hence  stiffening  pieces  will  be  needed  near  the  extremities 
of  the  girder.  Also,  since  the  shear  for  this  case  of  loading 
diminishes  uniformly  toward  zero  at  the  middle  they  will 
be  needed  from  each  end  up  to  a  distance  of  £££$  of  10  ft. 
from  the  middle. 

314a.  The  "Straight-Line"  Formula—  Mr.  T.  H.Johnson 
(Transac.  Am.  Soc.  C.  E.,  1886)  has  proposed  formulae  for 
breaking  load  in  Ibs.  per  sq.  inch  of  sectional  area  of  column, 
as  follows  (founded  on  observed  tests) : 

Wrought  iron  :   Hinged  ends,  p  =  42,000  —  157  j* ;         ^ 

7      5 

Flat  ends,      p  =  42,000  —  128  r-; 

Mild  steel :         Hinged  ends,  p  =  52,500  —  220r  ; 

fc 

Flat  ends,       p  =  52,500  —  179r  ; 

K 

where  I  is  the  length,  and  Jc  is  the  least  radius  of  gyration 
of  the  cross- section,  of  column. 

These  formulae  have  been  much  used. 


386  MECHANICS  OF  ENGINEERING. 


CHAPTER  VIL 


LINEAR  ARCHES  (OF  BLOCKWORK). 


815.  A  Blockwork  Arch  is  a  structure,  spanning  an  opening 
or  gap,  depending,  for  stability,  upon  the  resistance  to 
compresssion  of  its  blocks,  or  voussoirs,  the  material  of 
which,  such  as  stone  or  brick,  is  not  suitable  for  sustain- 
ing a  tensile  strain.  Above  the  voussoirs  is  usually 
placed  a  load  of  some  character,  (e.q.  a  roadway,)  whose 
pressure  upon  the  voussoirs  will  be  considered  as  vertical, 
only.  This  condition  is  not  fully  realized  in  practice, 
unless  the  load  is  of  cut  stone,  with  vertical  and  horizontal 
joints  resting  upon  voussoirs  of  corresponding  shape  (see 
Fig.  327),  but  sufficiently  so  to  warrant 
its  assumption  in  theory.  Symmetry 
of  form  about  a  vertical  axis  will  also 
be  assumed  in  the  following  treatment. 


316.  Linear  Arches. — For  purposes  of 
theoretical  discussion  the  voussoirs  of 
Fig.  327  may  be  considered  to  become 
PIG  327  infinitely  small  and  infinite  in  number, 

thus  forming  a  "  linear  arch,"  while  retaining  the  same 
shapes,  their  depth  "1  to  the  face  being  assumed  constant 
that  it  may  not  appear  in  the  formulae.  The  joints 
between  them  are  ~J  to  the  curve  of  the  arch,  i.e.,  adjacent 
voussoirs  can  exert  pressure  on  each  other  only  in  the 
direction  of  the  tangent-line  to  that  curve. 


LINEAR  ARCHES. 


387 


317.  Inverted  Catenary,  or  linear  Arch  Sustaining  its  Own 
Weight  Alone, — Suppose  the  infinitely  small  voussoirs  to 
have  weight,  uniformly  distributed  along  the  curve,  weigh- 
ing q  Ibs.  per  running  linear  unit.  The  equilibrium  of 
such  a  structure,  Fig.  328,  is  of  course  unstable  but  theo- 
retically possible.  Bequired  the  form  of  the  curve  when 
equilibrium  exists.  The  conditions  of  equilibrium  are, 
obviously  :  1st.  The  thrust  or  mutual  pressure  T  between 
any  two  adjacent  voussoirs  at  any  point,  A,  of  the  curve 
must  be  tangent  to  the  curve  ;  and  2ndly,  considering  a 
portion  BA  as  a  free  body,  the  resultant  of  H0  the  pres- 


BH0 


FIG.  328. 


Fie.  329. 


Fig.  330. 


sure  at  B  the  crown,  and  T  at  A,  must  balance  E  the  re- 
sultant of  the  II  vertical  forces  (i.e., weights  of  the  elementary 
voussoirs)  acting  between  B  and  A. 

But  the  conditions  of  equilibrium  of  a  flexible,  inexten- 
sible  and  uniformly  loaded  cord  or  chain,  are  the  very 
same  (weights  uniform  along  the  curve)  the  forces  being 
reversed  in,  direction.  Fig.  329.  Instead  of  compression 
we  have  tension,  while  the  II  vertical  forces  act  toward  in- 
stead of  away  from,  the  axis  X.  Hence  the  curve  of  equi- 
librium of  Fig.  328  is  an  inverted  catenary  (see  §  48)  whose 
equation  is 


(1) 


See  Fig.  330.  e  =  2.71828  the  Naperian  Base.  The  "par- 
ameter "  c  may  be  determined  by  putting  x  =  a,  the  half 
span,  and  y=  0  Y,  the  rise,  then  solving  for  c  by  successive 


388 


MECHANICS   OF   ENGINEERING. 


approximations.  The  "  horizontal  tlirust"  or  H0,  is  =  qc, 
while  if  s  =  length  of  arch  OA,  along  the  curve,  the  thrust 
T  at  any  point  A  is 


From  the  foregoing  it  may  be  inferred  that  a  series  of 

soirs  of  finite  dimensions,    arranged 

so  as  to  contain  the  catenary  curve, 

with  joints  "|  to  that  curve  and  of 

equal  weights  for  equal  lengths  of 

arc  will    be    in    equilibrium,    and 

moreover  in   stable   equilibrium  on 

account  ol  friction,  and  the   finite 

width  of  the  joints ;  see  Fig.  331. 


Flo.  331. 


318.  Linear  Arches  under  Given  Loading. — The  linear  arches 
to  be  considered  further  will  be  treated  as  without  weight 
themselves  but  as  bearing  vertically  pressing  loads  (each 
voussoir  its  own). 

Problem. — Given  the  form  of  the  linear  arch  itself,  it  is 
required  to  find  the  law  of  vertical  depth  of  loading  under 
which  the  given  linear  arch  will  be  in  equilibrium.  Fig. 
332,  given  the  curve  ABC,  i.e.,  the  linear  arch  itself,  re- 
quired the  form  of  the  curve  JM.ON,  or  upper  limit  of  load- 
ing, such  that  the  linear  arch  ABC  shall  be  in  equilibrium 
under  the  loads  lying  between  the  two  curves.  The  load- 
ing is  supposed  homogeneous  and  of  constant  depth  ~]  to 
paper ;  so  that  the  ordinates  z  between  the  two  curves  are 
proportional  to  the  load  per  horizontal  linear  unit.  Assume 
a  height  of  load  «..,  at  the  crown,  at  pleasure  ;  then  required 
the  z  of  any  point  m  as  a  function  of  a0  and  the  curve 
ABC. 


FIG.  332. 


FIG.  333. 


LINEAR   ARCHES. 


389 


Practical  Solution. — Since  a  linear  arch  under  vertical 
pressures  is  nothing  more  than  the  inversion  of  the  curve 
assumed  by  a  cord  loaded  in  the  same  way,  this  problerq 
might  be  solved  mechanically  by  experimenting  with  a 
light  cord,  Fig.  333,  to  which  are  hung  other  heavy  cords, 
or  bars  of  uniform  weight  per  unit  length,  and  at  equal 
horizontal  distances  apart  when  in  equilibrium.  By  varying 
the  lengths  of  the  bars,  and  their  points  of  attachment,  we 
may  finally  find  the  curve  sought,  MON.  (See  also  §  343.) 

Analytical  Solution. — Consider  the  structure  in  Fig.  334 
A  number  of  rods  of  finite  length,  in  the  same  plane,  are  in 
equilibrium,  bearing  the  weights  P,  Plf  etc.,  at  the  con- 


[•     - 


FIG.  334. 


FIG.  335. 


necting  joints,  each  piece  exerting  a  thrust  T  against  the 
adjacent  joint.  The  joint  A,  (the  "  pin  "  of  the  hinge),  im- 
agined separated  from  the  contiguous  rods  and  hence  free, 
is  held  in  equilibrium  by  the  vertical  force  P  (a  load)  and 
the  two  thrusts  T  and  T',  making  angles  =  d  and  0'  with 
the  vertical ;  Fig.  335  shows  the  joint  A  free.  From  -T(hor* 
izontal  comps.)=0,  we  have. 

I7  sin  0=T'  sind'. 

That  is,  the  horizontal  component  of  the  thrust  in  any  rod 
is  the  same  for  all ;  call  it  H».  .*. 


(i) 


3'JO 


MECHANICS   OF   ENGINEERING. 


Now  draw  a  line  As  *"i  to  T'  and  write   2  ( compons.  |  to 
As)=0 ;  whence  P  sin  0'=  T  sin  /?,  and  [see  (1)] 


(2) 


sin  6  sin  0' 


Let  the  rods  of  Fig.  334  become  infinitely  small  and  infi- 
nite in  number  and  the  load  continuous.  The  length  of 
each  rod  becomes  =ds  an  element  of  the  linear  arch,  ft  is 
the  angle  between  two  consecutive  cfo's,  6  is  the  angle  be- 
tween the  tangent  line  and  the  vertical,  while  P  becomes 
the  load  resting  on  a  single  dx,  or  horizontal  distance  be- 
tween the  middles  of  the  two  ds's.  That  is,  Fig.  336,  if 
f=  weight  of  a  cubic  unit  of  the 
loading,  P=ftdx.  (The  lamina  of 
arch  and  load  considered  is  unity, 
~|  to  paper,  in  thickness.)  H0=a, 
constant  =  thrust  at  crown  0 ; 
f)=0'}  and  sin  fi=ds-t-p,  (since  the 
angle  between  two  consecutive  tan- 
gents is  =  that  between  two  con- 
secutive radii  of  curvature).  Hence 
eq.  (2)  becomes 


-3 

p  sin2  0 


but  dx=ds  sin  0, 

H, 

|0  sin30 


Fie.  336. 


(3) 


Call  the  radius  of  curvature  at  the  crown  p0,  and  since 
there  z  =  ZQ  and  0*=90°,  (3)  gives  7?^ =-£{>;  hence  (3)  may 
be  written 


/>  sin3  0 

This  is  the  law  of  vertical  depth  of  loading  required.  For 
a  point  of  the  linear  arch  where  the  tangent  line  is  verti- 
cal, sin  0=0  and  z  would  =  oo  ;  i.e.,  the  load  would  be  in- 


LINEAR   ARCHES. 


391 


finitely  high.     Hence,  in  practice,  a  full  semi-circle,  for  in- 
stance, could  not  be  used  as  a  linear  arch. 

319.  Circular  Arc  as  Linear  Arch. — As  an  example  of  the 

preceding  problem  let  us  ap- 
ply eq.  (4)  to  a  circular  arc, 
Fig.  337,  as  a  linear  arch. 
Since  for  a  circle  p  is  con- 
eq.  (4)  reduces 


stant    = 
to 


(5) 


Hence  the  depth  of  loading 
must  vary  inversely  as  the  cube  of  the  sine  of  the  angle  0 
made  by  the  tangent  line  (of  the  linear  arch)  with  the  ver- 
tical. 

To  find  the  depth  z  by  construction.— Having  z0  given,  G 
being  the  centre  of  the  arch,  prolong  Ca  and  make  db  = 
«0 ;  at  b  draw  a  "1  to  Cb,  intersecting  the  vertical  through  a 
nt  some  point  d ;  draw  the  horizontal  dc  to  meet  Co  at 
some  point  c.  Again,  draw  ce  ~|  to  Cc,  meeting  ad  in  e ; 
then  ae  =  z  required  ;  a  being  any  point  of  the  linear  arch. 
For,  from  the  similar  right  triangles  involved,  we  have 

z0=ab=ad  sin  d=ac  sin  6.  sin  d=ae  sin  6  sin  6  sin  6 


ae= 


ae=z.  Q.E.D. 

[see  (5.)] 


320.  Parabola  as  Linear  Arch.  —  To  apply  eq.  4  §  318  to  a 
parabola  (axis  vertical)  as  linear  arch,  we  must  find  values 
of  p  and  p0  the  radii  of  curvature  at  any  point  and  the 
crown  respectively,  That  is,  in  the  general  formula, 


we  must  substitute  the  forms  for  the  first  and  second  dif- 
ferential co-efficients,  derived  from  the  equation  of  the 


392 


MECHANICS    OF    ENGINEERING. 


FIG.  338. 


FIG.  339. 


curve  (parabola)  in  Fig.  338,  i.e.  from  a?2  =  2  py  ;  whence 
we  obtain 


p 


Hence  o  ==- 


=  p  cosec.3#,  i.e.  p  =  P 


(6) 


At  the  vertex  6  =  90°  ,\  p0  =  p.     Hence  by  substituting 
for  p  and  j00  in  eq.  (4)  of  §  318  we  obtain 

z=s0=  constant  [Fig.  339]  .....  (7) 
for  a  parabolic  linear  arch.  Therefore  the  depth  of  homo- 
geneous loading  must  be  the  same  at  all  points  as  at  the 
crown  ;  i.e.,  the  load  is  uniformly  distributed  with  respect 
to  the  horizontal.  This  result  might  have  been  antici- 
pated from  the  fact  that  a  cord  assumes  the  parabolic 
form  when  its  load  (as  approximately  true  for  suspension 
bridges)  is  uniformly  distributed  horizontally.  See  §  46 
in  Statics  and  Dynamics. 

321.  Linear  Arch  for  a  Given  Upper  Contour  of  Loading,  the 
arch  itself  being  the  unknown  lower  contour.  Given  the 
upper  curve  or  limit  of  load  and  the  depth  20  at  crown,  re- 
quired the  form  of  linear  arch  which  will  be  in  equili- 
brium under  the  homogenous  load  between  itself  and  that 
upper  curve.  In  Fig.  340  let  MON  be  the  given  upper 
contour  of  load,  «0  is  given  or  assumed,  z'  and  z"  are  the 
respective  ordinates  of  the  two  curves  BA  C  and  M  ON. 
Required  the  eqation  of  BAG. 


LINEAR   ARCHES. 


393 


FIG.  340. 


FIG  341. 


As  before,  the  loading  is  homogenous,  so  that  the 
weights  of  any  portions  of  it  are  proportional  to  the 
corresponding  areas  between  the  curves.  (Unity  thick- 
ness T  to  paper.)  Now,  Fig.  341,  regard  two  consecutive 
ds's  of  the  linear  arch  as  two  links  or  consecutive  blocks 
bearing  at  their  junction  m  the  load  dP  =  f  (zf  +  z"}  dx  in 
which  f  denotes  the  heaviness  of  weight  of  a  cubic  unit  of 
the  loading.  If  T  and  T'  are  the  thrusts  exerted  on  these 
two  blocks  by  their  neighbors  (here  supposed  removed) 
we  have  the  three  forces  dP>  T  and  T',  forming  a  system 
in  equilibrium.  Hence  from  IX =0. 


T  cos  <p  =  T'  coe  <?' 


and 


2Y=0  gives  T'  sin  <p'—  T  sin  <p  =  dP 


(1) 


(2) 


From  (1)  it  appears  that  T  cos  <p  is  constant  at  all  points 
of  the  linear  arch  (just  as  we  found  in  §  318)  and  hence 
-=the  thrust  at  the  crown,  =  H,  whence  we  may  write 

T=H  -4-  cos  <p  and  T'=H  -T-  cos  <p'     .     .     .    (3) 
Substituting  from  (3)  in  (2)  we  obtain 


H  (tan  <p'  —  tan  <p)=dP 


(4) 


But  tan  <p  =  —  and  tan  (f,'= 
dx 


dx 


constant) 


while  dP  =  f  (z'  +  z")  dx.     Hence,  putting  for  convenience 
H  —  ;a2,  (where  a   =  side   of  an  imaginary  square  of  the 


394  MECHANICS  OF   ENGINEERING. 

loading,  whose  thickness  =  unity  and  whose  weight  —  H) 
we  have. 


as  a  relation  holding  good  for  any  point  of  the  linear  arch 
which  is  to  be  in  equilibrium  under  the  load  included 
between  itself  and  the  given  curve  whose  ordinates  are  z", 
Fig.  340. 

322.  Example  of  Preceding.  Tipper  Contour  a  Straight  Line.— 
Fig.  342.  Let  the  upper  contour  be  a  right  line  and  hor- 
izontal ;  then  the  «"  of  eq.  5  becomes  zero  at  all  points  of 
ON.  Hence  drop  the  accent  of  z'  in  eq.  (5)  and  we  have 


dx2     a2 
Multiplying  which  by  dz  we  obtain 

dz  d2z     I 


dx*      a2 


(6) 


This  being  true  of  the  z,  dz,  d2z  and  dx  of  each  element  of 
the  curve  O'B  whose  equation  is  desired,  conceive  it  writ- 
ten out  for  each  element  between  0'  and  any  point  m,  and 
put  the  sum  of  the  left-hand  members  of  these  equations 
=  to  that  of  the  right-hand  members,  remembering  that 
a2  and  dx2  are  the  same  for  each  element.  This  gives 


.    .    .    .    (7.) 


LINEAR  ARCHES. 

a.       N 


395 


FIG.  342.  PIG  343 

Integrating  (7.)  between  0'  and  any  point  m 


=  «r  log..  (;-+(-!)- 1)  . 

O  L*o  v^>  \  zo/          ' 


or  2=  -H 


(8.) 


(9.) 


This  curve  is  called  the  transformed  catenary  since  we  may 
obtain  it  from  a  common  catenary  by  altering  all  the  ordi- 
nates  of  the  latter  in  a  constant  ratio,  just  as  an  ellipse 
may  be  obtained  from  a  circle.  If  in  eq.  (9)  a  were  =  ZQ 
the  curve  would  be  a  common  catenary. 
.  Supposing  2d  and  the  co-ordinates  x±  and  zl  of  the  point 
B  (abutment)  given,  we  may  compute  a  from  eq.  8  by  put- 
ting x  =xl  and  «  =  «„  and  solving  for  a.  Then  the"  crown- 
thrust  H  =  f-a2  becomes  known,  and  a  can  be  used  in  eqs. 
(8)  or  (9)  to  plot  points  in  the  curve  or  linear  arch.  From 
eq.  (9)  we  have 

x  XT-      x  _*         -i  T-     —     — -2L-1 

area     )        /*          ?o  /»  f   a"         ^       _azQ\     a      «  QQ\ 

OO'mn  f  =Jo^=2  Jo  [e  ^+e  ^J~  ^  L6 "        J 

Fig.  343. 

Call  this  area,  A.  As  for  the  thrusts  at  the  different 
joints  of  the  linear  arch,  see  Fig.  343,  we  have  crown- 
thrust  =  H  =  fa?  .  .  .  ;  .  .  .  (11) 
and  at  any  joint  m  the  thrust 

?  =rVou^F2       •       •  •       (12; 


396  MECHANICS  OF   ENGINEERING. 

323.  Remarks.— The  foregoing  results  may  be  utilized 
with  arches  of  finite  dimensions  by  making  the  arch-ring 
contain  the  imaginary  linear  arch,  and  the  joints  T  to  the 
curve  of  the  same.  Questions  of  friction  and  the  resist- 
ance of  the  material  of  the  voussoirs  are  reserved  for  a 
succeeding  chapter,  (§  344)  in  which  will  be  advanced  a 
more  practical  theory  dealing  with  approximate  linear 
arches  or  "  equilibrium  polygons "  as  they  will  then  be 
called.  Still,  a  study  of  exact  linear  arches  is  valuable  on 
many  accounts.  By  inverting  the  linear  arches  so  far  pre- 
sented we  have  the  forms  assumed  by  flexible  and  inexten- 
sible  cords  loaded  in.  the  same  way. 


GRAPHICAL  STATICS.  397 


CHAPTEK   VTEL 


ELEMENTS    OF    GRAPHICAL,   STATICS. 


924.  Definition. — In  many  respects  graphical  processes 
have  advantages  over  the  purely  analytical,  which  recom- 
mend their  use  in  many  problems  where  celerity  is  desired 
without  refined  accuracy.  One  of  these  advantages  is  that 
gross  errors  are  more  easily  detected,  and  another  that 
the  relations  of  the  forces,  distances,  etc.,  are  made  so 
apparent  to  the  eye,  in  the  drawing,  that  the  general  effect 
of  a  given  change  in  the  data  can  readily  be  predicted  at 
a  glance. 

Graphical  Statics  w  the  system  of  geometrical  construc- 
tions by  which  problems  in  Statics  may  be  solved  by 
the  use  of  drafting  instruments,  forces  as  well  as  distances 
being  represented  in  amount  and  direction  by  lines  on  the 
paper,  of  proper  length  and  position,  according  to  arbi- 
trary scales  ;  so  many  feet  of  distance  to  the  linear  inch  of 
paper,  for  example,  for  distances ;  and  so  many  pounds  or 
tons  to  the  linear  inch  of  paper  for  forces. 

Of  course  results  should  be  interpreted  by  the  same 
scale  as  that  used  for  the  data.  The  parallelogram  of 
forces  is  the  basis  of  all  constructions  for  combining  and 
resolving  forces. 

325.    Force  Polygons  and  Concurrent  Forces  in  a  Plane. — If  a 

material  point  is  in  equilibrium  under  three  forces  PI  P3 
P3  (in  the  same  plane  of  course)  Fig.  344,  any  on©  of  them, 


398 


MECHANICS   OF   ENGINEERING. 


as  P19  must  be  equal  and  opposite  to  E  the  resultant  of 
the  other  two  (diagonal  of  their  parallelogram).  If  now 
we  lay  off  to  some  convenient  scale  a  line  in  Fig.  345  — 
Pl  and  ||  to  PI  in  Fig.  344  ;  and  then  from  the  pointed  end 

of  PI  a  line  equal  and  ||  to  P2  and 
laid  off  pointing  the  same  way,  we 
note  that  the  line  remaining  to 
i>  close  the  triangle  in  Fig.  345  must 
be  =  and  ||  to  P3,  since  that  tri- 
angle is  nothing  more  than  the 
left-hand  half -parallelogram  of 
FIG.  345.  Fig.  344.  Also,  in  345,  to  dose 
the  triangle  properly  the  directions  of  the  arrows  must 
be  continuous  Point  to  Butt,  round  the  periphery.  Fig. 
345  is  called  a  force  polygor  ;  of  three  sides  only  in  this 
case.  By  means  of  it,  given  any  two  of  the  three  forces 
which  hold  the  point  in  equilibrium,  the  third  can  be 
found,  being  equal  and  ||  to  the  side  necessary  to  "  close  " 
the  force  polygon. 

Similarly,  if  a  number  of  forces  in  a  plane  hold  a  mate- 
rial point  in  equilibrium,  Fig.  346,  their  force  polygon, 


Fia.344. 


FIG.  346. 


FIG.  347. 


Fig.  347,  must  close,  whatever  be  the  order  in  which  its 
sides  are  drawn.  For,  if  we  combine  PI  and  P2  into  a  re- 
sultant Oa,  Fig.  346,  then  this  resultant  with  P2  to  form  a 
resultant  Ob,  and  so  on ;  we  find  the  resultant  of  PI,  P2,  P3r 
and  P4  to  be  Oc,  and  if  a  fifth  force  is  to  produce  equilib- 
rium it  must  be  equal  and  opposite  to  Oc,  and  would  close 
the  polygon  OdabcO,  in  which  the  sides  are  equal  and  par- 


GRAPHICAL   STATICS. 


399 


allel  respectively  to  the  forces  mentioned.  To  utilize  this 
fact  we  can  dispense  with  all  parts  of  the  parallelograms  in 
Fig.  346  except  the  sides  mentioned,  and  then  proceed  as 
follows  in  Fig.  347  : 

If  P5  is  the  unknown  force  which  is  to  balance  the  other 
four  (i.e,  is  their  anti-resultant),  we  draw  the  sides  of  the 
force  polygon  from  A  round  to  B,  making  each  line  paral- 
lel and  equal  to  the  proper  force  and  pointing  the  same 
way  ;  then  the  line  EA  represents  the  required  P5  in 
amount  and  direction,  since  the  arrow  EA  must  follow 
the  continuity  of  the  others  (point  to  butt). 

If  the  arrow  BA  were  pointed  at  the  extremity  B,  then 
it  gives,  obviously,  the  amount  and  direction  of  the  result- 
ant of  the  four  forces  Pl  .  .  .  P4.  The  foregoing  shows 
that  if  a  system  of  Concurrent  Forces  in  a  Plane  is  in  equi- 
librium, its  force  polygon  must  close. 


326.  Non-Concurrent  Forces  in  a  Plane. — Given  a  system  of 
non-concurrent  forces  in  a  plane,  acting  on  a  rigid  body, 
required  graphic  means  of  finding  their  resultant  and  anti- 
resultant  ;  also  of  expressing  conditions  of  equilibrium. 
The  resultant  must  be  found  in  amount  and  direction  ;  and 
also  in  position  (i.e.,  its  line  of  action  must  be  determined). 
E.  g.,  Fig.  3^8  shows  a  curved  rigid  beam  fixed  in  a  vise 
at  T,  and  also  under  the  action  of  forces  Pl  P2  P3  and  P4 
(besides  the  action  of  the  vise);  required  the  resultant  of 

P1?  P2,  P3,andP4. 
By  the  ordinary 
parallelogram  of 
forces  we  com- 
bine P!  and  P2  at 
a,  the  intersection 
of  their  lines  of 

FIO  34?  action,  into  a  re- 

sultant E., ;  then  J?a  with  P3  at  b,  to  form  jftb;  and  finally  J?b 
with  P4  at  c  to  form  Pc  which  is  .*.  the  resultant  required, 
i.e.,  of  P,  .  .  .  .  P4 ;  and  c  .  .  .  F  is  its  line  of  action. 


400 


MECHANICS  OF   ENGINEERING. 


FIG- 


The  separate  force  triangles  (half-parallelograms)  by 
which  the  successive  partial  resultants  J?a,  etc.,  were  found, 
are  again  drawn  in  Fig.  349.  Now  since  JRC,  acting  in  the 

line  c..F,  Fig.  348, 
is  the  resultant  of 
PI  .  .  P4,  it  is  plain 
that  a  force  Pc' 
equal  to  RQ  and  act- 
ing along  c  .  .  P,but 
in  the  opposite  di- 
rection, would  balance  the  system  PI  .  .  .  P4,  (is  their  anti- 
resultant).  That  is,  the  forces  P,  P2  P3  P4  and  RQ'  would 
form  a  system  in  equilibrium.  The  force  RJ  then,  repre- 
sents the  action  of  the  vise  T  upon  the  beam.  Hence  re- 
place the  vise  by  the  force  RJ  acting  in  the  line  .  .  .  F  .  .  .  c  • 
to  do  which  requires  us  to  imagine  a  rigid  prolongation  of 
that  end  of  the  beam,  to  intersect  F  .  .  .  c.  This  is  shown  in 
Fig.  350  where  the  whole  beam  is  free,  in  equilibrium,  under 
the  forces  shown,  and  in  precisely  the  same  state  of  stress, 
part  for  part,  as  in  Fig.  348.  Also,  by  combining  in  one 
force  diagram,  in  Fig.  351,  all  the  force  triangles  of  Fig.  349 
(by  making  their  common  sides  coincide,  and  putting  RJ 
instead  of  Bc,  and  dotting  all  forces  other  than  those  of 
Fig.  350),  we  have  a  figure  to  be  interpreted  in  connection 
with  Fig.  350. 


SPACE  DIAGRAM 
FIG.  350. 


Hen&  we  note,  first,  that  in  the  figure  called  a  force-dia- 
gram,. P,  Pt  PS,  P4  and  Rc'  form  a  closed  polygon  and  that 


GRAPHICAL   STATICS.  401 

their  arrows  follow  a  continuous  order,  point  to  butt, 
around  the  perimeter  ;  which  proves  that  one  condition  of 
equilibrium  of  a  system  of  non-concurrent  forces  in  a  plane 
is  that  its  force  polygon  must  close.  Secondly,  note  that  db 
is  ||  to  Oa',  and  be  to  Ob' ;  hence  if  the  force-diagram  has 
been  drawn  (including  the  rays,  dotted)  in  order  to  deter- 
mine the  amount  and  direction  of  jR0',  or  any  other  one  force, 
we  may  then  find  its  line  of  action  in  the  space-diagram,  as 
follows:  (N.  B. — By  space  diagram  is  meant  the  figure  show- 
ing to  a  true  scale  the  form  of  the  rigid  body  and  the  lines 
of  action  of  the  forces  concerned).  Through  a,  the  intersec- 
tion of  Pl  and  P2,  draw  a  line  ||  to  Oa'  to  cut  P3  in  some  point 
b ;  then  through  b  a  line  ||  to  Ob'  to  cut  P4  at  some  point  o;  cF 
drawn  |j  to  Oc'  is  the  required  line  of  action  of  RQ',  the  anti- 
resultant  of  P0  P2,  P3,  and  P4. 

abc  is  called  an  equilibrium  polygon;  this  one  having  but 
two  segments,  ab  and  be  (sometimes  the  lines  of  action  of  Pl 
and  Rc'  may  conveniently  be  considered  as  segments.)  The 
segments  of  the  equilibrium  polygon  are  parallel  to  the  respect" 
ive  rays  of  the  force  diagram. 

Hence  for  the  equilibrium  of  a  system  of  non-concurrent 
forces  in  a  plane  not  only  must  its  force  polygon  close, 
but  also  the  first  and  last  segments  of  the  corre- 
sponding equilibrium  polygon  must  coincide  with 
the  resultants  of  the  first  two  forces,  and  of  the  last 
two  forces,  respectively,  of  the  system.  E.g.,  ab  coin- 
cides with  the  line  of  action  of  the  resultant  of  Pl  and  P2 ; 
be  with  that  of  P4  and  B/c.  Evidently  the  equil.  polygon 
will  be  different  with  each  different  order  of  forces  in 
the  force  polygon  or  different  choice  of  a  pole,  0.  But  if 
the  order  of  forces  be  taken  as  above,  as  they  occur  along 
the  beam,  or  structure,  and  the  pole  taken  at  the  "  butt "  of 
the  first  force  in  the  force  polygon,  there  will  be  only  one  ; 
(and  this  one  will  be  called  the  special  equilibrium  polygon 
in  the  chapter  on  arch-ribs,  and  the  "  true  linear  arch  "  in 
dealing  with  the  stone  arch.)  After  the  rays  (dotted  in 
Fig.  351)  have  been  added,  by  joining  the  pole  to  each 


402  MECHANICS    OF   ENGINEERING. 

vertex  with  which  it  is  not  already  connected,  the  final 
figure  may  be  called  the  force  diagram. 

It  may  sometimes  be  convenient  to  give  the  name  of 
rays  to  the  two  forces  of  the  force  pelygon  which  meet 
at  the  pole,  in  which  case  the  first  and  last  segments  of 
the  corresponding  equil.  polygon  will  coincide  with  the 
lines  of  action  of  those  forces  in  the  space-diagram  (as  we 
may  call  the  representation  of  the  body  or  structure  on 
which  the  forces  act).  This  "  space  diagram  "  shows  the 
real  field  of  action  of  the  forces,  while  the  force  diagram, 
which  may  be  placed  in  any  convenient  position  on  the 
paper,  shows  the  magnitudes  and  directions  of  the  forces 
acting  in  the  former  diagram,  its  lines  being  interpreted 
on  a  scale  of  so  many  Ibs.  or  tons  to  the  inch  of  paper  ;  in 
the  space-diagram  we  deal  with  a  scale  of  so  many/ee£  to 
the  inch  of  paper. 

We  have  found,  then,  that  if  any  vertex  or  corner  of  the 
closed  force  polygon  be  taken  as  a  pole,  and  rays  drawn 
from  it  to  all  the  other  corners  of  the  polygon,  and  a  cor- 
responding equil.  polygon  drawn  in  the  space  diagram,  the 
first  and  last  segments  of  the  latter  polygon  must  co-incide 
with  the  first  and  last  forces  according  to  the  order 
adopted  (or  with  the  resultants  of  the  first  two  anJ  last 
two,  if  more  convenient  to  classify  them  thus).  It  remains 
to  utilize  this  principle, 

327.  To  Find  the  Resultant  of  Several  Forces  in  a  Plane. — This 
might  be  done  as  in  §  326,  but  since  frequently  a  given  set 
of  forces  are  parallel,  or  nearly  so,  a  special  method  will 
now  be  given,  of  great  convenience  in  such  cases.  Fig.  352. 

Let  P,  P2  and 
P3  be  the  given 
forces      whose 
resultant  is  re- 
\2  quirsd.    Let  us 
first  find   their 
an^i  •  resultant, 
or  force  which 
FIG.  352.  FU>.  353.       will        balance 


GRAPHICAL  STATICS.  403 

them.  This  anti-resultant  may  be  conceived  as  decom- 
posed into  two  components  P  and  P'  one  of  which,  say  P, 
is  arbitrary  in  amount  and  position.  Assuming  P,  then, 
at  convenience,  in  the  space  diagram,  it  is  required  to  find 
P'.  The  five  forces  must  form  a  balanced  system ;  hence 
if  beginning  at  Olt  Fig.  353,  we  lay  off  a  line  0,A  =  P  by 
scale,  then  A\  —  and  ||  to  P,,  and  so  on  (point  to  butt),  the 
line  E0l  necessary  to  close  the  force  polygon  is  =  Pf  re- 
quired. Now  form  the  corresponding  equil.  polygon  in 
the  space  diagram  in  the  usual  way,  Yiz.:  through  a  the 
intersection  of  P  and  P1  draw  ab  \\  to  the  ray  6^  .  .  .  1 
(which  connects  the  pole  0^  with  the  point  of  the  last  force 
mentioned).  From  b,  where  ab  intersects  the  line  of  P2, 
draw  be,  \\  to  the  ray  0^  .  .  2,  till  it  intersects  the  line  of  P3. 
A  line  me  drawn  through  c  and  ||  to  the  P'  of  the  force 
diagram  is  the  line  of  action  of  P'. 

Now  the  resultant  of  P  and  P'  is  the  anti-resultant  of 
PI,  P2  and  P3; .'.  d,  the  intersection  of  the  lines  of  P  and 
P',  is  a  point  in  the  line  of  action  of  the  anti-resultant  re- 
quired, while  its  direction  and  magnitude  are  given  by  the 
line  BA  in  the  force  diagram  ;  for  BA  forms  a  closed  poly- 
gon both  with  Px  P2  P3,  and  with  PP'.  Hence  a  line 
through  d  \\  to  BA,  viz.,  de,  is  the  line  of  action  of  the  anti- 
resultant  (and  hence  of  the  resultant)  of  Plt  P2,  P3. 

Since,  in  this  construction,  P  is  arbitrary,  we  may  first 
choose  Oit  arbitrarily,  in  a  convenient  position,  i.e.,  in  such 
a  position  that  by  inspection  the  segments  of  the  result- 
ing equil.  polygon  shall  give  fair  intersections  and  not 
pass  off  the  paper.  If  the  given  forces  are  parallel  the 
device  of  introducing  the  oblique  P  and  P'  is  quite  neces- 
sary. 

328. — The  result  of  this  construction  may  be  stated  as 
follows,  (regarding  Oa  and  cm  as  segments  of  the  equil. 
polygon  as  well  as  ab  and  be):  If  any  two  segments  of  an 
equil.  polygon  be  prolonged,  their  intersection  is  a  point  in 
the  line  of  action  of  the  resultant  of  those  forces  acting  at 


404 


MECHANICS  OF  ENGINEERING* 


the  vertices  intervening  between  the  given  segments.    Here, 
the  resultant  of  PI  P2  P3  acts  through  d. 

329.  Vertical  Reaction  of  Piers,  etc. — Fig.  354.  Given  the 
vertical  forces  or  loads  P}  P2  and  P3  acting  on  a  rigid  body 
(beam,  or  truss)  which  is  supported  by  two  piers  having 
smooth  horizontal  surfaces  (so  that  the  reactions  must  be 
vertical),  required  the  reactions  VQ  and  Vn  of  the  piers. 
For  an  instant  suppose  VQ  and  Vn  known ;  they  are  in 


FIG.  354. 

equil.  with  Pl  P2  and  P3.  The  introduction  of  the  equal 
and  opposite  forces  P  and  P'  in  the  same  line  will  not  dis- 
turb the  equilibrium.  Taking  the  seven  forces  in  the 
order  P  V0  PI  P2  P3  Vn  and  P',  a  force  polygon  formed  with 
them  will  close  (see  (b)  in  Fig.  where  the  forces  which 
really  lie  on  the  same  line  are  slightly  separated).  With 
0,  the  butt  of  P,  as  a  pole,  draw  the  rays  of  the  force  dia- 
gram OA,  OB,  etc.  The  corresponding  equil.  polygon 
begins  at  a,  the  intersection  of  P  and  V0  in  (a)  (the  space 
diagram),  and  ends  at  n  the  intersection  of  P'  and  Fn. 
Join  an.  Now  since  P  and  P'  act  in  the  same  line,  an 
must  be  that  line  and  must  be  II  to  P  and  P'  of  the  force 
diagram.  Since  the  amount  and  direction  of  P  and  P'  are 
arbitrary,  the  position  of  the  pole  0  is  arbitrary,  while 
P!,  P2,  and  P3  are  the  only  forces  known  in  advance  in  the 
force  diagram. 

Hence  VQ  and  F"n  may  be  determined  as  follows:  Lay  off 
the  given  loads  Plf  P2,  etc.,  in  the  order  of  their  occur- 
rence in  the  space  diagram,  to  form  a  "  load-line  "  AD 


GRAPHICAL  STATICS 


405 


(set  (&.)  Pig.  354)  as  a  beginning  for  a  force-diagram  ;  take 
any  convenient  pole  0,  draw  the  rays  OA,  OB,  OC  and 
OD.  Then  beginning  at  any  convenient  point  a  in  the 
vertical  line  containing  the  unknown  VQt  draw  ab  ||  to  OA, 
be  |  4o  OB,  and  so  on,  until  the  last  segment  (dn  in  this 
case)  cuts  the  vertical  containing  the  unknown  Vn  in  some 
point  n.  Join  an  (this  is  sometimes  called  a  closing  line) 
and  draw  a  ||  to  it  through  0,  in  the  force -diagram.  This 
last  line  will  cut  tha  "  load-line  "  in  some  point  n',  and 
divide  it  in  two  parts  n'  A  and  Dn',  which  are  respectively 
VQ  and  Vu  required. 

Corollary. — Evidently,  for  a  given  system  of  loads,  in  given 
vertical  lines  of  action,  and  for  two  given  piers,  or  abut- 
ments, having  smooth  horizontal  surfaces,  the  location  of  the 
point  n'  on  the  load  line  is  independent  of  the  choice  of  a 
pole. 

Of  course,  in  treating  the  stresses  and  deflection  of  the 
•rigid  body  concerned,  P  and  P'  are  left  out  of  account,  as 
being  imaginary  and  serving  only  a  temporary  purpose. 

330.  Application  of  Foregoing  Principles  to  a  Roof  Truss.- 
¥ig.  355.  Wl  and  W2  are  wind  pressures,  P1  and  P2  are 
loads,  while  the  remaining  external  forces,  viz.,  the  re- 


FIG.  355. 


406  MECHANICS  OF  ENGINEERING. 

actions,  or  supporting  forces,  V0,  Vn  and  HM  may  be  found 
by  preceding  §§.  (We  here  suppose  that  the  right  abut- 
ment  furnishes  all  the  horizontal  resistance ;  none  at  the 
left). 

Lay  off  the  forces  (known)  Wly  W2,  Pl9  and  P2  in  the 
usual  way,  to  form  a  portion  of  the  closed  force  polygon 
To  close  the  polygon  it  is  evident  we  need  only  draw  a 
horizontal  through  5  and  limit  it  by  a  vertical  through  1. 
This  determines  Hu  but  it  remains  to  determine  nr  the 
point  of  division  between  V0  and  VD.  Select  a  convenient 
pole  Oit  and  draw  rays  from  it  to  1,  J2,  etc.  Assume  a  con- 
venient point  a  in  the  line  of  V0  in  the  space  diagram,  and 
through  it  draw  a  line  ||  to  Oi\  to  meet  the  line  of  Wl  in 
some  point  b  ;  then  a  line  ||  to  Oi2  to  meet  the  line  of  W2 
in  some  point  c ;  then  through  c  \\  to  Oi3  to  meet  the  line 
of  P!  in  some  point  d  ;  then  through  d  \\  to  0X4  to  meet  the 
line  of  P2  in  some  point  e,  (e  is  identical  with  d,  since  Pl 
and  P2  are  in  the  same  line)  ;  then  ef  \\  to  Ofi  to  meet  Hn 
in  some  point/;  then/gr  ||  to  Of>  to  meet  Fn  in  some 
point  g. 

abcdefg  is  an  equilibrium  polygon  corresponding  to  the 
pole  Oi. 

Now  join  ag,  the  "closing-line,"  and  draw  a  ||  to  it 
through  Oi  to  determine  n't  the  required  point  of  division 
between  V0  and  Vn  on  the  vertical  1  6.  Hence  V0  and  Fn 
are  now  determined  as  well  as  Ha. 

[The  use  of  the  arbitrary  pole  0t  implies  the  temporary 
employment  of  a  pair  of  opposite  and  equal  forces  in  the  line 
ag,  the  amount  of  either  being  =  0^']. 

Having  now  all  the  external  forces  acting  on  the  truss, 
and  assuming  that  it  contains  no  "  redundant  parts,"  i.e., 
parts  unnecessary  for  rigidity  of  the  frame-work,  we  proceed 
tc  find  the  pulls  and  thrusts  in  the  individual  pieces,  on 
the  following  plan.  The  truss  being  pin-connected,  no 
piece  extending  beyond  a  joint,  and  all  loads  being  con- 
sidered to  act  at  joints,  the  action,  pull  or  thrust,  of  each 
piece  on  the  joint  at  either  extremity  will  be  in  the  direction 
of  the  piece,  i.e.,  in  a  known  direction,  and  the  pin  of  each 


GKAPHICAL   STATICS.  407 

joint  is  in  equilibrium  under  a  system  of  concurrent  forces 
consisting  of  the  loads  (if  any)  at  the  joint  and  the  pulls 
or  thrusts  exerted  upon  it  by  the  pieces  meeting  there. 
Hence  we  may  apply  the  principles  of  §  325  to  each  joint 
in  turn.  See  Fig.  356.  In  constructing  and  interpreting 
the  various  force  polygons,  Mr.  B.  H  Bow's  convenient 
notation  will  be  used ;  this  is  as  follows  :  In  the  space 
diagram  a  capital  letter  [ABC,  etc.]  is  placed  in  each  tri- 
angular cell  of  the  truss,  and  also  in  each  angular  space  in 
the  outside  outline  of  the  truss  between  the  external  forces 
and  the  adjacent  truss -pieces.  In  this  way  we  can  speak  of 
the  force  Wi  as  the  force  EG,  of  W2  as  the  force  CE,  the 
stress  in  the  piece  aft  as  the  force  CD,  and  so  on.  That 
is,  the  stress  in  any  one  piece  can  be  named  from  the 
letters  in  the  spaces  bordering  its  two  sides.  Corresponding 
to  these  capital  letters  in  the  spaces  of  the  space-dia- 
gram, small  letters  will  be  used  at  the  vertices  of  the  closed 
force-polygons  (one  polygon  for  each  joint)  in  such  a  way 
that  the  stress  in  the  piece  CD,  for  example,  shall  be  the 
force  cd  of  the  force  polygon  belonging  to  any  joint  in 
which  that  piece  terminates ;  the  stress  in  the  piece  FO 
by  the  force  fg  in  the  proper  force  polygon,  and  so  on. 

In  Fig.  356  the  whole  truss  is  shown  free,  in  equili- 
brium under  the  external  forces.  To  find  the  pulls  or 
thrusts  (i.e.,  tensions  or  compressions)  in  the  pieces,  con- 
sider that  if  all  but  two  of  the  forces  of  a  closed  force 
polygon  are  known  in  magnitude  and  direction,  while  the 
directions,  only,  of  those  two  are  known,  the  whole  force 
polygon  may  be  drawn,  thus  determining  the  amounts  of 
those  two  forces  by  the  lengths  of  the  corresponding 
sides. 

We  must  .*.  begin  with  a  joint  where  no  more  than  two 
pieces  meet,  as  at  a ;  [call  the  joints  a,  /9,  f,  d,  and  the  cor- 
corresponding  force  polygons  a',  ft  etc.  Fig.  356.]  Hence 
at  a!  (anywhere  on  the  paper)  make  ab  \\  and  =  (by  scale) 
to  the  known  force  AB  (i.e.,  F0)  pointing  it  at  the  upper  end, 
and  from  this  end  draw  be  =  and  ||  to  the  known  force  BG 
(i.e.,  W,)  pointing  this  at  the  lower  end. 


403 


MECHANICS   OF   ENGINEERING. 


FIG.  356. 

To  close  the  polygon  draw  through  c  a  ||  to  the  piece 
CD,  and  through  a  a  ||  to  AD ;  their  intersection  deter- 
mines d,  and  the  polygon  is  closed.  Since  the  arrows 
must  be  point  to  butt  round  the  periphery,  the  force  with 
which  the  piece  CD  acts  on  the  pin  of  the  joint  a  is  a 
force  of  an  amount  =  cd  and  in  a  direction  from  c  toward 
d ;  hence  the  piece  CD  is  in  compression  ;  whereas  the 
action  of  the  piece  DA  upon  the  pin  at  a  is  from  d  toward 
a  (direction  of  arrow)  and  hence  DA  is  in  tension.  Notice 
that  in  constructing  the  force  polygon  «'  a  right-handed 
(or  clock-wise)  rotation  has  been  observed  in  considering 
in  turn  the  spaces  ABC  and  D,  round  the  joint  a.  A 
similar  order  will  be  found  convenient  in  each  of  the  other 
joints. 

Knowing  now  the  stress  in  the  piece  CD,  (as  well  as  ii^ 
DA)  all  but  two  of  the  forces  acting  on  the  pin  at  the  joint 
(3  are  known,  and  accordingly  we  begin  a  force  polygon,  ft', 
for  that  joint  by  drawing  dc,=  and  ||  to  the  dc  of  polygon 
a',  but  pointed  in  the  opposite  direction,  since  the  action  of 
OD  on  the  joint  ft  is  equal  and  opposite  to  its  action  on 
the  joint  a  (this  disregards  the  weight  of  the  piece). 
Through  c  draw  ce  =  and  ||  to  the  force  CE(i.e.,  W2)  and 


GRAPHICAL   STATICS.  409 

pointing  the  same  way ;  then  ef,  =  and  ||  to  the  load  EF 
(i.e.  Pj)  and  pointing  downward.  Through  f  draw  a  ||  to 
the  piece  FG  and  through  d,  a  ||  to  the  piece  GD,  and  the 
polygon  is  closed,  thus  determining  the  stresses  in  the 
pieces  FG  and  GD.  Noting  the  pointing  of  the  arrows, 
we  readily  see  that  FG  is  in  compression  while  GD  is  in 
tension. 

Next  pass  to  the  joint  d,  and  construct  the  polygon  d', 
thus  determining  the  stress  gh  in  GH  and  that  ad  in  AD  ; 
this  last  force  ad  should  check  with  its  equal  and  oppo- 
site ad  already  determined  in  polygon  a'.  Another  check 
consists  in  the  proper  closing  of  the  polygon  7-',  all  of 
whose  sides  are  now  known. 

[A  compound  stress-diagram  may  be  formed  by  super- 
posing the  polygons  already  found  in  such  a  way  as  to 
make  equal  sides  co-incide ;  but  the  character  of  each 
stress  is  not  so  readily  perceived  then  as  when  they  are 
kept  separate]. 

In  a  similar  manner  we  may  find  the  stresses  in  any  pin- 
connected  frame-work  (in  one  plane  and  having  no  redun- 
dant pieces)  under  given  loads,  provided  all  the  support- 
ing forces  or  reactions  can  be  found.  In  the  case  of  a 

braced-arch  (truss)  as 
shown  in  Fig.  357,  hinged 
to  the  abutments  at  t)oth 
ends  and  not  free  to  slide 
laterally  upon  them,  the 
reactions  at  0  and  B  de- 
357.  pend,  in  amount  and  direc- 

tion, not  only  upon  the  equations  of  Statics,  but  on  the 
form  and  elasticity  of  the  arch-truss.  Such  cases  will  be 
treated  later  under  arch -ribs,  or  curved  beams. 

332.  The  Special  EquiL  Polygon.  Its  Relation  to  the  Stresses 
in  the  Rigid  Body.— Eeproducing  Figs.  350  and  351  in  Figs. 
358  and  359,  (where  a  rigid  curved  beam  is  in  equilibrium 
under  the  forces  P,,  P2,  P3,  P4  and  P'c)  we  call  a  .  .  b  .  ,  v 


410 


MECHANICS   OF   ENGINEERING. 


the  special  equil.  polygon  because  it  corresponds  to  a  force 
diagram  in  which  the  same  order  of  forces  has  been  ob- 
served as  that  in  which  they  occur  along  the  beam  (from 
left  to  right  here).  From  the  relations  between  the  force 


PIG,  359. 


diagram  and  equil.  polygon,  this  special  equil.  polygon  in 
the  space  diagram  has  the  following  properties  in  connec- 
tion with  the  corresponding  rays  (dotted  lines)  in  the  force 
diagram. 

The  stresses  in  any  cross-section  of  the  portion  O'A  of 
the  beam,  are  due  to  Pl  alone ;  those  of  any  cross-section 
on  AB  to  Pl  and  P2,  i.e.,  to  their  resultant  H^  whose  mag- 
nitude is  given  by  the  line  Oa'  in  the  force  diagram,  while 
its  line  of  action  is  ab  the  first  segment  of  the  equil.  poly- 
gon. Similarly,  the  stresses  in  B  G  are  due  to  Pl9  P2  and 
P3,  i.e.,  to  their  resultant  7?b  acting  along  the  segment  be, 
its  magnitude  being  =Obr  in  the  force  diagram.  E.g.,  if 
the  section  at  m  be  exposed,  considering  O'ABm  as  a  free 
body,  we  have  (see  Fig.  360)  the  elastic  stresses  (or  inter- 


PIG.  361. 


nal  forces)  at  m  balancing  the  exterior  or  "  applied  forces  w 
Plt  P2  and  P3.     Obviously,  then,  the  stresses  at  m  are  just 


GRAPHICAL   STATICS.  411 

the  same  as  if  Eb  the  resultant  of  Pl9  P2  and  P3,  acted  upon 
an  imaginary  rigid  prolongation  of  the  beam  intersecting 
be  (see  Fig.  361). 72b  might  be  called  the  "anti-stress-result- 
ant "  for  the  portion  BG  of  the  beam.  We  may  /.  state 
the  following :  If  a  rigid  body  is  in  equilibrium  under  a  sys- 
tem of  Non-Concurrent  Forces  in  a  plane,  and  the  special  equi- 
librium polygon  has  been  drawn,,  then  each  ray  of  the  force 
diagram  is  the  anti-stress-resultant  of  that  portion  of  the  beam 
which  corresponds  to  the  segment  of  the  equilibrium  polygon 
to  which  the  ray  is  parallel ;  and  its  line  of  action  is  the  seg- 
ment just  mentioned, 

Evidently  if  the  body  is  not  one  rigid  piece,  but  com- 
posed of  a  ring  of  uncemented  blocks  (or  voussoirs),  it  may 
be  considered  rigid  only  so  long  as  no  slipping  takes  place 
or  disarrangement  of  the  blocks;  and  this  requires  that  the 
"  anti-stress-resultant "  for  a  given  joint  between  two 
blocks  shall  not  lie  outside  the  bearing  surface  of  the 
joint,  nor  make  too  small  an  angle  with  it,  lest  tipping  or 
slipping  occur.  For  an  example  of  this  see  Fig.  362,  show- 
ing a  line  of  three  blocks  in  equilibrium  under  five  forces. 

The  pressure  borne  at  the 
joint  MN9  is  —  R&  in  the 
|Pj  force -diagram  and  acts  in 
the  line    ab.       The    con- 
struction    supposes     all 
the  forces    given    except 
FIG.  362.  one,  in  amount  and  posi- 

tion, and  that  this  one  could  easily  be  found  in  amount,  as 
being  the  side  remaining  to  close  the  force  polygon,  while 
its  position  would  depend  ok  ihe  equil.  polygon.  But  in 
practice  the  two  forces  PT  and  J?'c  are  generally  unknown, 
hence  the  point  0,  or  pole  of  the  force  diagram,  can  not 
be  fixed,  nor  the  special  equil.  polygon  located,  until  other 
considerations,  outside  of  those  so  far  presented,  are 
brought  into  play.  In  the  progress  of  such  a  problem,  as 
will  be  seen,  it  will  be  necessary  to  use  arbitrary  trial  po- 
sitions for  the  pole  0,  and  corresponding  trial  equilibrium 
polygons. 


ALLtiiA^IU? 


CHAPTEE  IX. 


GRAPHICAL  STATICS  OF  VERTICAL  FORCES, 


333.  Remarks. — (With  the  exception  of  §  378  a)  in  prob- 
lems to  be  treated  subsequently  (either  the  stiff  arch -rib, 
or  the  block-work  of  an  arch-ring,  of  masonry)  when  the 
body  is  considered  free  all  the  forces  holding  it  in  equiL 
will  be  vertical  (loads,  due  to  gravity)  except  the  reactions 
at  the  two  extremities,  as  in  Fig.  363 ;  but  for  convenience 
each  reaction  will  be  replaced  by  its  horizontal  and  verti- 
cal components  (see  Fig.  364).  The  two  ZTs  are  of  course 
<>qual,  since  they  are  the  only  horizontal  forces  in  the 
system.  Henceforth,  aU  equil.  polygons  under  discussion 
le  understood  to  imply  this  kind  of  system  of  forces.  Plt 


FIG.  363. 


FIG.  364. 


FIG.  364a. 


etc.,  will  represent  the  "  loads  " ;  VQ  and  Vn  the  vertical 
components  of  the  abutment  reactions  ;  H  the  value  of 
either  horizontal  component  of  the  same.  (We  here  sup- 
pose the  pressures  TQ  and  TD  resolved  along  the  horizon- 
tal and  vertical.) 


GRAPHICAL   STATICS. 


334,  Concrete  Conception  of  an  Equilibrium  Polygon, — Any 
equilibrium  polygon  has  this  property,  due  to  its  mode 
of  construction,  viz.:  If  the  ab  and  be  of  Fig.  358  were  im- 
ponderable straight  rods,  jointed  at  b  without  friction,  they 
would  be  in  equilibrium  under  the  system  of  forces  there 
given.  (See  Fig.  364a).  The  rod  ab  suffers  a  compression 
equal  to  the  E&  of  the  force  diagram,  Fig.  359,  and  be  a 
compression  =  J?b.  In  some  cases  these  rods  might  be  in 
tension,  and  would  then  form  a  set  of  links  playing  the 
part  of  a  suspension-bridge  cable.  (See  §  44). 

335,  Example  of  Equilibrium  Polygon  Drawn  to  Vertical  Loads 
— Fig.  365.  [The  structure  bearing  the  given  loads  is  not 
shown,  but  simply  the  imaginary  rods,  or  segments  of  an 
equilibrium  polygon,  which  would  support  the  given  loads 
in  equilibrium  if  the  abutment  points  A  and  B,  to  which 
the  terminal  rods  ar^  hinged,  were  firm.  In  the  present 
case  this  equilibrium  is  unstable  since  the  rods  form  a 
standing  structure ;  but  if  they  were  hanging,  the  equilibri- 
um would  be  stable.  Still,  in  the  present  case,  a  very  light 
bracing,  or  a  little  friction  at  all  joints  would  make  the 
equilibrium  stable. 


FIG.  365. 


Given  three  loads  P19  P2,  and  P3,  and  two  "  abutment 
verticals  "  A'  and  B't  in  which  we  desire  the  equil.  poly- 
gon to  terminate,  lay  off  as  a  "load-line"  to  scale,  P,,  P2, 
and  P3  end  to  end  in  their  order.  Then  selecting  any  pole, 


414  MECHANICS  OF   ENGINEERING. 

0,  draw  the  rays  01,  02,  etc.,  of  a  force  diagram  (the 
and  P's,  though  really  on  the  same  vertical,  are  separated 
slightly  for  distinctness  ;  also  the  ZTs,  which  both  pass 
through  0  and  divide  the  load-line  into  V0  and  Vn).  We 
determine  a  corresponding  equilibrium  polygon  by  draw- 
ing through  A  (any  point  in  A')  a  line  ||  to  0  .  .  1,  to  inter- 
sect P!  in  some  point  b ;  through  b  a  ||  to  0  .  .  2,  and  so  on> 
until  B'  the  other  abutment-vertical  is  struck  in  some 
point  B.  AB  is  the  "  abutment-line  "  or  "  closing -line." 

By  choosing  another  point  for  0,  another  equilibrium 
polygon  would  result.  As  to  which  of  the  infinite 
number  (which  could  thus  be  drawn,  for  the  given  loads 
and  the  A  and  B'  verticals)  is  the  special  equilibrium  poly- 
gon for  the  arch-rib  or  stone-arch,  or  other  structure,  on 
which  the  loads  rest,  is  to  be  considered  hereafter.  In 
any  of  the  above  equilibrium  polygons  the  imaginary 
series  of  jointed  rods  would  be  in  equilibrium. 

336,  Useful  Property  of  an  Equilibrium  Polygon  for  Vertical 
Loads.— (Particular  case  of  §  328).  See  Fig.  366.  In  any 
equil.  polygon,  supporting  vertical  loads,  consider  as  free 
any  number  of  consecutive  segments,  or  rods,  with  the 
loads  at  their  joints,  e.  g.,  the  5th  and  6th  and  portions  of 

the  4th  and  7th  which,  we  sup- 
pose cut  and  the  compressive 
forces  in  them  put  in,  T4  and 
T7t  in  order  to  consider  4567 
as  a  free  body.  For  equil., 
according  to  Statics,  the  lines 
of  action  of  1\  and  T7  (the  com- 
see.  pression  in  those  rods)  must  in- 

tersect in  a  point,  (7,  in  the  line  of  action  of  the  resultant 
of  P4,  P5,  and  P6 ;  i.e.,  of  the  loads  occurring  at  the  inter- 
vening vertices.  That  is,  the  point  C  must  lie  in  the  ver- 
tical containing  the  centre  of  gravity  of  those  loads.  Since 
the  position  of  this  vertical  must  be  independent  of  the 
particular  equilibrium  polygon  used,  any  other  (dotted 
lines  in  Fig.  366)  for  the  same  loads  will  give  the  same  re- 


GRAPHICAL    STATICS.  415 

suits.  Hence  the  vertical  CD,  containing  the  centre  of 
gravity  of  any  number  of  consecutive  loads,  is  easily  found 
by  drawing  the  equilibrium  polygon  corresponding  to 
any  convenient  force  diagram  having  the  proper  load-line. 
This  principle  can  be  advantageously  applied  to  finding 
a  gravity -line  of  any  plane  figure,  by  dividing  the  latter 
into  parallel  strips,  whose  areas  may  be  treated  as  loads 
applied  in  their  respective  centres  of  gravity.  If  the  strips 
are  quite  numerous,  the  centre  of  gravity  of  each  may  be 
considered  to  be  at  the  centre  of  the  line  joining  the  mid- 
dles of  the  two  long  sides,  while  their  areas  may  be  taken 
as  proportional  to  the  lengths  of  the  lines  drawn  through 
these  centres  of  gravity  parallel  to  the  long  sides  and  lim- 
ited by  the  end-curves  of  the  strips.  Hence  the  "  load- 
line  "  of  the  force  diagram  may  consist  of  these  lines,  or. 
of  their  halves,  or  quarters,  etc.,  if  more  convenient  (§  376). 


USEFUL  RELATIONS  BETWEEN  FORCE  DIA- 
GRAMS AND  EQUILIBRIUM  POLYGONS, 
(for  vertical  loads.) 

337.  R6sum6  of  Construction.— Fig.  367.  Given  the  loads 
P.lt  etc.,  their  verticals,  and  the  two  abutment  verticals  A' 
and  B',  in  which  the  abutments  are  to  lie  ;  we  lay  off  a 
load-line  1  ...  4,  take  any  convenient  pole,  0,  for  a  force- 
diagram  and  complete  the  latter.  For  a  corresponding 
equilibrium  polygon,  assume  any  point  A  in  the  vertical 
A',  for  an  abutment,  and  draw  the  successive  segments 
Al,  2,  etc.,  respectively  parallel  to  the  inclined  lines  of  the 
force  diagram  (rays),  thus  determining  finally  the  abut- 
ment B,  in  B\  which  (B)  will  not  in  general  lie  in  the  hor- 
izontal through  A. 

Now  join  AB,  calling  AB  the  abutment-line,  and  draw  a 
parallel  to  it  through  0,  thus  fixing  the  point  nf  on  the 


416 


MECHANICS   OF   ENGINEERING. 


FIG.  367. 


FIG.  368. 


load-line.  This  point  n't  as  above  determined,  is  indepen- 
dent of  the  location  of  the  pole,  -0,  (proved  in  §  329)  and 
divides  the  load-line  into  two  portions  (  V0  =  1 .  .  .  n',  and 
V'n  =  n'  . . .  4)  which  are  the  vertical  pressures  which  two 
supports  in  the  verticals  A'  and  B'  would  sustain  if  the 
given  loads  rested  on  a  horizontal  rigid  bar,  as  in  Fig.  368. 

See  §  329.  Hence  to  find  the  point  n'  we  may  use  any 
convenient  pole  0. 

[N.  B.—  The  forces  V0  and  Vn  of  Fig.  367  are  not  identi- 
cal with  F'0  and  F'n,  but  may  be  obtained  by  dropping  a 
"]  from  0  to  the  load-line,  thus  dividing  the  load-line 
into  two  portions  which  are  V0  (upper  portion)  and  Vn. 
However,  if  A  and  B  be  connected  by  a  tie-rod,  in  Fig. 
367,  the  abutments  in  that  figure  will  bear  vertical  press- 
ures only  and  they  will  be  the  same  as  in  Fig.  368,  while 
the  tension  in  the  tie-rod  will  be  =  On'.'] 


Theorem, — The  vertical  dimensions  of  any  two  equili- 
brium polygons,  drawn  to  the  same  loads,  load-verticals,  and 
abutment-verticals,  are  inversely  proportional  to  their  H's  (or 
"pole  distances  ").  We  here  regard  an  equil.  polygon  and 
its  abutment-line  as  a  closed  figure.  Thus,  in  Fig.  369, 
we  have  two  force-diagrams  (with  a  common  load-line,  for 
convenience)  and  their  corresponding  equil.  polygons,  for 
the  same  loads  and  verticals.  From  §  337  we  know  that 
On'  is  II  to  AB  and  00n'  is  ||  to  A0Bn.  Let  CD  be  any  ver- 
tical cutting  the  first  segments  of  the  two  equil.  polygons. 


GRAPHICAL   STATICS. 


417 


the  intercepts  thus  determined  by  2'  and  «'0,  respect- 


ively. From  the 
parallelisms  just 
mentioned,  and 
others  more  famil- 
iar, we  have  the 
triangle  0  In'  sim» 
ilar  to  the  triangle 
Az'  (shaded),  and 
the  triangle  00In' 
similar  to  the  tri- 
angle A0z£.  Hence 


FIG.  369. 


the  proportions  between  j  In'  _z'       ••  In'  _z"0  ) 
bases  and  altitudes  (  ~H      h  H0      h   ) 

.'.  z'  :  z'0  :  :  H0  :  H.          The  same  kind  of  proof  may  easily 
be  applied  to  the  vertical  intercepts  in  any  other  segments, 


e.g. 


and  zf 


Q.  E.  D. 


339.    Corollaries  to  the  foregoing.     It  is  evident  that : 
(1.)     If  the  pole  of  the  force-diagram  be  moved  along  a 
vertical  line,  the  equilibrium  polygon  changing  its  form 
in  a  corresponding  manner,  the  vertical  dimensions  of  the 
equilibrium  polygon  remain  unchanged ;  and 

(2.)  If  the  pole  move  along  a  straight  line  which  con- 
tains the  point  ^',  the  direction  of  the  abutment-line 
remains  constantly  parallel  to  the  former  line,  while  the 
vertical  dimensions  of  the  equilibrium  polygon  change  in 
inverse  proportion  to  the  pole  distance,  or  H,  of  the  force- 
diagram.  \_H  is  the  "1  distance  of  the  pole  from  the  load- 
line,  and  is  called  the  pole-distance]. 

§  340.  Linear  Arch  as  Equilibrium  Polygon. — (See  §  316.) 
If  the  given  loads  are  infinitely  small  with  infinitely  small 
horizontal  spaces  between  them,  any  equilibrium  polygon 
becomes  a  linear  arch.  Graphically  we  can  not  deal  with 
these  infinitely  small  loads  and  spaces,  but  from  §  336  it 
is  evident  that  if  we  replace  them,  in  successive  groups. 


418 


MECHANICS  OF    ENGINEERING. 


Fm  370 


by  finite  forces,  each  of  which  =  the  sum  of  those  com- 

]  I  I  1.  J  *  H  J  *  '  t,vl  I  t  t  /f  applied  through  the  cen- 
tre of  gravity  of  that 
group,  we  can  draw  an 
equilibrium  polygon 
whose  segments  will  be 
tangent  to  the  curve  of 
the  corresponding  linear 
arch,  and  indicate  its  posi- 
tion with  sufficient  exactness  for  practical  purposes.  (See 
Fig.  370).  The  successive  points  of  tangency  A,  m,  n,  etc., 
lie  vertically  under  the  points  of  division  between  the 
groups.  This  relation  forms  the  basis  of  the  graphical 
treatment  of  voussoir,  or  blockwork,  arches. 

341,  To  Pass  an  Equilibrium  Polygon  Through  Three  Arbitrary 
Points. — (In  the  present  case  the  forces  are  vertical.  For 
a  construction  dealing  with  any  plane  system  of  forces  see 
construction  in  §  378#.)  Given  a  system  of  loads,  it  is  re- 
quired to  draw 
an  equilibrium 


them  through 
any  three  points, 
two  of  which 
may  be  consid- 

Fio.  371.  -I, 

e  r  e  d  as  abut- 
ments, outside  of  the  load- verticals,  the  third  point  being 
between  the  verticals  of  the  first  two.  See  Fig.  371.  The 
loads  Plt  etc.,  are  given,  with  their  verticals,  while  A,  p, 
and  B  are  the  three  points.  Lay  off  the  load-line,  and 
with  any  convenient  pole,  Olt  construct  a  force-diagram, 
then  a  corresponding  preliminary  equilibrium  polygon 
beginning  at  A.  Its  right  abutment  B},  in  the  vertical 
through  5,  is  thus  found.  Ol  n'  can  now  be  drawn  ||  to  ABlf 
to  determine  n'.  Draw  n'O  \\  to  BA.  The  pole  of  the 
required  equilibrium  polygon  must  lie  on>i'0  (§  337} 


GRAPHICAL   STATICS. 


419 


Draw  a  vertical  through  p.  The  H  of  the  required  equili- 
brium polygon  must  satisfy  the  proportion  H  :  H^  :  :  r$  : 
pni.  (See  §  338).  Hence  construct  or  compute  If  from 
the  proportion  and  draw  a  vertical  at  distance  H  from 
the  load-line  (on  the  left  of  the  load-line  here)  ;  its  inter- 
section with  n'  0  gives  0  the  desired  pole,  for  which  a 
force  diagram  may  now  be  drawn.  The  corresponding 
equilibrium  polygon  beginning  at  the  first  point  A  will 
also  pass  through  p  and  B  ;  it  is  not  drawn  in  the  figure. 

342.     Symmetrical  Case  of  the  Foregoing  Problem.—  If  two 

points  A  and  B  are  on  a  level,  the  third,  p,  on  the  middle 

vertical  between  them  ;    and  the  loads  (an  even  number) 

symmetrically  disposed  both  in  position  and  magnitude,  about 

,  we  may  proceed  more  simply,  as  follows  :  (Fig.  372). 

From  symmetry  n' 
must  occur  in  the  mid- 
dle of  the  load-line,  of 
which  we  need  lay  off 
only  the  upper  half. 
Take  a  convenient  pola 
#!,  in  the  horizontal 
through  n',  and  draw  a  half  force  diagram  and  a  corres- 
ponding half  equilibrium  polygon  (both  dotted).  The  up- 
per segment  be  of  the  latter  must  be  horizontal  and  being 
prolonged,  cuts  the  prolongation  of  the  first  segment  in  a 
point  dy  which  determines  the  vertical  CD  containing  the 
centre  of  gravity  of  the  loads  occurring  over  the  half-span 
on  the  left.  (See  §  336).  In  the  required  equilibrium  poly- 
gon the  segment  containing  the  point  p  must  be  horizon- 
tal, and  its  intersection  with  the  first  segment  must  lie  in 
CD.  Hence  determine  this  intersection,  6y,  by  drawing  the 
vertical  CD  and  a  horizontal  through  p  ;  then  join  CA> 
which  is  the  first  segment  of  the  required  equil.  polygon. 
A  parallel  to  CA  through  1  is  the  first  ray  of  the  corres- 
ponding force  diagram,  and  determines  the  pole  0  on  tbe 
horizontal  through  n'.  Completing  the  force  diagram  for 


PlG.  372. 


420 


MECHANICS   OF   ENGINEERING. 


this  pole  (half  of  it  only  here),  the  required  equil.  poly- 
gon is  easily  finished  afterwards. 

343,  To  Find  a  System  of  Loads  Tinder  Which  a  Given  Equi- 
librium Polygon  Would  be  in  Equilibrium,  —  Fig.  373.  Let  AB 
be  the  given  equilibrium  polygon.  Through  any  point  0 
as  a  pole  draw  a  parallel  to  each 
segment  of  the  equilibrium  polygon. 
Any  vertical,  as  V,  cutting  these 
lines  will  have,  intercepted  upon  it, 
a  load-line  1,  2,  3,  whose  parts  1  .  .  2, 
2  .  .  3,  etc.,  are  proportional  to  the 
successive  loads  which,  placed  on 
the  corresponding  joints  of  the  equilibrium  polygon  would 
be  supported  by  it  in  equilibrium  (unstable). 

One  load  may  be  assumed  and  the  others  constructed. 
A  hanging,  as  well  as  a  standing,  equilibrium  polygon 
snay  be  dealt  with  in  like  manner,  but  will  be  mstable  equi- 
librium.    The  problem  in  §  44  may  be  solved  in  this  way. 


FIG.  373. 


ARCHES  OF 


421 


CHAPTEE  X. 


RIGHT  ARCHES  OF  MASONRY. 


344, — In  an  ordinary  "right"  stone-arch  (i.e.,  one  in 
which  the  faces  are  ~|  to  the  axis  of  the  cylindrical  soffit, 
or  under  surface),  the  successive  blocks  forming  the  arch- 
ring  are  called  voussoirs,  the  joints  between  them  being 
planes  which,  prolonged,  meet  generally  in  one  or  more 
horizontal  lines  ;  e.g.,  those  of  a  three-centred  arch  in  three 
j|  horizontal  lines ;  those  of  a  circular  arch  in  one,  the  axis 
of  the  cylinder,  etc.  Elliptic  arches  are  sometimes  used.  The 
inner  concave  surface  is  called  the  soffit,  to  which  the  radiat- 
ing joints  between  the  voussoirs  are  made  perpendicular. 
The  curved  line  in  which  the  soffit  is  intersected  by  a  plane 


FIG.  374. 

T  to  the  axis  of  the  arch  is  the  Intrados.  The  curve  in  the 
same  plane  as  the  intrados,  and  bounding  the  outer  ex- 
tremities of  the  joints  between  the  voussoirs,  is  called  the 
Extrados. 

Fig.  374  gives  other  terms  in  use  in  connection  with  a 


422 


MECHAXICS    OF 


stone  arch,  and  explains  those  already  given.     AB  is  the 
u  springing-line." 

3450  Mortar  and  Friction. — As  common  mortar  hardens 
very  slowly,  no  reliance  should  be  placed  on  its  tenacity 
as  an  element  of  stability  in  arches  of  any  considerable 
size ;  though  hydraulic  mortar  and  thin  joints  of  ordinary 
mortar  can  sometimes  be  depended  on.  Friction,  however, 
between  the  surfaces  of  contiguous  voussoirs,  plays  an 
essential  part  in  the  stability  of  an  arch,  and  will  there- 
fore be  considered. 

The  stability  of  voussoir-arches  must  .*.  be  made  to 
depend  on  the  resistance  of  the  voussoirs  to  compresssion 
and  to  sliding  upon  each  other ;  as  also  of  the  blocks 
composing  the  piers,  the  foundations  of  the  latter  being 
firm. 

346.  Point  of  Application  of  the  Resultant  Pressure  between 
two  consecutive  voussoirs ;  (or  pier  blocks).  Applying 
Navier's  principle  (as  in  flexure  of  beams)  that  the  press- 
ure per  nnit  area  on  a  joint  varies  uniformly  from  the 
extremity  under  greatest  compression  to  the  point  of  least 
compression  (or  of  no  compression);  and  remembering 
that  negative  pressures  (i.e.,  tension)  can  not  exist,  as  they 
might  in  a  curved  beam,  we  may  represent  the  pressure 
per  unit  area  at  successive  points  of  a  joint  (from  the  intra- 
dos  toward  the  extrados,  or  vice  versa)  by  the  ordinates  of 
a  straight  line,  forming  the  surface  of  a  trapezoid  or  tri- 
angle, in  which  figure  the  foot  of  the  ordinate  of  the  cen- 
tre of  gravity  is  the  point  of  application  of  the  resultant 
pressure.  Thus,  where  the  least  compression  is  supposed 


FIG.  375. 


Fio.  376. 


FIG.  377. 


Fi&.  378. 


MASONRY   ARCHES. 


423 


to  occur  at  the  intrados  A,  Fig.  375,  the  pressures  vary  as 
the  ordinates  of  a  trapezoid,  increasing  to  a  maximum  value 
at  Bt  in  the  extrados.  In  Fig.  376,  where  the  pressure  is  zero 
at  B,  and  varies  as  the  ordinates  of  a  triangle,  the  result- 
ant pressure  acts  through  a  point  one-third  the  joint- 
length  from  A.  Similarly  in  Fig.  377,  it  acts  one-third 
the  joint-length  from  B.  Hence,  when  the  pressure  is  not 
zero  at  either  edge  the  resultant  pressure  acts  within  the 
middle  third  of  the  joint.  Whereas,  if  the  resultant  press- 
ure falls  without  the  middle  third,  it  shows  that  a  portion 
Am  of  the  joint,  see  Fig.  378,  receives  no  pressure,  i.e.,  the 
joint  tends  to  open  along  Am. 

Therefore  that  no  joint  tend  to  open,  the  resultant  press- 
ure must  fall  within  the  middle  third. 

It  must  be  understood  that  the  joint  surfaces  here  dealt 
with  are  rectangles,  seen  edgewise  in  the  figures. 

347.  Friction. — By   experiment   it   has   been  found   the 
angle  of  friction  (see  §  156)  for  two  contiguous  voussoirs 
of  stone  or  brick  is  about  30°  ;   i.e.,  the  coefficient  of  fric- 
tion is  /  =  tan.  30°.     Hence  if  the  direction  of  the  press- 
ure exerted  upon  a  voussoir  by  its  neighbor  makes  an 
angle  a  less  than  30°  with  the  normal  to  the  joint  surface^ 
there  is  no  danger  of  rupture  of  the  arch  by  the  sliding 
of  one  on  the  other.     (See  Fig.  379). 

348.  Eesistance  to  Crushing. — When  the  resultant  pressure 
falls  at  its  extreme  allowable  limit,  viz. :   the  edge  of  the 
middle  third,  the  pressure  per 

unit  of  area  at  n,  Fig.  380,  is 
double  the  mean  pressure  per 
unit  of  area.  Hence,  in  de- 
signing an  arch  of  masonry, 
we  must  be  assured  that  at 
every  joint  (taking  10  as  a 
factor  of  safety) 

(  Double  the  mean  press-  )          ,  T     -,        ,-,        \  i    n 
\  ure  per  nnit  of  area         I  must  be  less  than  /»  ° 


FIG.  379. 


FIG.  380. 


424  MECHANICS   OF    ENGINEERING. 

C  being  the  ultimate  resistance  to  crushing,  of  the  material 
employed  (§  201)  (Modulus  of  Crushing). 

Since  a  lamina  one  foot  thick  will  always  be  considered 
in  what  follows,  careful  attention  must  be  paid  to  the  units 
employed  in  applying  the  above  tests. 

EXAMPLE. — If  a  joint  is  3  ft.  by  1  foot,  and  the  resultant 
pressure  is  22.5  tons  the  mean  pressure  per  sq.  foot  is 

p=22.5-v-3=7.5  tons  per  sq.  foot 

.'.  its  double=15  tons  per  sq.  foot=208.3  Ibs.  sq.  inch, 
which  is  much  less  than  yio  of  G  for  most  building  stones ; 
see  §  203,  and  below. 

At  joints  where  the  resultant  pressure  falls  at  the  middle, 
the  max.  pressure  per  square  inch  would  be  equal  to  the 
mean  pressure  per  square  inch  ;  but  for  safety  it  is  best  to 
assume  that,  at  times,  (from  moving  loads,  or  vibrations) 
it  may  move  to  the  edge  of  the  middle  third,  causing  the 
max.  pressure  to  be  double  the  mean  (per  square  inch). 

Gen;  Gillmore's  experiments  in  1876  gave  the  following 
results,  among  many  others  : 

NAME  OF  BUILDING  STONE.  C  IN  LBS.  PER  SQ.  INCH. 

Berea  sand-stone,  2-inch  cube,       -        ...        8955 

4     "         "      -  11720 

Limestone,  Sebastopol,  2-inch  cube  (challc\  -        -         1075 

Limestone  from  Caen,  France,      -                 -  3650 

Limestone  from  Kingston,  N.  Y.,    -  13900 

Marble,  Vermont,  2-inch  cube,                   -  8000  to  13000 

Granite,  New  Hampshire,  2-inch  cube,  15700  to  24000 

349.  The  Three  Conditions  of  Safe  Equilibrium  for  an  arch  of 
nncemented  voussoirs. 

Kecapitulating  the  results  of  the  foregoing  paragraphs, 
we  may  state,  as  follows,  the  three  conditions  which  must 
be  satisfied  at  every  joint  of  arch-ring  and  pier,  for  each 
of  any  possible  combination  of  loads  upon  the  structure : 
<  (1).  The  resultant  pressure  must  pass  within  the  middle- 
third, 

(2).  The  resultant  pressure  must  not  make  an  angle  > 
30°  with  the  normal  to  the  joint. 

(3).  The  m^an  pressure  per  unit  of  area  on  the  surface 


ARCH    OF    MASONRY. 


425 


of  the  joint  must  not  exceed 
ing  of  the  material. 


of  the  Modulus  of  crush- 


350. The  True  Linear-Arch,  or  Special  Equilibrium  Polygon; 
and  the  resultant  pressure  at  any  joint.  Let  the  weight 
of  each  voussoir  and  its  load  be  represented  by  a  vertical 
force  passing  through  the  centre  of  gravity  of  the  two,  as 
in  Fig.  381.  Taking  any 
two  points  A  and  B,  A 
being  in  the  first  joint  and 
B  in  the  last  ;  also  a  third 
point,  jt>,  in  the  crown 
joint  (supposing  such  to 
be  there,  although  gener- 
ally a  key-stone  occupies 
the  crown),  through  these  FIG.  ssi. 

three  points  can  be  drawn  [§  341]  an  equilibrium  polygon 
for  the  loads  given  ;  suppose  this  equil.  polygon  nowhero 
passes  outside  of  the  arch-ring  (the  arch-ring  is  the  por- 
tion between  the  intrados,  mn,  and  the  (dotted)  extrados 
m'n'}  intersecting  the  joints  at  b,  c,  etc.  Evidently  if  such 
be  the  case,  and  small  metal  rods  (not  round)  were  insert- 
ed at  A,  b,  c,  etc.,  so  as  to  separate  the  arch  -stones  slight- 
ly, the  arch  would  stand,  though  in  unstable  equilibrium, 
the  piers  being  firm  ;  and  by  a  different  choice  of  A,  p,  and 
By  it  might  be  possible  to  draw  other  equilibrium  poly- 
gons with  segments  cutting  the  joints  within  the  arch- 
ring,  and  if  the  metal  rods  were  shifted  to  these  new  inter- 
sections the  arch  would  again  stand  (in  unstable  equilib- 
rium). 

In  other  words,  if  an  arch  stands,  it  may  be  possible  to 
draw  a  great  number  of  linear  arches  within  the  limits  of 
the  arch-ring,  since  three  points  determine  an  equilibrium 
polygon  (or  linear  arch)  for  given  loads.  The  question 
arises  then  :  ivhich  linear  arch  is  the  locus  of  the  actual  re- 
sultant  pressures  at  the  successive  joints  ? 

[Considering  the  arch-ring  as  an  elastic  curved  beam 
inserted  in  firm  piers  (i.e.,  the  blocks  at  the  springing-line 


426  MECHANICS   OF   ENGINEERING. 

are  incapable  of  turning)  and  having  secured  a  close  fit  at 
all  joints  before  the  centering  is  lowered,  the  most  satisfac- 
tory answer  to  this  question  is  given  in  Prof.  Greene's 
"  Arches,"  p.  131 ;  viz.,  to  consider  the  arch-ring  as  an 
arch  rib  of  fixed  ends  and  no  hinges ;  see  §  380  of  next 
chapter ;  but  the  lengthy  computations  there  employed 
(and  the  method  demands  a  simple  algebraic  curve  for  the 
arch)  may  be  most  advantageously  replaced  by  Prof. 
Eddy's  graphic  method  ("  New  Constructions  in  Graphical 
Statics,"  published  in  Yan  Nostrand's  Magazine  for  1877), 
which  applies  to  arch  curves  of  any  form. 

This  method  will  be  given  in  a  subsequent  chapter,  on 
Arch  Ribs,  or  Curved  Beams  ;  but  for  arches  of  masonry  a 
much  simpler  procedure  is  sufficiently  exact  for  practical 
purposes  and  will  now  be  presented]. 

If  two  elastic  blocks 
of  an  arch-ring  touch  at 
one  edge,  Fig.  382,  their 
adjacent  sides  making  a 
small  angle  with  each 

382.  FIG.  383.          other,  and  are  then  grad- 

ually pressed  more  and  more  forcibly  together  at  the  edge 
m,  as  the  arch-ring  settles,  the  centering  being  gradually 
lowered,  the  surface  of  contact  becomes  larger  and  larger, 
from  the  compression  which  ensues  (see  Fig.  383);  while 
the  resultant  pressure  between  the  blocks,  first  applied  at 
the  extreme  edge  m,  has  now  probably  advanced  nearer  the 
middle  of  the  joint  in  the  mutual  adjustment  of  the  arch- 
Btones.  With  this  in  view  we  may  reasonably  deduce  the 
following  theory  of  the  location  of  the  true  linear  arch 
(sometimes  called  the  "  line  of  pressures  "  and  "  curve  of 
pressure  ")  in  an  arch  under  given  loading  and  with/rm 
piers.  (Whether  the  piers  are  really  unyielding,  under  the 
oblique  thrusts  at  the  springing-line,  is  a  matter  for  sub- 
sequent investigation. 

351,  Location  of  the  True  Linear  Arch. — Granted  that  the 
voussoirs  have  been  closely  fitted  to  each  other  over  the 


ARCH   OF   MASONRY.  427 

centering  (sheets  of  lead  are  sometimes  used  in  the  joints 
to  make  a  better  distribution  of  pressure);  and  that  the 
piers  are  firm  ;  and  that  the  arch  can  stand  at  all  without 
the  centering  ;  then  we  assume  that  in  the  mutual  accom- 
modation between  the  voussoirs,  as  the  centering  is  low- 
ered, the  resultant  of  the  pressures  distributed  over  any 
joint,  if  at  first  near  the  extreme  edge  of  the  joint,  advances 
nearer  to  the  middle  as  the  arch  settles  to  its  final  posi- 
tion of  equilibrium  under  its  load ;  and  hence  the  f ollow~ 
ing 

352.  Practical  Conclusions. 

I.  If  for  a  given  arch  and  loading,  with  firm   piers,  an 
equilibrium  polygon  can  be  drawn  (by  proper  selection  of 
the  points  A,  p,  and  B,  Fig.  381)  entirely  within  the  mid- 
dle third  of  the  arch  ring,  not  only  will  the  arch  stand,  but 
the  resultant  pressure  at  every  joint  will  be  within  the 
middle  third  (Condition  1,  §  349) ;  and  among  all  possible 
equilibrium  polygons  which  can  be  drawn  within  the  mid- 
dle third,  that  is  the  "  true  "  one  which  most  nearly  coin- 
cides with  the  middle  line  of  the  arch-ring. 

II.  If  (with  firm  piers,  as  before)  no  equilibrium  poly- 
gon can  be  drawn  within  the  middle  third,  and  only  one 
within  the  arch-ring  at  all,  the  arch  may  stand,  but  chip- 
ping and  spawling  are  likely  to  occur  at  the  edges  of  the 
joints.     The  design  should  .*.  be  altered. 

III.  If  no  equilibrium   polygon  can  be  drawn  within 
the  arch -ring,  the  design  of  either  the  arch  or  the  loading 
must   be   changed  ;  since,  although  the  arch  may  stand, 
from  the  resistance  of  the  spandrel  walls,  such  a  stability 
must  be  looked  upon  as  precarious  and  not  countenanced 
in  any   large  important  structure.     (Very  frequently,  in 
small  arches  of  brick  and  stone,  as  they  occur  in  buildings, 
the  cement  is  so  tenacious  that  the  whole  structure  is  vir- 
tually a  single  continuous  mass). 

When  the  "  true  "  linear  arch  has  once  been  determined, 
the  amount  of  the  resultant  pressure  on  any  joint  is  given 
by  the  length  of  the  proper  ray  in  the  force  diagram. 


428 


MECHANICS   OF   ENGINEERING. 


ARRANGEMENT  OF  DATA  FOR  GRAPHIC 
TREATMENT. 

353.  Character  of  Load. — In  most  large  stone  arch  bridges 
fche  load  (permanent  load)  does  not  consist  exclusively  of 
masonry  up  to  the  road-way  but  partially  of  earth  filling 
above  the  masonry,  except  at  the  faces  of  the  arch  where 
the  spandrel  walls  serve  as  retaining  walls  to  hold  the 
earth.  (Fig.  384).  If  the  intrados  is  a  half  circle  or  half- 


\ 


\U 


FIG.  884. 


FIG.  385. 


ellipse,  a  compactly -built  masonry  backing  is  carried  up 
beyond  the  springing-line  to  AB  about  60°  to  45°  from  the 
crown,  Fig.  385 ;  so  that  the  portion  of  arch  ring  below 
AB  may  be  considered  as  part  of  the  abutment,  and  thus 
AB  is  the  virtual  springing-line,  for  graphic  treatment. 

Sometimes,  to  save  filling,  small  arches  are  built  over 
the  haunches  of  the  main  arch,  with  earth  placed  over 
them,  as  shown  in  Fig.  386.  In  any  of  the  preceding  cases 


FIG.  386. 


FIG.  387. 


it  is  customary  to  consider  that,  on  account  of  the  bond- 
ing of  the  stones  in  the  arch  shell,  the  loading  at  a  given 
distance  from  the  crown  is  uniformly  distributed  over  the 
width  of  the  roadway. 


ARCHES   OF   MASONRY.  429 

354,  Reduced  Load-Contour, — In  the  graphical  discussion 
of  a  proposed  arch  we  consider  a  lamina  one  foot  thick, 
this  lamina  being  vertical  and  ~|  to  the  axis  of  the  arch ; 
i.e.,  the  lamina  is  ||  to  the  spandrel  walls.     For  graphical 
treatment,  equal  areas  of  the  elevation  (see  Fig.  387)  of 
this  lamina  must  represent  equal   weights.     Taking  the 
material  of  the  arch-ring  as  a  standard,  we   must  find  for 
each  point  p  of  the  extrados  an  imaginary  height  z  of  the 
^rch-ring  material,  which   would  give  the  same  pressure 
(per  running  horizontal  foot)  at  that  point  as  that  due  to 
the  actual  load  above  that  point.     A  number   of  such  or- 
dinates,  each  measured  vertically  upward  from   the  extra- 
dos  determine   points  in  the  "Reduced   Load-Contour,"  i.e., 
the  imaginary  line,  AM,  the  area  between  which   and  the 
extrados  of  the  arch-ring  represents  a  homogeneous  load 
of  the  same  density  as  the  arch-ring,  and  equivalent  to  the 
actual  load  (above  extrados),  vertical  by  vertical. 

355.  Example  of  Reduced  Load-Contour. — Fig.    388.     Given 
an   arch-ring  of  granite  (heaviness  =170  Ibs.  per  cubic 
foot)  with  a  dead  load  of  rubble  (heav.  =  140)  and  earth 
(heav.  =  100),  distributed  as  in  figure.     At  the  point  p,  of 
the  extrados,  the  depth  5  feet  of  rubble  is  equivalent  to  a 
depth  of  [j*  x5]=4.1  ft.  of  granite,  while  the  6  feet  of  earth 
is  equivalent  to  [|^x6]=3.5   feet  of  granite.     Hence  the 
Reduced  Load-Contour  has  an  ordinate,  above  p,  of   7.6  feet. 
That  is,  for  each  of  several  points  of  the  arch -ring  extrados 
reduce  the  rubble  ordinate  in  the  ratio  of  170  :  140,  and 
the  earth  ordinate  in  the  ratio  170  :  100  and  add  the  re- 
sults, setting  off  the  sum  vertically  from  the  points  in  the 
extrados*.     In  this  way  Fig.  389  is  obtained  and  the  area 

*This  is  most  conveniently  done  by  graphics,  thus  :  On  a  right-line  set  off  17  equal 
parts  (of  any  convenient  magnitude.)  Call  this  distance  OA.  Through  0  draw  another 
right  line  at  any  convenient  angle  (30°  to  60°)  with  OA,  and  on  it  from  O 

set  off  OB  equal  to  14  (for  the  rubble ;  or  10  for  the  earth)  of  the  game  equal 
parts.  Join  AS.  From  0  toward  A  set  off*  all  the  rubble  ordinates  to  be  reduced, 
feach  being  set  off  from  0)  and  through  the  other  extremity  of  each  dravr  a  line  par- 
allel to  AS.  The  reduced  ordinates  will  be  the  respective  lengths,  from  0,  along  OB, 
(to  the  intersections  of  these  parallels  with  OB. 

*  With  the  dividers. 


430 


MECHANICS   OF   ENGINEERING. 


FIG.  388. 


FIG.  389. 


there  given  is  to  be  treated  as  representing  homogeneous 
granite  one  foot  thick.  This,  of  course,  now  includes  the 
arch-ring  also.  AB  is  the  "  reduced  load- contour." 

356.  Live  Loads. — In  discussing  a  railroad  arch  bridge 
the  "  live  load"  (a  train  of  locomotives,  e.g.,  to  take  an  ex- 
treme case)  can  not  be  disregarded,  and  for  each  of  its  po- 
sitions we  have  a  separate  Reduced  Load-Contour. 

EXAMPLE. — Suppose  the  arch  of  Fig.  388  to  be  12  feet 
wide  (not  including  spandrel  walls)  and  that  a  train  of  lo- 
comotives weighing  3,000  Ibs.  per  running  foot  of  the  track 
covers  one  half  of  the  span.  Uniformly  *  distributed  later- 
ally over  the  width,  12  ft.,  this  rate  of  loading  is  equiva- 
lent to  a  masonry  load  of  one  foot  high  and  a  heaviness  of 
250  Ibs.  per  cubic  ft.,  i.e.,  is  equivalent  to  a  height  of  1.4 
ft.  of  granite  masonry  [since  ^  X  1.0=1.4]  over  the  half 
span  considered.  Hence  from  Fig.  390  we  obtain  Fig.  391 
in  an  obvious  manner.  Fig.  391  is  now  ready  for  graphic 
treatment. 


FIG.  390.  FIG.  391. 

357.  Piers  and  Abutments. — In  a  series  of  equal  arches 
the  pier  between  two  consecutive  arches  bears  simply  the 
weight  of  the  two  adjacent  semi-arches,  plus  the  load  im- 

*  If  the  earth-filling  is  shallow,  the  laminae  directly  under  the  track  prob» 
ably  receive  a  greater  pressure  than  the  others. 


ARCHES   OF  MASONRY. 


431 


mediately  above  the  pier,  and  /.  does  not  need  to  be  as 
large  as  the  abutment  of  the  first  and  last  arches,  since 
these  latter  must  be  prepared  to  resist  the  oblique  thrusts 
of  their  arches  without  help  from  the  thrust  of  another  on 
the  other  side. 

In  a  very  long  series  of  arches  it  is  sometimes  customary 
to  make  a  few  of  the  intermediate  piers  large  enough  to 
act  as  abutments.  These  are  called  "  abutment  piers,"  and 
in  case  one  arch  should  fall,  no  others  would  be  lost  except 
those  occurring  between  the  same  two  abutment  piers  as 
the  first.  See  Fig.  392.  A  is  an  abutment-pier. 


nrvnnn 


FIG.  3W 


GRAPHICAL  TREATMENT  OF  ARCH. 

358, — Having  found  the  "  reduced  load-contour,"  as  in 
preceding  paragraphs,  for  a  given  arch  and  load,  we  are 
ready  to  proceed  with  the  graphic  treatment,  i.e.,  the  first 
given,  or  assumed,  form  and  thickness  of  arch-ring  is  to  be 
investigated  with  regard  to  stability.  It  may  be  necessary 
to  treat,  separately,  a  lamina  under  the  spandrel  wall,  and 
one  under  the  interior  loading.  The  constructions  are 
equally  well  adapted  to  arches  of  all  shapes,  to  Gothic  as 
well  as  circular  and  elliptical. 

359.— Case  I.  Symmetrical  Arch  and  Symmetrical  Loading. — . 
(The  "  steady  "  (permanent)  or  "  dead  "  load  on  an  arch  is 
usually  symmetrical).  Fig.  393.  From  symmetry  we  need, 


FIG.  393. 


FIG.  394. 


FIG.  395. 


MECHANICS   OF   ENGINEERING. 

deal  with  only  one  half  (say  the  left)  of  the  arch  and  load. 
Divide  this  semi-arch  and  load  into  six  or  ten  divisions 
by  vertical  lines ;  these  divisions  are  considered  as  trape- 
zoids  and  should  have  the  same  horizontal  width  =  b  (a 
convenient  whole  number  of  feet)  except  the  last  one,  LKN, 
next  the  abutment,  and  this  is  a  pentagon  of  a  different 
width  bl9  (the  remnant  of  the  horizontal  distance  LC).  The 
weight  of  masonry  in  each  division  is  equal  to  (the  area 
of  division)  X  (unity  thickness  of  lamina)  x  (weight  of  a  cu- 
bic unit  of-  arch-ring).  For  example  for  a  division  having 
an  area  of  20  sq.  feet,  and  composed  of  masonry  weighing 
160  Ibs.  per  cubic  foot,  we  have  20x1x160=3,200  Ibs., 
applied  through  the  centre  of  gravity  of  the  division. 
The  area  of  a  trapezoid,  Fig.  394,  is  ^(/&i+^2)>  and  its  cen- 
tre of  gravity  may  be  found,  Fig.  395,  by  the  construction 
of  Prob.  6,  in  §  26 ;  or  by  §  27a.  The  weight  of  the  pen- 
tagon LN,  Fig.  393,  and  its  line  of  application  (through 
centre  of  gravity)  may  be  found  by  combining  results  for 
the  two  trapezoids  into  which  it  is  divided  by  a  vertical 
through  K.  See  §  21. 

Since  the  weights  of  the  respective  trapezoids  (except, 
ing  LN}  are  proportional  to  their  middle  vertical  in- 
tercepts [such  as  ^(^+^2)  Fig.  394]  these  intercepts  (trans- 
ferred with  the  dividers)  may  be  used  directly  to  form  the 
load-line,  Fig.  396,  or  proportional  parts  of  them  if  more 
convenient.  The  force  scale,  which  this  implies,  is  easily 
computed^  and  a  proper  length  calculated  to  represent  the 
weight  of  the  odd  division  LN ;  i.e.,  1  ...  2  on  the  load- 
line. 

Now  consider  A,  the  middle  point  of  the  abutment  joint, 
Fig.  396,  as  the  starting  point  of  an  equilibrium  polygon 
(or  abutment  of  a  linear  arch)  for  a  given  loading,  and  re- 
quire that  this  equilibrium  polygon  shall  pass  through  pt 
the  middle  of  the  crown  joint,  and  through  the  middle  of 
the  abutment  joint  on  the  right  (not  shown  in  figure). 

Proceed  as  in  §  342,  thus  determining  the  polygon  Ap 
for  the  half -arch.  Draw  joints  in  the  arch -ring  through 
those  points  where  the  extrados  is  intersected  by  the  ver- 


ARCHES    OF    MASONRY. 


433 


FIG.  396. 


FIG.  397. 


tical  separating  the  divisions  (not  the  gravity  verticals). 
The  points  in  which  these  joints  are  cut  by  the  segments 
of  the  equilibrium  polygon,  Fig.  397,  are  (very  nearly,  if 
the  joint  is  not  more  than  60°  from  p,  the  crown)  the  points 
of  application  in  these  joints,  respectively,  of  the  resultant 
pressures  on  them,  (if  this  is  the  "  true  linear  arch  "  for 
this  arch  and  load)  while  the  amount  and  direction  of  each 
such  pressure  is  given  by  the  proper  ray  in  the  force-dia- 
gram. 

If  at  any  joint  so  drawn  the  linear  arch  (or  equilibrium 
polygon)  passes  outside  the  middle  third  of  the  arch-ring, 
the  point  A,  or  p,  (or  both)  should  be  judiciously  moved 
(within  the  middle  third)  to  find  if  possible  a  linear  arch 
which  keeps  within  limits  at  all  joints.  If  this  is  found 
impossible,  the  thickness  of  the  arch -ring  may  be  increased 
at  the  abutment  (giving  a  smaller  increase  toward  the 
crown)  and  the  desired  result  obtained  ;  or  a  change  in  the 
distribution  or  amount  of  the  loading,  if  allowable,  may 
gain  this  object.  If  but  one  linear  arch  can  be  drawn 
within  the  middle  third,  it  may  be  considered  the  "  true  " 
one ;  if  several,  the  one  most  nearly  co-inciding  with  the 
middles  of  the  joints  (see  §§  351  and  352)  is  so  considered. 

360.— Case  II.  TJnsymmetrical  Loading  on  a  Symmetrical  Arch; 
(e.g.,  arch  with  live  load  covering  one  half -span  as  in  Figs. 
390  and  391).  Here  we  must  evidently  use  a  full  force 
diagram,  and  the  full  elevation  of  the  arch -ring  and  load. 


434 


MECHANICS  OF   ENGINEERING. 


See  Fig.  398.      Select  three  points  A,  p,  and  B,  as  follows, 
to  determine  a  trial  equilibrium  polygon : 

Select  A  at  the  lower  limit  of  the  middle  third  of  the 


FIG.  398. 


abutment-joint  at  the  end  of  the  span  which  is  the  more 
heavily -loaded  ;  in  the  other  abutment-joint  take  B  at  the 
upper  limit  of  the  middle  third ;  and  take  p  in  the  middle 
of  the  crown-joint.  Then  by  §  341  draw  an  equilibrium 
polygon  (i.e.,  a  linear  arch)  through  these  three  points  for 
the  given  set  of  loads,  and  if  it  does  not  remain  within  the 
middle  third,  try  other  positions  for  A,  p,  and  B,  within 
the  middle  third.  As  to  the  "  true  linear  arch  "  alterations 
of  the  design,  etc.,  the  same  remarks  apply  as  already 
given  in  Case  I.  Yery  frequently  it  is  not  necessary  to 
draw  more  than  one  linear  arch,  for  a  given  loading,  for 
even  if  one  could  be  drawn  nearer  the  middle  of  the  arch- 
ring  than  the  first,  that  fact  is  almost  always  apparent  on 
mere  inspection,  and  the  one  already  drawn  (if  within 
middle  third)  will  furnish  values  sufficiently  accurate  for 
the  pressures  on  the  respective  joints,  and  their  direction 
angles. 

360a. — The  design  for  the  arch-ring  and  loading  is  not 
to  be  considered  satisfactory  until  it  is  ascertained  that  for 
the  dead  load  and  any  possible  combination  of  live-load 
(in  addition)  the  pressure  at  any  joint  is 


ARCHES   OF   MASONRY.  435 

(1.)     Within  the  middle  third  of  that  joint ; 

(2.)  At  an  angle  of  <  30°  .with  the  normal  to  joint- 
surface. 

(3.)  Of  a  mean  pressure  per  square  inch  not  >  than  V^ 
of  the  ultimate  crushing  resistance.  (See  §  348.) 

§  361.  Abutments, — The  abutment  should  be  compactly 
and  solidly  built,  and  is  then  treated  as  a  single  rigid  mass. 
The  pressure  of  the  lowest  voussoir  upon  it  (considering 
a  laming  one  foot  thick)  is  given  by  the  proper  ray  of  the 
force  diagram  (0  ..  1,  e.  g.,  in  Fig.  396)  in  amount  and  direc- 
tion. The  stability  of  the  abutment  will  depend  on  the 
amount  and  direction  of  the  resultant  obtained  by  com- 
bining that  pressure  Pa  with  the  weight  G  of  the  abutment 
and  its  load,  see  Fig.  399.  Assume  a  probable  width  RS 
f°r  the  abutment  and  compute  the  weight  G 
of  the  corresponding  abutment  OBRS  and 
MNBO,  and  find  the  centre  of  gravity  of  the 
whole  mass  C.  Apply  G  in  the  vertical 
through  (7,  and  combine  it  with  Pa  at  their  in- 
tersection  D.  The  resultant  P  should  not  cut 
the  base  ES  in  a  point  beyond  the  middle  third 
(or,  if  this  rule  gives  too  massive  a  pier,  take 
such  a  width  that  the  pressure  per  square 
inch  at  S  shall  not  exceed  a  safe  value  as 
FIG.  399.  computed  from  §  362.)  After  one  or  two 
trials  a  satisfactory  width  can  be  obtained, 
"We  should  also  be  assured  that  the  angle  PD  G  is  less 
than  30°.  The  horizontal  joints  above  US  should  also  be 
tested  as  if  each  were,  in  turn,  the  lowest  base,  and  if 
necessary  may  be  inclined  (like  mn)  to  prevent  slipping. 
On  no  joint  should  the  maximum  pressure  per  square  inch 
be  >  than  l/w  the  crushing  strength  of  the  cement.  Abut- 
ments of  firm  natural  rock  are  of  course  to  be  preferred 
where  they  can  be  had.  If  water  penetrates  under  an 
abutment  its  buoyant  effort  lessens  the  weight  of  the  lat- 
ter to  a  considerable  extent, 


436 


MECHANICS    OF   ENGINEERING. 


362,  Maximum  Pressure  Per  Unit  of  Area  When  the  Resultant 
Pressure  Falls  at  Any  Given  Distance  from  the  Middle ;  according 
to  Navier's  theory  of  the  distribution  of  the  pressure  ;  see 
§  346.  Case  I.  Let  the  resultant  pressure  P,  Fig.  400,  (a), 


FIG.  400. 


FIG.  401. 


fall  within  the  middle  third,  a  distance  =  nd  ( <  l/&  d) 
from  the  middle  of  joint  (d  =  depth  of  joint.)  Then  we 
have  the  following  relations  : 

p  (the  mean  press,  per.  sq.  in.),pin  (max.  press,  persq.  in.), 
and  pn  (least  press,  per  sq.  in.)  are  proportional  to  the  lines 
h  (mid.  width),  a  (max.  base),  and  c  (min.  base)  respectively, 
of  a  trapezoid,  Fig.  400,  (b),  through  whose  centre  of  gravity 
P  acts.  But  (§  26) 

^=4-.^Z_c  i.e.,  n=/6  ±Z*  or  a=h  (Qn+l) 
6  a-f  c  h 

-'•  Pm—P  (6»H-1).     Hence  the  following  table : 


press.  pm=   2 


% 


then  the  max. 

times  the  mean  pressure. 


Case  II.  Let  P  fall  outside  the  mid.  third,  a  distance = 
nd  (>  yk  d)  from  the  middle  of  joint.  Here,  since  the 
joint  is  not  considered  capable  of  withstanding  tension, 
we  have  a  triangle,  instead  of  a  trapezoid.  Fig.  401.  First 
compute  the  mean  press,  per  sq.  in. 

*>  =  — — _  x  :„   ,'. — ? —  or   from   this  table :    (lamina  ona 
(1— 2n)  18  d  inches 

foot  thick). 


&SCHES   OF  MASONRY. 


For  nd  = 

A<* 

A^ 

^d 

ft* 

&d 

A^ 

P  = 

i  p 

10  "  d 

i  P 

8  "  d 

1       P 

6   "  d' 

1       P 

4   "  d 

1       P 

2   "  <2 

infinity. 

(d  in  inches  and  P  in  Ibs. ;  with  arch  lamina  1  ft.  in  thickness.) 
Then  the  maximum  pressure  (at  A,  Fig.  401)  pm)  =  %p, 
becomes  known,  in  Ibs.  per  sq.  in. 

362a.  Arch-ring  under  Non- vertical  Forces. — An  example  of 
this  occurs  when  a  vertical  arch-ring  is  to  support  the  pressure 
of  a  liquid  on  its  extrados.  Since  water-pressures  are  always 
at  right  angles  to  the  surface  pressed  on,  these  pressures  on  the 
extradosal  surface  of  the  arch-ring  form  a  system  of  non-paral- 
lel forces  which  are  normal  to  the  curve  of  the  extrados  at 
their  respective  points  of  application  and  lie  in  parallel 
vertical  planes,  parallel  to  the  faces  of  the  lamina.  We  here 
assume  that  the  extradosal  surface  is  a  cylinder  (in  the  most 
general  sense)  whose  rectilinear  elements  are  ~1  to  the  faces  of 
the  lamina.  If,  then,  we  divide  the  length  of  the  extrados, 
from  crown  to  each  abutment,  into  from  six' to  ten  parts,  the 
respective  pressures  on  the  corresponding  surfaces  are  obtained 
by  multiplying  the  area  of  each  by  the  depth  of  its  centre  of 
gravity  from  the  upper  free  surface  of  the  liquid,  and  this 
product  by  the  weight  of  a  unit  of  volume  of  the  liquid ;  and 
each  such  pressure  may  be  considered  as  acting  through  the 
centre  of  the  area.  Finally,  if  we  find  the  resultant  of  each 
of  these  pressures  and  the  weight  of  the  corresponding  portion 
of  the  arch-ring,  these  resultants  form  a  series  of  non-vertical 
forces  in  a  plane,  for  which  an  equilibrium  polygon  can  then 
be  passed  through  three  assumed  points  by  §  3T8a,  these  three 
points  being  taken  in  the  crown-joint  and  the  two  abutment- 
joints.  As  to  the  "  true  linear  arch"  see  §  359. 

As  an  extreme  theoretic  limit  it  is  worth  noting  that  if  the 
extrados  and  intrados  of  the  arch-ring  are  concentric  circles ;  if 
the  weights  of  the  voussoirs  are  neglected ;  and  if  the  rise  of 
th«  arch  is  very  small  compared  with  the  depth  of  the  crown 
^/elow  the  water  surface,  then  the  circular  centre-line  of  the 
wrch-ring  is  the  "  true  linear  arch" 


438  MECHANICS   OF 


CHAPTEE  XL 


ARCH-RIBS. 


364.  Definitions  and  Assumptions. — An  arch-rib   (or  elastic- 
arch,  as  distinguished  from  a  block-work  arch)  is  a  rigid 
curved  beam,   either   solid,  or   built  up   of  pieces  like  a 
truss  (and  then  called  a  braced  arch)  the  stresses  in  which, 
under  a  given  loading  and  with  prescribed  mode  of  sup- 
port it  is  here  proposed  to  determine.     The  rib  is  sup- 
posed symmetrical  about  a  vertical  plane   containing  its 
axis  or  middle  line,  and  the  Moment  of  Inertia  of  any  cross 
section  is  understood  to  be  referred  to  a  gravity  axis  of 
the  section,  which  (the  axis)  is  perpendicular  to  the  said 
vertical  plane.     It  is  assumed  that  in  its  strained  condi- 
tion under  a  load,  the  shape  of  the  rib   differs   so   little 
from  its  form  when  unstrained  that  the  change  in  the  ab- 
scissa or  ordinate  of  any  point  in  the  rib   axis  (a  curve) 
may  be  neglected  when  added  (algebraically)  to  the  co- 
ordinate itself  ;  also  that  the  dimensions  of  a  cross-section 
are  small  compared  with  the  radius   of  curvature  at  any 
part  of  the  curved  axis,  and  with  the  span. 

365.  Mode  of  Support. — Either  extremity  of  the  rib  may  be 
hinged  to  its  pier  (which  gives  freedom  to  the  end-tangent- 
line  to  turn  in  the  vertical  plane  of  the  rib  when  a  load  is 
applied);  or  may  be  fixed,  i.e.,  so  built-in,   or  bolted  rigid- 
ly to  the  pier,  that  the   end-tangent-line  is  incapable  of 
changing  its  direction  when  a  load  is   applied.     A  hinge 
may  be  inserted  anywhere  along  the  rib,  and  of  course 


ARCH   RIBS. 


439 


destroys  the  rigidity,  or  resistance  to  bending  at  that 
point.  (A.  hinge  having  its  pin  horizontal  "]  to  the  axis  of 
the  rib  is  meant).  Evidently  no  more  than  three  such 
hinges  could  be  introduced  along  an  arch- rib  between  two 
piers  ;  unless  it  is  to  be  a  hanging  structure,  acting  as  a 
suspension-cable. 

366.  Arch  Rib  as  a  Free  Body. — In  considering  the  whole 
rib  free  it  is  convenient,  for  graphical  treatment,  that  no 
section  be  conceived  made  at  its  extremities,  if  fixed  ;  hence 
in  dealing  with  that  mode  of  support  the  end  of  the  rib 
will  be  considered  as  having  a  rigid  prolongation  reach- 
ing to  a  point  vertically  above  or  below  the  pier  junction, 
an  unknown  distance  from  it,  and  there  acted  on  by  a  force 
of  such  unknown  amount  and  direction  as  to  preserve  the 
actual  extremity  of  the  rib  and  its  tangent  line  in  the  same 
position  and  direction  as  they  really  are.  As  an  illustra- 
tion of  this  Fig.  402 
shows  free  an  arch  rib. 
ONB,  with  its  extremi- 
ties 0  and  B  fixed  in  the 
piers,  with  no  hinges, 
and  bearing  two 
loads  P;  and  P2.  The 
other  forces  of  the  sys- 
tem holding  it  in  equi-  FIG.  402. 
librium  are  the  horizontal  and  vertical  components,  of  the 
pier  reactions  (H,  V,  Hn,  and  Fn),  and  in  this  case  of  fixed 
ends  each  of  these  two  reactions  is  a  single  force  not  in- 
tersecting the  end  of  the  rib,  but  cutting  the  vertical 
through  the  end  in  some  point  F  (on  the  left ;  and  in  G  on 
the  right)  at  some  vertical  distance  c,  (or  cT),  from  the  end. 
Hence  the  utility  of  these  imaginary  prolongations  OQFt 
and  BRG,  the  pier  being  supposed  removed.  Compare 
Figs.  348  and  350. 

The  imaginary  points,  or  hinges,  F  and  #,  will  be  called 
abutments  being  such  for  the  special  equilibrium  polygon 


440  MECHANICS  OF   ENGINEEBLtfG. 

(dotted  line),  while  0  and  B  are  the  real  ends  of  the  curved 
beam,  or  rjb. 

In  this  system  of  forces  there  are  five  unknowns,  viz.:  V9 
Vnt  H  =  Hny  and  the  distances  c  and  d.  Their  determina- 
tion by  analysis,  even  if  the  rib  is  a  circular  arc,  is  ex- 
tremely intricate  and  tedious  ;  but  by  graphical  statics 
(Prof,  Eddy's  method ;  see  §  350  for  reference),  it  is  com- 
paratively simple  and  direct  and  applies  to  any  shape  of 
rib,  and  is  sufficiently  accurate  for  practical  purposes. 
This  method  consists  of  constructions  leading  to  the  loca- 
tion of  the  "  special  equilibrium  polygon "  and  its  force 
diagram.  In  case  the  rib  is  hinged  to  the  piers,  the  re- 
actions of  the  latter  act  through  these  hinges,  Fig.  403, 
i.e.,  the  abutments  of  the  special 
equilibrium  polygon  coincide  with 
the  ends  of  the  rib  0  and  B,  and  for 
a  given  rib  and  load  the  unknown 
quantities  are  only  three  V,  Vn9  and 
H\  (strictly  there  are  four  ;  but  IX 
=  0  gives  Hn  =  H).  The  solution  FlG<  403. 

by  analytics  is  possible  only  for  ribs  of  simple  algebraic 
curves  and  is  long  and  cumbrous ;  whereas  Prol  Eddy'a 
graphic  method  is  comparatively  brief  and  simple  and  ia 
applicable  to  any  shape  of  rib  whatever. 

367.  Utility  of  the  Special  Equilibrium  Polygon  and  its  force 
diagram.  The  use  of  locating  these  will  now  be  illustrated 
[See  §  332].  As  proved  in  §§  332  and  334  the  compres- 
sion in  each  "  rod  "  or  segment  of  the  "  special  equilibrium 
polygon"  is  the  anti-stress  resultant  of  the  cross  sections  in 
the  corresponding  portion  of  the  beam,  rib,  or  other  struc- 
ture, the  value  of  this  compression  (in  Ibs.  or  tons)  being 
measured  by  the  length  of  the  parallel  ray  in  the  force 
diagram.  Suppose  that  in  some  way  (to  be  explained  sub- 
sequently) the  special  equilibrium  polygon  and  its  force 
diagram  have  been  drawn  for  the  arch-rib  in  Fig.  404  hav- 
ing fixed  ends,  0  and  B,  and  no  hinges  ;  required  the  elastia 
stresses  in  any  cross-section  of  the  rib  as  at  m.  Let  the 


RTBS. 


441 


FIG.  404. 

of  the  force-diagram  on  the  right  be  200  Ibs.  to  the 
inch,  say,  and  that  of  the  space -diagram  (on  the  left)  30  ft. 
to  the  inch. 

The  cross  section  ra  lies  in  a  portion  TK,  of  the  rib,  cor- 
responding to  the  rod  or  segment  be  of  the  equilibrium 
polygon;  hence  its  anti-stress-resultant  is  a  force  R2  acting 
in  the  line  6c,  and  of  an  amount  given  in  the  force-diagram. 
Now  R2  is  the  resultant  of  V,  H,  and  Plt  which  with  the 
elastic  forces  at  m  form  a  system  in  equilibrium,  shown  in 
Fig,  405  ;  the  portion  FO  Tm  being  considered  free.  Hence 


FlO.405, 


Fia.  406. 


taking  the  tangent  line  and  the  normal  at  m  as  axes  we 
should  have  2  (tang,  comps.)  =  0 ;  2  (norm,  comps.)  =  0 ; 
and  2  (moms,  about  gravity  axis  of  the  section  at  m)  =  0, 
and  could  thus  find  the  unknowns  pl9  p2,  and  J",  which  ap- 
pear in  the  expressions  p±F  the  thrust,  ^-  the  moment  of 


442  MECHANICS    OF    ENGINEERING. 

the  stress-couple,  and  J  the  shear.  These  elastic  stresses 
are  classified  as  in  §  295,  which  see.  pl  and  p2  ar®  Ibs.  per 
square  inch,  «/is  Ibs.,  e  is  the  distance  from  the  horizontal 
gravity  axis  of  the  section  to  the  outermost  element  of 
area,  (where  the  compression  or  tension  is  p2  Ibs.  per  sq. 
in.,  as  due  to  the  stress-couple  alone)  while  1  is  the  "  mo- 
ment of  inertia  "  of  the  section  about  that  gravity  axis. 
[See  §§  247  and  295  ;  also  §  85].  Graphics,  however,  gives 
us  a  more  direct  method,  as  follows  :  Since  7?2,  in  the  line 
be,  is  the  equivalent  of  V,  H,  and  Pl9  the  stresses  at  m  will 
be  just  the  same  as  if  S2  acted  directly  upon  a  lateral  pro- 
longation of  the  rib  at  T  (to  intersect  6cFig.  405)  as  shown 
in  Fig.  406,  this  prolongation  Tb  taking  the  place  of  TOF 
in  Fig.  405.  The  force  diagram  is  also  reproduced  here. 
Let  a  denote  the  length  of  the  "1  from  m's  gravity  axis 
upon  be,  and  z  the  vertical  intercept  between  m  and  be. 
For  this  imaginary  free  body,  we  have, 

from  JT(tang.  compons.)—  0,  Rz  cos  a=plF 
and  from  Jf(norm.  compons.)  =0,fi<2  sin  a=J 


while  from  J  (moms  about)  )  we  h        £ 

the  gravity  axis  01  m)=0,      j  e 

But  from  the  two  similar  triangles  (shaded  ;  one  of  them 
is  in  force  diagram)  a  :z  ::  H\R2  .'.  R2a=Hz,  whence  we 
may  rewrite  these  relations  as  follows  (with  a  general  state- 
ment), viz.: 

If  the  Special  Equilibrium  Polygon  and  Its  Force  Diagram  Have 
Seen  Drawn  for  a  given  arch-rib,  of  given  mode  of  support, 
and  under  a  given  loading,  then  in  any  cross-section  of  the 
rib,  we  have  (F  =  area  of  section): 

The  projection  of  improper 


(L) 


ray  (of  the  force  diagram)  up 
on  the  tangent  line  of  the  rill 
drawn  at  the  given  section. 


ARCH    EIBS.  443 

(2.)     The  Shear,  i.e.,  J,  = 

,  .  ,   -,          ,  ,1  The  projection  of  the  proper 

(upon  which  dependsthe  *    J  £  \ 

ray  (of  the  force  diagram)  up- 


shearing    stress   in   the- 
web).    (See  §§  253  and 


(3.)    The  Moment  of   the 
stress  couple,  i.e., 


on  the  normal  to  the  rib  curve 
at  the  given  section. 


256). 

The  product  (Hz)  of  the  H 
(or  pole-distance)  of  the  force- 


diagram  by  the  vertical  dis- 
tance of  the  gravity  axis  of  the 


section  from  the  spec,  equilib- 
rium polygon. 

By  the  "  proper  ray  "  is  meant  that  ray  which  is  parallel 
to  the  segment  (of  the  equil.  polygon)  immediately  under 
or  above  which  the  given  section  is  situated.  Thus  in 
Eig.  404,  the  proper  ray  for  any  section  on  TK  is  H2  >  °n 
KB,  J?3 ;  on  TO,  HI.  The  projection  of  a  ray  upon  any 
given  tangent  or  normal,  is  easily  found  by  drawing  through 
each  end  of  the  ray  a  line  "1  to  the  tangent  (or  normal) ; 
the  length  between  these  "|'s  on  the  tangent  (or  normal)  is 
the  force  required  (by  the  scale  of  the  force  diagram).  We 
may  thus  construct  a  shear  diagram,  and  a  thrust  diagram 
for  a  given  case,  while  the  successive  vertical  intercepts 
between  the  rib  and  special  equilibrium  polygon  form  a 
moment  diagram.  For  example  if  the  z  of  a  point  m  is  }4 
inch  in  a  space  diagram  drawn  to  a  scale  of  20  feet  to  the 
inch,  while  H  measures  2.1  inches  in  a  force  diagram  con- 
structed on  a  scale  of  ten  tons  to  the  inch,  we  have,  for  the 
moment  of  the  stress-couple  at  m,  M=Hz=  [2.1x10]  tons 
X[^x20]  ft. =210  ft.  tons. 

368. — It  is  thus  seen  how  a  location  of  the  special  equili- 
brium polygon,  and  the  lines  of  the  corresponding  force- 
diagram,  lead  directly  to  a  knowledge  of  the  stresses  in  all 
the  cross-sections  of  the  curved  beam  under  consideration, 
bearing  a  given  load ;  or,  vice  versa,  leads  to  a  statement 
of  conditions  to  be  satisfied  by  the  dimensions  of  the  rib, 
for  proper  security. 

It  is  here  supposed  that  the  rib  has  sufficient  lateral 


444  MECHANICS    OF    ENGINEERING. 

bracing  (with  others  which  lie  parallel  with  it)  to  prevent 
buckling  sideways  in  any  part  like  a  long  column.  Before 
proceeding  to  the  complete  graphical  analysis  of  the  differ- 
ent cases  of  arch-ribs,  it  will  be  necessary  to  devote  the 
next  few  paragraphs  to  developing  a  few  analytical  rela- 
tions in  the  theory  of  flexure  of  a  curved  beam,  and  to 
giving  some  processes  in  "  graphical  arithmetic." 

369.  Change  in  the  Angle  Between  Two  Consecutive  Rib  Tan- 
gents when  the  rib  is  loaded,  as  compared  with  its  value 
before  loading.  Consider  any  small  portion  (of  an  arch 
rib)  included  between  two  consecutive  cross- sections ;  Fig. 
407.  KHG  W  is  its  unstrained  form.  Let  EA,  =  ds,  be 
the  original  length  of  this  portion  of  the  rib  axis.  The 
length  of  all  the  fibres  (||  to  rib-axis)  was  originally  =ds 
(nearly)  and  the  two  consecutive  tangent-lines,  at  E  and  A, 
made  an  angle  =  dd  originally,  with  each  other.  While 
under  strain,  however,  all  the  fibres  are  shortened  equally 
an  amount  dh,  by  the  uniformly  distributed  tangential 
thrust,  but  are  unequally  shortened  (or  lengthened,  accord- 
ing as  they  are  on  one  side  or  ttie  other  of  the  gravity  axis 
E,  or  A,  of  the  section)  by  the  system  of  forces  making 
what  we  call  the  "  stress  couple,"  among  which  the  stress 
at  the  distance  e  from  the  gravity  axis  A  of  the  section  is 
called  p.,,  per  square  inch ;  so  that  the  tangent  line  at  A' 
now  takes  the  direction  A'D  "]  to  H'A'G'  instead  of  A'G 
(we  suppose  the  eection  at  E  to  remain  fixed,  for 


ARCH   RIBS. 


445 


ence,  since  the  change  of  angle  between  the  two  tangents 
depends  on  the  stresses  acting,  and  not  on  the  new  posi- 
tion in  space,  of  this  part  of  the  rib),  and  hence  the  angle 
between  the  tangent-lines  at  E  and  A  (originally  =  d6]  is 
now  increased  by  an  amount  CA'D  =  d<p  (or  G'A'R  =  dip); 
G'H'  is  the  new  position  of  GH.  We  obtain  the  value  of 
d(f>  as  follows :  That  part  (c^)  of  the  shortening  of  the 
fibre  at  G,  at  distance  e  from  A  due  to  the  force  p$F,  is 

§  201  eq.  (1),  cUj  =  -^.  But,  geometrically,  eM,  also  = 


Eed<p=p2ds 


(i.) 


But,  letting  M  denote  the  moment  of  the  stress-couple 
at  section  A  (M  depends  on  the  loading,  mode  of  support, 
etc.,  in  any  particular  case)  we  know  from  §  295  eq.  (6)  that 

M=^-9  and  hence  by  substitution  in  (1)  we  have 


(2) 


&M 

EI 


[If  the  arch-rib  in  question  has  less  than  three  hinges, 
the  equal  shortening  of  the  fibres  due  to  the  thrust  (of 
the  block  in  last  figure)  p^F,  will  have  an  indirect  effect  on 
the  angle  d<p.  This  will  be  considered  later.] 

370.  Total  Change    i.e.  A^  1  in  the  Angle  Between  the  End 

Tangents  of  a  Rib,  before  and  after  loading.      Take  the  ex- 
ample in  Fig.  408  of  a  rib  fixed  at  one  end  and  hinged  at 


FIG.  408. 


446 


MECHANICS   OF   ENGINEERING. 


the  other.  When  the  rib  is  unstrained  (as  it  is  supposed 
to  be,  on  the  left,  its  own  weight  being  neglected ;  it  is  not 
supposed  sprung  into  place,  but  is  entirely  without  strain) 
then  the  angle  between  the  end-tangents  has  some  value 

B 

6'—   I    dO=  the  sum  of  the  successive  small  angles  dd  for 

e/o 

each  element  ds  of  the  rib  curve  (or  axis).  After  loading, 
[on  the  right,  Fig.  408],  this  angle  has  increased  having 
now  a  value 


£. 


=  o'+  rl 

Jo 


There  must  oe  no  hinge  between  0 


FIG.  409. 


§  371.  Example  of  Equation  (I.)  in  Anal- 
ysis. —  A  straight,  homogeneous,  pris- 
matic beam,  Fig.  409,  its  own  weight 
neglected,  is  fixed  obliquely  in  a  wall. 
After  placing  a  load  P  on  the  free  end, 
required  the  angle  between  the  end- 
tangents.  This  was  zero  before  load- 
ing .*.  its  value  after  loading  is 


=0+,'=0+ 


By  considering  free  a  portion  between  0  and  any  ds  of  the 
beam,  we  find  that  M=Px=mom.  of  the  stress  couple. 
The  flexure  is  so  slight  that  the  angle  between  any  ds  and 
its  dx  is  still  practically  =a  (§  364),  and  .•.  ds=dx  sec  «. 
Hence,  by  substitution  in  eq.  (I.)  we  have 


...  <p'=P(°°*,")P  [Compare  with  §  237]. 

2-C/J: 


ARCH   BIBS. 


447 


It  is  now  apparent  that  if  both  ends  of  an  arch  rib  are 
Jixed,  when  unstrained,  and  the  rit}  be  then  loaded  ^within 
elastic  limit,  and  deformation  slight)  we  must  have 

/(Mds-r-EI}=  zero,  since  <p'=Q. 
. 

372.  Projections  of  the  Displacement  of  any  Point  of  a  Loaded 
Hib  Relatively  to  Another  Point  and  the  Tangent  Line  at  the  Lat- 
ter.— (There  must  be  no  hinge  between  0  and  B).  Let  0 
be  the  point  whose  displacement  is  considered  and  B  the 
other  point.  Fig.  410.  If  j5's  tangent-line  is  fixed  while 
the  extremity  0  is  not  supported  in  any  way  (Fig.  410) 
then  a  load  P  put  on,  0  is  displaced  to  a  new  position  On. 


FIG.  410. 


FIG.  411. 


FIG.  412. 


With  0  as  an  origin  and  OB  as  the  axis  of  X,  the  projec- 
tion of  the  displacement  00n  upon  Xt  will  be  called  dx, 
that  upon  Yt  Ay. 

In  the  case  in  Fig.  410,  O's  displacement  with  respect  to 
B  and  its  tangent-line  B  T,  is  also  its  absolute  displacement 
in  space,  since  neither  B  nor  B  T  has  moved  as  the  rib 
changes  form  under  the  load.  In  Fig.  411,  however,  the 
extremities  0  and  B  are  both  hinged  to  piers,  or  supports, 
the  dotted  line  showing  its  form  when  deformed  under  a 
load.  The  hinges  are  supposed  immovable,  the  rib  being 
free  to  turn  about  them  without  friction.  The  dotted  line 
is  the  changed  form  under  a  load,  and  the  absolute  dis- 
placement of  0  is  zero  ;  but  not  so  its  displacement  rela- 
tively to  B  and  .Z?'s  tangent  BT,  for  BT  has  moved  to  a 
new  position  BT'.  To  find  this  relative  displacement  con- 
ceive the  new  curve  of  the  rib  superposed  on  the  old  in 
a  way  that  B  and  B  T  may  coincide  with  their  original  po- 


448 


MECHANICS   OF   ENGINEERING. 


sitions,  Fig.  412.  It  is  now  seen  that  O's  displacement 
relatively  to  B  and  BT  is  not  zero  but  =00n,  and  has  a 
small  Ax  but  a  comparatively  large  Ay.  In  fact  for  this 
case  of  hinged  ends,  piers  immovable,  rib  continuous  be* 
tweeu.  them,  and  deformation  slight,  we  shall  write  Ax  = 
zero  as  compared  with  Ay,  the  axis  X passing  through  OB). 

373.  Values  of  the  X  and  Y  Projections  of  O's  Displacement  Rela- 
tively to  B  and  B's  Tangent ;  the  origin  being  taken  at  0. 
Fig.  413.  Let  the  co- 
ordinates of  the  dif- 
ferent points  E,  D,  G, 
etc.,  of  the  rib,  re- 
ferred to  0  and  an 
arbitrary  X  axis,  be 
x  and  y,  their  radial 
distances  from  0  be- 
ing u  (i.e.,  u  for  G,  u' 
for  D,  etc.;  in  gener-  i  $x 
al,  u).  OEDG  is  the  _L  , 
unstrained  form  of  the  * 

rib,  (e.g.,  the  form  it  Fl'G  413 

would  assume  if  it  lay  flat  on  its  side  on  a  level  platform, 
under  no  straining  forces),  while  OnE"D'GB  is  its  form 
under  some  loading,  i.e.,  under  strain.  (The  superposi- 
tion above  mentioned  (§  372)  is  supposed  already  made  if 
necessary,  so  that  BT  is  tangent  at  B  to  both  forms). 
Now  conceive  the  rib  OB  to  pass  into  its  strained  condi- 
tion by  the  successive  bending  of  each  ds  in  turn.  The 
straining  or  bending  of  the  first  ds,  BG,  through  the  small 
angle  dip  (dependent  on  the  moment  of  the  stress  couple 
at  G  in  the  strained  condition)  causes  the  whole  finite  piece 
06yto  turn  about  G  as  a  centre  through  the  same  small 
angle  d<p ;  hence  the  point  0  describes  a  small  linear  arc 
O0'=dv,  whose  radius  =  u  the  hypothenuse  of  the  x  and 
y  of  (7,  and  whose  value  .'.  is  ov=ud(p. 

Next   let  the  section  2),  now  at  D',   turn  through   its 
proper  angle  dip'  (dependent  on  its  stress-couple)  carrying 


ARC  El   BIBS.  449 

with  it  the  portion  D'0f,  into  the  position  D'O",  making 
0'  describe  a  linear  arc  O'O"  =(dv}'  =u'd<p't  in  which  u'  = 
the  hypothenuse  on  the  x'  and  y'  (of  D),  (the  deformation 
is  so  slight  that  the  co-ordinates  of  the  different  points 
referred  to  0  and  X  are  not  appreciably  affected).  Thus, 
each  section  having  been  allowed  to  turn  through  the  an- 
gle proper  to  it,  0  finally  reaches  its  position,  On,  of  dis- 
placement. Each  successive  8v,  or  linear  arc  described  by 
0,  has  a  shorter  radius.  Let  dx,  (dx)'t  etc.,  represent  the 
projections  of  the  successive  (dv)'s  upon  the  axis  X\  and 
similarly  %,  (%)'  etc.,  upon  the  axis  F.  Then  the  total  X 
projection  of  the  curved  line  0  ....  Oa  will  be 

Jx=    Jdx  and  similarly  dy  =  jdy     .     .     .     (1) 

But  d  v  =  u  d  (p,  and  from  similar  right-triangles, 
J  x  :  dv  :  :  y  :  u  and  dy  :  dv  :  :  x  :  u  .*.  dx  —  yd<p  and  dy=xd<f>  ; 
whence,  (see  (1)  and  (2)  of  §369) 

.    .      (ID 


and  Jy=fy=«ty    =  •    •    •    (HL) 


If  the  rib  is  homogeneous  E  is  constant,  and  if  it  is  of 
constant  cross-section,  all  sections  being  similarly  cut  by 
the  vertical  plane  of  the  rib's  axis  (i.e.,  if  it  is  a  "  curved 
prism  ")  /,  the  moment  of  inertia  is  also  constant. 

374.  Recapitulation  of  Analytical  Relations,   for    reference* 
(Not  applicable  if  there  is  a  hinge  between  0  and  E) 

Total  Change  in  Angle  between  )  _     /**Mds 


eween     _     /* 
tangent-lines    0   and   B      ]~  J0 

The  X-Projection  of  O's  Displacement  ~) 

Relatively  to  B  and  B's  tangent-  I         ^Myds  /TT  \ 

line  ;    (the  origin  being   at    0)  L  =  /       *LT-  •    •    •  (-LL) 
and  the  axes  X  and  Y  T  to         c/0     « 

each  other) 


450 


MECHANICS  OF   ENGINEERING. 


The  Y-Projection  of  O's  Displacement,  )  _     s*-mi 
etc.,  as  above.  }  ~Jo~l 


EI 


•  (III.) 


Here  x  and  y  are  the  co-ordinates  of  points  in  the  rib- 
curve,  cfc  an  element  of  that  curve,  M  the  moment  of  the 
stress-couple  in  the  corresponding  section  as  induced  by 
the  loading,  or  constraint,  of  the  rib. 

(The  results  already  derived  for  deflections,  slopes,  etc., 
for  straight  beams,  could  also  be  obtained  from  these 
formulae,  I.,  II.  and  III.  In  these  formulae  also  it  must 
be  remembered  that  no  account  has  been  taken  of  the 
shortening  of  the  rib-axis  by  the  thrust,  nor  of  the  effect 
of  a  change  of  temperature.) 

374a.  Eesumfe  of  the  Properties  of  Equilibrium  Polygons  and 
their  Force  Diagrams,  for  Systems  of  Vertical  Loads. — See  §  §  335 
to  343.  Given  a  system  of  loads  or  vertical  forces,  P]?  P2, 

etc.,  Fig.  414,  and 
^Ip,  two  abutment  verti- 
2  cals,  F'  and  G' ;  if 
P2,  we  lay  off,  vertically, 
to  form  a  "  load- 
line,"  1 ..  2  =  Plf  2. .. 
PS  3=P2,etc.,  select  any 
Pole,  019  and  join  Ol 
...  1,  Oi  .  .  .  2,  etc. ; 
also,  beginning  at 
any  point  Fl  in  the 
vertical  P',  if  we  draw  PL..  a\\  to  Oi..l  to  intersect  the 
line  of  P!  ;  then  ab  \\  to  Ol . .  2,  and  so  on  until  finally  a 
point  GI,  in  G',  is  determined ;  then  the  figure  PL  .abc  G^  is 
an  equilibrium  polygon  for  the  given  loads  and  load  verti- 
cals, and  Oi . . .  1234  is  its  "  force  diagram."  The  former 
is  so  called  because  the  short  segments  F&  ab,  etc.,  if 
considered  to  be  rigid  and  imponderable  rods,  in  a  vertical 
plane,  hinged  to  each  other  and  the  terminal  ones  to  abut- 
ments Fl  and  (r},  would  be  in  equilibrium  under  the  given 
loads  hung  at  the  joints.  An  infinite  number  of  equilib- 


414. 


AKCH-3IBS.  451 

rium   polygons  may  be  drawn  for  the  given  loads  and 

abutment-verticals,  by  choosing  different  poles  in  the  force 

.  diagram.     [One  other  is  shown  in  the  figme  ;  02  is  its 

pole.  (.Z^  Gl  and  F2  (^  are  abutment  lines.)]      For  all   of 

these  the  following  statements  are  true : 

s     (1.)    A  line  through  the  pole,  ||  to  the  abutuieutline  cuts 

/  the  load-line  in  the  same  point  n',  whichever  equilibrium 

)    P°lyg°n  be  used  (  /.  any  one  will  serve  to  determine  n1). 

(2.)     If  a  vertical  CD  be  drawn,  giving  an  intercept  z'  in 

I    each  of  the  equilibrium  polygons,  the  product  Hz'  is  the 

same  for  all  the  equilibrium  polygons.     That  is,  (see  Fig. 

414)  for  any  two  of  the  polygons  we  have 

U.'.H,::  z2'  :  «/  ;  or  H2  z,'  =  H,  z,'. 

(3.)  The  compression  in  each  rod  is  given  by  that 
"  ray  "  (in  the  force  diagram)  to  which  it  is  parallel. 

(4.)  The  "  pole  distance  "  H,  or  -|  let  fall  from  the  pole 
upon  the  load-line,  divides  it  into  two  parts  which  are  the 
•  vertical  components  of.  the  compressions  in  the  abutment- 
rods  respectively  ( the  other  component  being  horizontal)  ; 
//  is  the  horizontal  component  of  each  (and,  in  fact,  of 
each  of  the  compressions  in  all  the  other  rods).  The 
compressions  in  the  extreme  rods  may  also  be  called  the 
abutment  reactions  (oblique)  and  are  given  by  ike  extreme 
rays. 

(5.)  Three  Points  [not  all  in  the  same  segment  (or  rod)] 
determine  an  equilibrium  polygon  for  given  loads.  Hav- 
ing given,  then,  three  points,  we  may  draw  the  eq uilibrium 
polygon  by  §341. 

375.  Summation  of  Products.  Before  proceeding  to  treat 
graphically  any  case  of  arch-ribs,  a  few  processes  in 
graphical  arithmetic,  as  it  may  be  called,  must  be  pre- 
sented, and  thus  established  for  future  use. 

To  make  a  summation  of  products  of  two  factors  in  each 
by  means  of  an  equilibrium  polygon. 


452 


MECHANICS   OF   ENGINEERING. 


Construction.     Suppose  it  required  to  make  the  summa- 
tion 2  (x  z)  i.  e.t  to  sum  the  series 

#1  «i+  #2  32  +  #3  %  +    .     .     .     by  graphics. 

Having  first  arranged  the  terms  in  the  order  of  magni- 
tude of  the  as's,  we  proceed  as  follows :  Supposing,  for 
illustration,  that  two  of  the  s's  (%  and  «4)  are  negative 
(dotted  in  figure)  see  Fig.  415.  These  quantities  x  and  z 
may  be  of  any  nature  whatever,  anything  capable  of  being 
represented  by  a  length,  laid  off  to  scale. 

First,  in  Fig 
416,  lay  off  the 
s's  in  their 
order,  end  to 
end,  on  a  ver- 
tical load-line 
taking  care  to 
lay  off  %  and 
34  upward  in 
their  turn. 
Take  any  con- 
FIG.  415.  FIG.  416.  venient  pole 

0 ;  draw  the  rays  0  ...  1,  0  ...  2,  etc.;  then,  having  pre- 
viously drawn  vertical  lines  whose  horizontal  distances 
from  an  extreme  left-hand  vertical  F'  are  made  =  xlt  x}, 
a?3,  etc.,  respectively,  we  begin  at  any  point  F,  in  the  verti- 
cal F',  and  draw  a  line  ||  to  0  ...  1  to  intersect  the  x}  ver- 
tical in  some  point  ;  then  V  2'  H  to  0  ...  2,  and  so  on,  fol- 
lowing carefully  the  proper  order.  Produce  the  last  seg- 
ment (6'  ...  G  in  this  case)  to  intersect  the  vertical  F'  in 
some  point  K.  Let  KF  =k  (measured  on  the  same  scale 
as  the  x's),  then  the  summation  required  is 

2Y5  (xz)  =  Hk. 

H  is  measured  on  the  scale  of  the  s's,  which  need  not  be 
the  same  as  that  of  the  #'s  ;  in  fact  the  «'s  may  not  be  the 
same  kind  of  quantity  as  the  x's. 

[PROOF. — From  similar  triangles  R:  «L ::  xl :  klf  -.:  x^^HTc^ ; 
and  "  "  "         H:  «, ::  x2 :  L,  .:  x.z.2=Bk2 , 


ARCH-RIBS. 


453 


and  so  on.     But  H(k1+k2+etG.)=HxFK=Hk']. 

376.  Gravity  Vertical. — From  the  same  construction  in 
Fig.  415  we  can  determine  the  line  of  action  (or  gravity 
vertical)  of  the  resultant  of  the  parallel  vertical  forces  zl9 
«2,  etc.  (or  loads);  by  prolonging  the  first  and  last  segments 

to  their  intersection  at 
(j.  The  resultant  of  the 
system  of  forces  or  loads 
acts  through  C  and  is 
vertical  in  this  case  ;  its 
value  being  —  2'  (z)y 
that  is,  it  =  the  length 
1  ...  7  in  the  force  dia- 
gram, interpreted  by  the 
proper  scale.  It  is  now 
supposed  that  the  z's 
represent  forces,  the  x'a 
being  their  respective 
lever  arms  about  F.  If 
the  ?'s  represent  the 
areas  of  small  finite  por- 
tions of  a  large  plane 
figure,  we.  may  find  a 
giavity-line  (through  C) 
of  that  figure  by  the 
above  construction;  each 
z  being-applied  through 
the  centre  of  gravity  of 
its  own  portion. 

Calling   the    distance 
x  between  the  verticals 
through    G  and    F,   we 
k3  have     also   x .  I   (z)    = 
2'  (xz)  because   2  (z)  is 
the  resultant  of  the  |[  z's. 
This  is  also  evident  from 
the  proportion  (similar 
triangles) 
H  :  (1  .  .  7)  ::  x  :  k. 


MECHANICS   OF   ENGINEERING. 

376  a,  .Moment  of  Inertia  (of  Plane  Figure)  by  Graphics,—  Fig. 
416  a.  IN  =  ?  First,  for  the  portion  on  right.  Divide  OR 
?nto  equal  parts  each  =  Ax.  Let  zly  z2,  etc.,  be  the  middle 
ordinates  of  the  strips  thus  obtained,  and  xly  etc.  their 
abscissas  (of  middle  points). 

Then  we  have  approximately 


IN  for 

=4x[(z1x1)xl+(z2x2)x2+  ...]..(!) 

But  by  §375  we  may  construct  the  products  z^x^x^  etc., 
taking  a  convenient  .H*9  (see  Fig.  416,  (6)),  and  obtain  kly  k2t 
etc.,  such  that  z^  =  H'klt  z2x2  =  H'Jc2,  etc.  Hence  eq.  (1) 
becomes  : 

7Nfor  OR  a,pprox.=H'4x['>-lxi'^-k2x2+  ...]•••  (2) 

'&.  -  *  V*, 

By  a  second  use  of  §  375  (seu^i'g.  416  c)  we  construct  lt 
such  that  k&t  +  k.2x2'+  .  .  .  .  =  H"l  \R"  taken  at  con- 
venience]. .'.  from  eq.  (2)  we  have  finally,  (approx.), 

JN  for  OR=H'H"lAx  ____  (3) 

For  or?».mple  if  OR  ~-  4  in.,  with  four  strips,  Ax  would  = 
1  in.;  and  if  lir  •-  <*  in.,  B."  =  2  in.,  and  I  =  5.2  in.,  then 

7N  for  OR  =  2x2x5.2x1.0=20.8  biquad.  inches. 

The  Iy  for  0  L,  on  the  left  of  N,  is  found  in  a  similar 
manner  and  added  to  7N  for  OR  to  obtain  the  total  7N.  The 
position  of  a  gravity  axis  is  easily  found  by  cutting  the 
shape  out  of  sheet  metal  and  balancing  on  a  knife  edge  ;  or 
may  be  obtained  graphically  by  §  336  ;  or  376. 

377.  Construction  for  locating  a  line  vm(Fig.  417)  at  (a),  in 
the  polygon  FG  in  such  a  position  as  to  satisfy  the  two 
following  conditions  with  reference  to  the  vertical  inter- 
cepts at  1,  2,  3,  4,  and  5,  between  it  and  the  given  points 
1.  2,  3,  etc.,  of  the  perimeter  of  the  polygon. 


ARCH-KIBS. 


455 


Condition  I. — (Calling  these  intercepts  u^  u2,  etc.,  and  their 
horizontal  distances  from  a  given  vertical  F,  xly  x.2,  etc.) 

2  (u)  is  to  =  0 ;  i.e.,  the  sum  of  the  positive  u'&  must  be 
numerically  ~  that  of  the  negative  (which  here  are  at  1 
and  5).  An  infinite  number  of  positions  of  vm  will  satisfy 
condition  I. 


Condition  II — S  (ux)  is 


(d) 


to  =  0 ;  i.e.,  the  sum  of  the 
moments  of  the  positive  w's 
about  F  must  =  that  of  the 

G(a)  negative  -M'S.  i.e.,  the  moment 
of  the  resultant  of  the  posi- 

m  tive  w's  must  =  that  of  the 
resultant  of  the  negative ; 

G       and     /.     (Condit.    I    being 

m  already  satisfied)  these  two 
resultants  must  be  directly 

G(C)  opposed  and  equal.  But  the 
ordinates  u  in  (a)  are  indi- 

m  vidually  equal  to  the  differ- 
ence of  the  full  and  dotted 
ordinates  in  (b)  with  the 
same  #'s  .*.  the  conditions 
may  be  rewritten  : 

2  (dotted  ords.  in  (b)). 

II.  I  [each  full  ord.  in  (b) 
X  its  x~\   =  2  [each  dotted 
ord.   in  (b)   x  its  x~]  i.e.,  the 
centres  of  gravity  of  the  full 
and  of  the  dotted  in  (b)  must  lie  in  the  same  vertical 

Again,  by  joining  vG,  we  may  divide  the  dotted  ordi- 
nates of  (b)  into  two  sets  which  are  dotted,  and  broken,  re- 
spectively, in  (c)  Then,  finally,  drawing  in  (d\ 

ft,  the  resultant  of  full      ords.  of  (c) 
T,    ".          "          "  broken   "       "   " 
T',  "  "          "  dotted    "       "   " 


FIG.  417. 


4o6  MECHANICS    OF    ENGINEEK1NG. 

we  are  prepared  to  state  in  still  another  and  final  form  the 
conditions  which  vm  must  fulfil,  viz. : 

(I.)  T+T'  must  =  E\  and  (II.)  The  resultant  of  T 
and  T'  must  act  in  the  same  vertical  as  H. 

In  short,  the  quantities  T,  T',  and  J%  must  form  a  bal- 
anced system,  considered  as  forces.  All  of  which  amounts 
practically  to  this :  that  if  the  verticals  in  which  T  and  T' 
act  are  known  and  E  be  conceived  as  a  load  supported  by 
a  horizontal  beam  (see  foot  of  Fig.  417,  last  figure)  resting 
on  piers  in  those  verticals,  then  T  and  T'  are  the  respec- 
tive reactions  of  those  piers.  It  will  now  be  shown  that  the 
verticals  of  T  and  T'  are  easily  found,  being  independent  of 
the  position  of  vm ;  and  that  both  the  vertical  and  the  mag- 
nitude of  R,  being  likewise  independent  of  vm,  are  deter- 
mined with  facility  in  advance.  For,  if  v  be  shifted  up 
or  down,  all  the  broken  ordinates  in  (c)  or  (d)  will  change 
in  the"  same  proportion  (viz.  as  vF  changes),  while  the 
dotted  ordinates,  though  shifted  along  their  verticals,  do» 
not  change  in  value  ;  hence  the  shifting  of  v  affects  neither 
the  vertical  nor  the  value  of  T't  nor  the  vertical  of  J7. 
The  value  of  T,  however,  is  proportional  to  vF.  Similar- 
ly, if  m  be  shifted,  up  or  down,  T'  will  vary  proportionally 
to  mG,  but  its  vertical,  or  line  of  action,  remains  the  same. 
T  is  unaffected  in  any  way  by  the  shifting  of  m.  ft,  d&- 
pending  for 'iL,:' ~T°1  IIP  and  position  on  the  full  ordinates  of 
(c)  Fig.  417,  is  independent  01  the  location  of  vm.  We 
may  /.  proceed  as  follows  : 

1st.  Determine  R  graphically,  in  amount  and  position, 
by  means  of  §  376. 

2ndly.  Determine  the  verticals  of  T  and  T'  by  any  trial 
position  of  vm  (call  it  v2m.2),  and  the  corresponding  trial 
values  of  T  and  T'  (call  them  T2  and  T'2). 

3rdly.  By  the  fiction  of  the  horizontal  beam,  construct 
(§  329)  or  compute  the  true  values  of  T  and  T',  and  then 
determine  the  true  distances  vF  and  mG  by  the  propor- 
tions 

vF  :  v2F  :  :  T  :  T,  and  mG  :  m,2G '  : :  T'  :  F» 


ARCH-BIBS. 


457 


Example  of  this.      Fig.  418.      (See  Fig.  417  for  s  and  t.) 

From  A  toward  B  in  (e)  Fig.  418,  lay  off  the  lengths  (or 
lines  proportional 
to  them)  of  the  full 
ordinates  1,  2,  etc., 
of  (/).  Take  any 
pole  Oj,  and  draw  the 
equilibrium  poly- 
gon (/")'  and  pro- 
long its  extreme  seg- 
ments to  find  C  and 
thus  determine  J?'s 
vertical.  R  is  repre- 
sented by  AB.  In 
(g)  [same  as  (/)  but 
shifted  to  avoid 
complexity  of  lines], 
draw  a  trial  v.2m.2  and 
join  v2G2.  Deter- 
mine the  sum  T2  of 
the  broken  ordi-  SFIG.  4is. 

nates  (between  v2G2  ana  F2G2)  and  its  vertical  line  of  ap- 
plication, precisely  as  in  dealing  with  E ;  also  T\  that  of 
the  dotted  ordinates  (five)  and  its  vertical.  Now  the  true 
<-  T=Rt+(8+t)  and  _the  true-  T'=fis+(s^t).  Hence  com- 
pute v~F=(T+T2)  v7F2  and  mG^T'+T'^  m,2G2,  and  by 
laying  them  off  vertically  upward  from  F  and  G  respec- 
tively we  determine  v  and  m,  i.e.,  the  line  vm  to  fulfil  the 
conditions  imposed  at  the  beginning  of  this  article,  rela- 
ting to  the  vertical  ordinates  intercepted  between  vm  and 
given  points  on  the  perimeter  of  a  polygon  or  curve. 

Note  (aX  If  the  verticals  in  which  the  intercepts  lie  are 
equidistant  and  quite  numerous,  then  the  lines  of  action 
of  T.2  and  T'2  will  divide  the  horizontal  distance  between 
F  and  G  into  three  equal  parts.  This  will  be  exactly  true 
in  the  application  of  this  construction  to  §  390. 

Note  (b).  Also,  if  the  verticals  are  symmetrically  placed 
about  a  vertical  line,  (as  will  usually  be  the  case)  v.2m2  is 


458 


MECHANICS    OF   ENGINEERING. 


best  drawn  parallel  to  FG,  for  then  T2  and  T\  will  be 
equal  and  equi-distant  from  said  vertical  line. 

378.  Classification  of  Arch-Ribs,  or  Elastic  Arches,  according 
to  continuity  and  modes  of  support.  In  the  accompany- 
ing figures  i~h.Q/ull  curves  show  the  unstrained  form  of  the 
rib  (before  any  load,  even  its  own  weight,  is  permitted  to 
come  upon  it)  ;the  dotted  curve  shows  its  shape  (much  ex- 
aggerated) when  bearing  a  load.  For  a  given  loading 
Three  Conditions  must  be  given  to  determine  the  special 
equilibrium  polygon  (§§  366  and  367). 

Class  A. — Continuous  rib,  free  to  slip  laterally  on  the 
piers,  which  have  smooth  horizontal  surfaces,  Fig.  420. 

This  is  chiefly  of  theoretic  interest,  its  consideration 
being  therefore  omitted.  The  pier  reactions  are  neces- 
sarily vertical,  just  as  if  it  were  a  straight  horizontal 
beam. 

Class  B,  Rib  of  Three  Hinges,  two  at.  the  piers  and  one 
intermediate  (usually  at  the  crown)  Fig.  421.  Fig.  36  also 
is  an  example  of  this.  That  is,  the  rib  is  discontinuous 
and  of  two  segments*  Since  at  each  hinge  the  moment  of 
the  stress  couple  must  be  be  zero,  the  special  equilibrium 
polygon  must  pass  through  the  hinges.  Hence  as  three 
points  fully  determine  an  equilibrium  polygon  for  given 
load,  the  special  equilibrium  is  drawn  by  §  341. 


H  B 


FIG.  420. 


FIG.  421, 


[§  378a  will  contain  a  construction  for  arch-ribs  of  three 
hinges,  when  the  forces  are  not  all  vertical.] 

Class  C.  Rib  of  Two  Hinges,  these  being  at  the  piers,  the 
rib  continuous  between.  The  piers  are  considered  im- 
movable, i.e.,  the  span  cannot  change  as  a  consequence  of 
loading.  It  is  also  considered  that  the  rib  is  fitted  to  its 


AliCII   1UBS. 


459 


hinges  at  a  definite  temperature,  and  is  then  under  no  con- 
straint from  the  piers  (as  if  it  lay  flat  on  the  ground),  not 
even  its  own  weight  being  permitted  to  act  when  it  is  fi- 
nally put  into  position.  When  the  "  false  works " 
or  temporary  supports  are  removed,  stresses  are  in- 
duced in  the  rib  both  by  its  loading,  including  its 
own  weight,  and  by  a  change  of  temperature.  Stresses 
due  to  temperature  may  be  ascertained  separately  and 
then  combined  with  those  due  to  the  loading.  [Classes 
A  and  B  are  not  subject  to  temperature  stresses.]  Fig. 

422  shows  a  rib  of  two  hinges, 
at  ends.  Conceive  the  dotted 
curve  (form  and  position  un- 
der strain)  to  be  superposed 
^^.,  on  the  continuous  curve 

I  /       ^— — --""  P-?  (form  before  strain)  in '  such 

6,T~  a  way  that  B  and  its  tangent 

FIG-  422-  line    (which    has   been   dis- 

placed from  its  original  position)  may  occupy  their  pre- 
vious position.  This  gives  us  the  broken  curve  OnB.  00n 
is  .'.  O's  displacement  relatively  to  B  and  j^'s  tangent. 
Now  the  piers  being  immovable  OnB  (right  line)  =OB  ;  i.e., 
the  X  projection  (or  Ax)  of  00n  upon  OB  (taken  as  an  axis 
of  X)  is  zero  compared  with  its  Ay.  Hence  as  one  condi- 
tion to  fix  the  special  equilibrium  polygon  for  a  given  load- 
ing we  have  (from  §  373) 


r 

t/o 


(1) 


The  other  two  are  that  the  ]  must  pass  through  0  .  (2) 
special  equilibrium  polygon  (  "  "  "  B  .  (3) 

Class  D.  Rib  with  Fixed  Ends  and  no  hinges,  i.e.,  continu- 
ous. Piers  immovable.  The  ends  may  be  fixed  by  being 
inserted,  or  built,  in  the  masonry,  or  by  being  fastened  to 
large  plates  which  are  bolted  to  the  piers.  [The  St.  Louis 
Bridge  and  that  at  Coblenz  over  the  Rhine  are  of  this 
class.]  Fig.  423.  In  this  class  there  being  no  hinges  we 


4(JO 


MECHANICS   OF   ENGINEERING. 


FIG.  423. 


have  no  point  given  in  advance  through  which  the  special 
equilibrium  polygon  must  pass.  However,  since  O's  dis- 
placement relatively  (and  absolutely)  to  B  and  J5's  tangent 
is  zero,  both  Ax  and  Ay  [see  §  373]  —  zero.  Also  the  tan- 
gent-lines both  at  0  and  B  being 
fixed  in  direction,  the  angle  be- 
tween them  is  the  same  under 
loading,  or  change  of  temperature, 
as  when  the  rib  was  first  placed 
in  position  under  no  strain  and  at 
a  definite  temperature. 
Hence  the  conditions  for  locating  the  special  equilibrium 
polygon  are 

/*B  Mds  _  Q .      /*  Myds  _  Q  t     n*Mxds  __  Q 
Jo   El  '   Jo  ~E2~         '  Jo    El 

In  the  figure  the  imaginary  rigid  prolongations  at   the 
ends  are  shown  [see  §  366]. 

Other  designs  than  those  mentioned  are  practicable 
(such  as :  one  end  fixed,  the  other  hinged ;  both  ends  fixed 
and  one  hinge  between,  etc.),  but  are  of  unusual  occur- 
rence. 

378a.  Bib  of  Three  Hinges,  Forces  not  all  Vertical.*  If  the 
given  rib  of  three  hinges  upholds  a  roof,  the  wind-press- 
ure on  which  is  to  be  considered  as  well  as  the  weights  of 
the  materials  composing  the  roof-covering,  the  forces  will 
not  all  be  vertical.  To  draw  the  special  equil.  polygon  in 

such  a  case  the  following 
construction  holds  :  Re- 
quired to  draw  an  equilib- 
rium polygon,  for  any 
plane  system  of  forces, 
through  three  arbitrary 
points,  A,  p  and  B ;  Fig. 
B423a.  Find  the  line  of 
action  of  Elt  the  resultant 
of  all  the  forces  occurring 
between  A  and  p;  also, 

*  See  p.  117  of  the  author's  "Notes  and  Examples  in  Mechanics"  fora 
detailed  example  of  the  following  construction. 


FIG.  428a. 


ARCH-RIBS.  401 

that  of  JR:,  the  resultant  of  all  forces  between  ,p  and  B  ; 
also  the  line  of  action  of  R,  the  resultant  of  R^  and  2t2,  [see 
§  328.]  Join  any  point  M  in  R  with  A  and  also  with  B, 
and  join  the  intersections  N  and  0.  Then  A  N  will  be  the 
direction  of  the  first  segment,  0  B  that  of  the  last,  and 
NO  itself  is  the  segment  corresponding  to  p  (in  the  de- 
,sired  polygon)  of  an  equilibrium  polygon  for  the  given 
forces.  See  §  328.  If  A  N' p  0'  B  are  the  corresponding 
segments  (as  yet  unknown)  of  the  desired  equil.  polygon, 
we  note  that  the  two  triangles  MNO  and  M'N' 'O ',  having 
their  vertices  on  three  lines  which  meet  in  a  point  [i.e.,  R 
meets  Rl  and  R2  in  (7'],  are  homological  [see  Prop.  VII.  of 
Introduc.  to  Modern  Geometry,  in  Chauvenet's  Geometry,] 
and  that  .  • .  the  three  intersections  of  their  corresponding 
sides  must  lie  on  the  same  straight  line.  Of  those  inter^ 
sections  we  already  have  A  and  B,  while  the  third  must  be 
at  (7,  found  at  the  intersection  of  AB  and  NO.  Hence  by 
connecting  C  and  p,  we  determine  N'  and  0'.  Joining 
WA  and  O'B,  the  first  ray  of  the  required  force  diagram  will 
be  II  to  NA,  while  the  last  ray  will  be  ||  to  O'B,  and  thus 
the  pole  of  that  diagram  can  easily  be  found  and  the  cor- 
responding equilibrium  polygon,  beginning  at  A,  will  pass 
through  p  and  B. 
(This  general  case  includes  those  of  §§  341  and  342.) 

379.  Arch-Rib  of  two  Hinges;  by  Prof.  Eddy's  Method,* 
[It  is  understood  that  the  hinges  are  at  the  ends.]  Ke- 
quired  the  location  of  the  special  equilibrium  polygon.  We 
here  suppose  the  rib  homogeneous  (i.e.,  the  modulus  of 
Elasticity  E  is  the  same  throughout),  that  it  is  a  "  curved 
prism  "  (i.e.,  that  the  moment  of  inertia  /  of  the  cross- 
section  is  constant),  that  the  piers  are  on  a  level,  and  that- 
the  rib-curve  is  symmetrical  about  a  vertical  line.  Fig. 

424.     For  each  point  m  of  the  rib 
curve  we  have  an  x  and  y  (both 
known,  being  the  co-ordinates  of 
the  point),  and  also  a  z  (intercept 
T-  between   rib  and  special  equilib- 
ria. 424.  rium  polygon)  and  a  z'  (intercept 

*  r.  25  of  Prof.  Eddy's  book  ;  see  reference  in  preface  of  this  work. 


462 


MECHANICS    OF    ENGINEEKING. 


between  the  spec.  eq.  pol.  and  the  axis  X  (which  is  OB). 
The  first  condition  given  in  §  378  for  Class  C  may  be 
transformed  as  follows,  remembering  [§  367  eq.  (3)]  that 
M  =  Hz  at  any  point  ra  of  the  rib  (and  that  El  is  con- 
stant). 

-A-     TB  Myds  =  0,  i.e.,  M.    fzyds  =  0 .  • .   fzyds  =  0 
tii  c/o  MJ!  cjo  c/o 


>  \  - '  -  f(y  ~  *')2/ds=o;  i-6-'  C  yyds  =  C  yz'ds  • 

)         c/o  c/o  e/o 


but 

z=y  - 

In  practical  graphics  we  can  not  deal  with  infinitesimals ; 
hence  we  must  substitute  As  a  small  finite  portion  of  the 
rib-curve  for  ds\  eq.  (1)  now  reads  2*  yy  As  =  2'0B  yz'  As. 
But  if  we  take  all  the  As's  equal,  As  is  a  common  factor 
and  cancels  out,  leaving  as  a  final  form  for  eq.  (1) 

S*(yy)  =  I*  (yz')     .     .         .     (1)' 

The  other  two  conditions  are  that  the  special  equilibrium 
polygon  begins  at  0  and  ends  at  B.  (The  subdivision  of 
the  rib-curve  into  an  even  number  of  equal  As's  will  be  ob- 
served in  all  problems  henceforth.) 

379a.  Detail  of  the  Construction,     Given  the  arch-rib  0  B9 
Fig.  425,  with  specified  loading.     Divide  the  curve  into 


2345678 

!  U -I  J I ! 


4'-4-4-4-~i-t- 


FIG.   426. 


ARCH  KIBS. 


463 


eight  equal  ^s's  and  draw  a  vertical  through  the  middle 
of  each.  Let  the  loads  borne  by  the  respective  /te's  be, 
Plt  P2,  etc.,  and  with  them  form  a  vertical  load-line  A  C  to 
some  convenient  scale.  With  any  convenient  pole  0" 
draw  a  trial  force  diagram  Q^  AC,  and  a  corresponding 
trial  equilibrium  polygon  F  G,  beginning  at  any  point  in 
the  vertical  jF.  Its  ordinates  z/',  z£*  etc.,  are  propor- 
tional to  those  of  the  special  equil.  pol.  sought  (whose 
abutment  line  is  OB)  [§  374a  (2)].  We  next  use  it  to  de- 
termine n'  [see  §  374a].  We  know  that  OB  is  the  "  abut- 
ment-line1" of  the  required  special  polygon,  and  that  .  * . 
its  pole  must  lie  on  a  horizontal  through  n'.  It  remains 
to  determine  its  S,  or  pole  distance,  by  equation  (1)'  just 
given,  viz.  :  2f  yy  =  Sfyz'.  First  by  §  375  find  the  value 
of  the  summation  ^i(yy),  which,  from  symmetry,  we  may 
write  —  2  m 

j/}  Hence,  Fig.  426,  we  obtain 

fr, 


i 

2 

r 
j 

\ 

s. 

f 

\ 

x 

L 

/ 

r 

^ 

u 

\ 

i 

/ 

^ 

^x 

\ 

r 

\ 

V 

6 

f 

\ 

\ 

L 

—  • 

^—  - 

^ 

Vr* 

-% 

*" 

n 

3  Next,  also  by  §  375,  see  Fig. 
427,  using  the  same  pole  dis- 

4  tance  H0   as  in  Fig. .  426,  we 
find 


2/i*i"  +2/2*2"  - 
«•'  H&". 

£  Again,  since  I\    (yz")  = 

+  2/7*7"  +  2/6*e"  +  y&"  which 
"  from    symmetry  (of  rib) 


we  obtain,  Fig.  428, 

21  (y«")  =  BJk,",  (same    P 

and  .\ 

Z{  (y*")=H0  (W+kt").     If  now  we  find  that  &,"+ V'=! 


464  MECHANICS  OF  ENGINEERING. 

the  condition  2\  (yy]  =  Z\  (yz")  is  satisfied,  and  the  polo 
distance  of  our  trial  polygon  in  Fig.  425,  is  also  that  of 
the  special  polygon  sought;  i.e.,  the  2"  's.are  identical  in 
value  with  the  z"s  of  Fig.  424.  In  general,  of  course,  we 
do  not  find  that  &/'+  V  =  2&.  Hence  the  a"  's  must  all 
be  increased  in  the  ratio  2k:  (&i''+V)  to  become  equal  to 
the  3"s.  That  is,  the  pole  distance  H  of  the  spec,  equil* 
polygon  must  be 

TT_  &/'+&/'  jjtt  (in  which  H"  =  the  pole  distance  of  the 
2;£  trial  polygon)  since  from  §339  the   ordi- 

nates  of  two  equilibrium  polygons  (for  the  same  loads) 
are  inversely  as  their  pole  distances.  Having  thus  found 
the  H  of  the  special  polygon,  knowing  that  the  pole  must 
lie  on  the  horizontal  through  n'9  Fig.  425,  it  is  easily 
drawn,  beginning  at  0.  As  a  check,  it  should  pass  through 
S. 

For  its  utility  see  §  367,  but  it  is  to  be  remembered  that 
the  stresses  as  thus  found  in  the  different  parts  of  tha 
rib  under  a  given  loading,  must  afterwards  be  combined 
with  those  resulting  from  change  of  temperature  and  the 
shortening  of  the  rib  axis  due  to  the  tangential  thrusts,. 
before  the  actual  stress  can  be  declared  in  any  part. 

[NOTE.  —  If  the  "  moment  of  inertia,"  /,  of  the  rib-sec- 
tion is  different  at  different  sections,  i.e.,  if  /  is  variable, 

foreq.(l)'  we  may  write  ^JlWl'J.         ,     .     .     (1)' 


(where  n  =  -,  70  being  the  moment  of  inertia  of  a  particu* 

*o 

lar  section  taken  as  a  standard  and  /that  at  any  section 
of  rib)  and  in  Fig.  426,  use  the  ^  of  each  Js  instead  of  y 


in  the  vertical  "  load-line,"  and  —  for  a"  in  Pigs.  427  and 
428]. 


AKCH-RIBS. 


465 


380.  Arch  Rib  of  Fixed  Ends  and  no  Hinges.— Example  of 
Class  D.  Prof.  Eddy's  Method.*  As  Before,  E  and  /  are 
constant  along  the  rib  Piers  immovable.  Bib  curve 
symmetrical  about  a  vertical  line.  Fig.  429  shows  such  a 
rib  under  any  loading.  Its  span  is  OB,  which  is  taken  as 
an  axis  X.  The  co-ordinates  of  any  point  ra'  of.  the  rib 
curve  are  x  and  y,  and  z  is  the  vertical  intercept  between 
m'  and  the  special  equilibrium  polygon  (as  yet  unknown, 
but  to  be  constructed).  Prof.  Eddy's  method  will  now  be 

given  for  finding  tha  spe- 
v»  cial  equil.  polygon.  The 
three  conditions  it 
must  satisfy  (see  §  378, 
Class  D,  remembering 
that  E  and  /  are  constant 
and  that  M  =  ITz  from 
§  367)  are 


FIQ.  429. 


rlds=  0  ;  fxzds=  ;  0  and    Cyzds=0     .    .      (1) 

e/o  e/o  e/o 

Now  suppose  the  auxiliary  reference  line  (straight)  vm 
to  have  been  drawn  satisfying  the  requirements,  with 
respect  to  the  rib  curve  that 


/*«'cfe=0 ;  and   fxz'da=0    .     .    .    . 

e/o  e/o 


(2) 


in  which  z'  is  the  vertical  distance  of  any  point  mf  from 
vm  and  x  the  abscissa  of  m'  fronl  0. 

From  Fig.  429,  letting  z"  denote  the  vertical  intercept 
(corresponding  to  any  m')  between  the  spec,  polygon  and 
the  auxiliary  line  vm9  we  have  z=z/—xf/,  hence  the  three 
conditions  in  (1.)  become 

Clz'-z")ds=0',  i.e.,  see  eqs.  (2)    A*  z"ds=0  .    .  .   (3) 


*  P.  14  of  Prof  Eddy's  book  ;  see  reference  in  preface  of  this  work 


466  MECHANICS   OF   ENGINEERING. 

Cx  (zr—z")ds=Q  ;  i.e.,  see  eqs.  (2)    C  xz"ds=o 

c/o  e/  o 


provided  vm  has  been  located  as  prescribed. 

For  graphical  purposes,  having  subdivided  the  rib  curve 
into  an  even  number  of  small  equal  Js's,  and  drawn  a  verti- 
cal through  the  middle  of  each,  we  first,  by  §  377,  locate 
rm'to  satisfy  the  conditions 

1?(*')=0  and  I?(a*')=0          .         .         (6) 

(see  eq.  (2)  ;  the  As  cancels  out)  ;  and  then  locate  the 
special  equilibrium  polygon,  with  vm  as  a  reference-line, 
by  making  it  satisfy  the  conditions. 

20V)=0  .  (7);    S%ar)=0  .  (8);   !•„%*")  =I»(y*')  .   (9) 

(obtained  from  eqs.  (3),  (4),  (5)  by  putting  ds=4s,  and  can- 
celling). 

Conditions  (7)  and  (8)  may  be  satisfied  by  an  infinite 
number  of  polygons  drawn  to  the  given  loading.  Any  one 
of  these  being  drawn,  as  a  trial  polygon,  we  determine  for  it 
the  value  of  the  sum  2%(yz")  by  §  375,  and  compare  it  with 
the  value  of  the  sum  2'J(y«')  which  is  independent  of  the 
special  polygon  and  is  obtained  by  §  375.  [N.B.  It  must 
be  understood  that  the  quantities  (lengths)  x,  y,  z,  z',  and  2", 
here  dealt  with  are  those  pertaining  to  the  verticals  drawn 
through  the  middles  of  the  respective  ^/s's,  which  must  be 
sufficiently  numerous  to  obtain  a  close  result,  and  not  to 
the  verticals  in  which  the  loads  act,  necessarily,  since  these 
latter  may  be  few  or  many  according  to  circum  stances,  see 
Fig.  429],  If  these  sums  are  not  equal,  the  pole  distance 
of  the  trial  equil.  polygon  must  be  altered  in  the  proper 
ratio  (and  thus  change  the  «'"s  in  the  inverse  ratio)  neces- 
sary to  make  these  sums  equal  and  thus  satisfy  condition 
(9).  The  alteration  of  the  s'"s,  all  in  the  same  ratio,  will 


ABOH-BLBS. 


467 


not  interfere  with  conditions  (7)  and  (8)  which  are  already 

satisfied. 


381.  Detail  of  Construction  of  Last  Problem.  Symmetrical  Arch- 
Rib  of  Fixed  Ends. — As  an  example  take  a  span  of  the  St. 
Louis  Bridge  (assuming  /constant)  with  "  live  load"  cov- 
ering the  half  span  on  the  left,  Fig.  430,  where  the  vertical 


FIG.  430. 


scale  is  much  exaggerated  for  the  sake  of  distinctness*. 
Divide  into  eight  equal  Js's.  (In  an  actual  example  sixteen 
or  twenty  should  be  taken.)  Draw  a  vertical  through  the 

*  Each  arch-rib  of  the  St.  Louis  bridge  is  a  built  up  or  trussed  rib  of  steel  about  B3C 
ft.  spau  and  52  ft  rise,  in  the  form  of  a  segment  of  a  circle  .  Its  moment  of  inertia, 
however,  is  not  strictly  constant,  the  portions  near  each  pier,  of  a  length  equal  to  one 
twelfth  of  the  span,  having  a  value  of  /one-half  greater  than  that  of  the  remainder  of 
the  arc. 


468  MECHANICS    OF    ENGINEERING. 

middle  of  each  As.     P^ ,  etc.,  are  the  loads  coming  upon 
the  respective  zte's. 

First,  to  locate  vm,  by  eq.  (6) ;  from  symmetry  it  must 
be  horizontal.  Draw  a  trial  vm  (not  rhown  in  the  figure), 
and  if  the  (+  3r)'s  exceed  the  (—  z')'s  by  an  amount  «0',  the 

true  vm  will  lie  a  height  —  «0'  above  the  trial  vm  (or  below, 

97/ 

if  vice  versa) ;  n  =  the  number  of  zfs's. 

Now  lay  off  the  load-line  on  the  right  (to  scale), 
take  any  convenient  trial  pole  0'"  and  draw  a  correspond- 
ing trial  equil.  polygon  F"'G'".  In  F"'G'",  by  §377, 
locate  a  straight  line  v'"m'"  so  as  to  make  ^(3"')  =  0  and 
2*(xz"')  =  0  (see  Note  (b)  of  §  377). 

[We  might  now  redraw  F"' '  G-'"  in  such  a  way  as  to  bring 
v"'m'ff  into  a  horizontal  position,  thus :  first  determine  a 
point  n'"  on  the  load-line  by  drawing  O'"n'"  \  to  v"rm'"t 
take  a  new  pole  on  a  horizontal  through  n'",  with  the  same 
II'" ,  and  draw  a  corresponding  equil.  polygon  ;  in  the  lat- 
ter v"fm'"  would  be  horizontal.  We  might  also  shift  this 
new  trial  polygon  upward  so  as  to  make  v'"m'"  and  vm 
coincide.  It  would  satisfy  conditions  (7)  and  (8),  having 
the  same  «""s  as  the  first  trial  polygon ;  but  to  satisfy  con- 
dition (9)  it  must  have  its  «""s  altered  in  a  certain  ratio, 
which  we  must  now  find.  But  we  can  deal  with  the  individ- 
ual z""s  just  as  well  in  their  present  positions  in  Fig.  430.] 
The  points  .Z^and  L  in  vm,  vertically  over  E'"  and  L'"  in 
v'"m'",  are  now  fixed  ;  they  are  the  intersections  of  the  special 
polygon  required,  with  vm. 

The  ordinates  between  vrf'm'"  and  the  trial  equilibrium 
polygon  have  been  called  z'"  instead  of  z" ;  they  are  pro- 
portional to  the  respective  z'"s  of  the  required  special 
polygon. 

The  next  step  is  to  find  in  what  ratio  the  (2/r/)'s  need  to 
be  altered  (or  H"r  altered  in  inverse  ratio)  in  order  to  be~ 
come  the  (s")'s ;  i.e.,  in  order  to  fulfil  condition  (9),  viz. : 


AliCH-RlBS. 


.    (9) 

This  may  be  done  pre- 
cisely as  for  the  rib  with 
two  hinges,  but  the  nega- 
tive (s'")'8  must  be  prop- 
erly considered  (§  375) 
See  Fig.  431  for  the  de- 
tail. Negative  z"s  or  «""s 
point  upward. 

From  Fig.  431a 


l>»  .'.  from  symmetry 

From  Fig.  4316  we  have 


FIG.  431. 

and  from  Fig.  431c 


[The  same  pole  distance  H0  is  taken  in  all  these  construc- 
tions] .-.  Z\(yz'"}=Hn(k,+k,\ 

If,  then,  H0  (h+k)  =  2H0k  condition  (9)  is  satisfied  by  the 
z""s.  If  not,  the  true  pole  distance  for  the  special  equil. 
polygon  of  Fig.  430  will  be 

H= 

With  this  pole  distance  and  a  pole  in  the  horizontal  througi 
»'"  (Fig.  430)  the  force  diagram  may  be  completed  for  the 
required  special  polygon  ;  and  this  latter  may  be  con- 
structed as  follows  :  Beginning  at  the  point  E,  in  vm, 
through  it  draw  a  segment  ||  to  the  proper  ray  of  the  force 
diagram.  In  our  present  figure  (430)  this  "  proper  ray  " 
would  be  the  ray  joining  the  pole  with  the  point  of  meet- 
ing of  P2  and  P3  on  the  load-line.  Having  this  one  seg- 


470  MECHANICS  OP  ENGINEERING. 

ment  of  the  special  polygon  the  others  are  added  in  an 
obvious  manner,  and  thus  the  whole  polygon  completed. 
It  should  pass  through  L,  but  not  0  and  B. 

For  another  loading  a  different  special  equil.  polygon 
would  result,  and  in  each  case  we  may  obtain  the  thrust, 
shear,  and  moment  of  stress  couple  for  any  cross-section  of 
the  rib,  by  §  367.  To  the  stresses  computed  from  these, 
should  be  added  (algebraically)  those  occasioned  by  change 
of'temperature  and  by  shortening  of  the  rib  as  occasioned 
by  the  thrusts  along  the  rib.  These  "  temperature 
stresses,"  and  stresses  due  to  rib-shortening,  will  be  con- 
sidered in  a  subsequent  paragraph.  They  have  no  exist- 
ence for  an  arch-rib  of  three  hinges. 

[NOTE. — If  the  moment  of  inertia  of  the  rib  is  variable 


z" 


we  put  —  for  2'  and  —  for  z"  in  equation  (6),  (7),  (8),  and 
n  n 

(9),  n  having  the  meaning  given  in  the  Note  in  §  379  a, 
which  see ;  and  proceed  accordingly]. 


381a.  Exaggeration  of  Vertical  Dimensions  of  Both  Space  and 
Force  Diagrams. — In  case,  as  often  happens,  the  axis  of  the 
given  rib  is  quite  a  flat  curve,  it  is  more  accurate  (for  find- 
ing M)  to  proceed  as  follows  : 

After  drawing  the  curve  in  its  true  proportions  and  pass- 
ng  a  vertical  through  the  middle  of  each  of  the  equal 
zfs's,  compute  the  ordinate  (y)  of  each  of  these  middle  points 
from  the  equation  of  the  curve,  and  multiply  each  y  by 
four  (say).  These  quadruple  ordinates  are  then  laid  off 
from  the  span  upward,  each  in  its  proper  vertical.  Also 
multiply  each  load,  of  the  given  loading,  by  four,  and  then 
with  these  quadruple  loads  and  quadruple  ordinates,  and 
the  upper  extremities  of  the  latter  as  points  in  an  exagge- 
rated rib-curve,  proceed  to  construct  a  special  equilibrium 
polygon,  and  the  corresponding  force  diagram  by  the 
proper  method  (  for  Class  /?,  C,  or  D,  as  the  case  may  be) 
for  this  exaggerated  rib -curve. 

The  moment,  Hz,  thus  found  for  any  section  of  the  ex- 


ARCH-RIBS.  471 

aggerated  rib-curve,  is  to  be  divided  by  four  to  obtain  the 
moment  in  the  real  rib,  in  the  same  vertical  line.  To  find 
the  thrust  and  shear,  however,  for  sections  of  the  real  rib, 
besides  employing  tangents  and  normals  of  the  real  rib  we 
must  draw,  and  use,  another  force  diagram,  obtained  from 
the  one  already  drawn  (for  the  exaggerated  rib)  by  re- 
ducing its  vertical  dimensions  (only),  in  the  ratio  of  four 
to  one.  [Of  course,  any  other  convenient  number  besides 
four,  may  be  adopted  throughout.] 

382.  Stress  Diagrams. — Take  an  arch-rib  of  Class  D,  §  378, 
i.e.,  of  fixed  ends,  and  suppose  that  for  a  given  loading  (in- 
cluding its  own  weight)  the  special 
equil.  polygon  and  its  force  diagram 
have  been  drawn  [§  381].  It  is  re- 
quired to  indicate  graphically  the 
variation  of  the  three  stress-elements 
for  any  section  of  the  rib,  viz.,  the 
thrust,  shear,  and  mom.  of  stress- 
couple.  /  is  constant.  If  at  any 
point  m  of  the  rib  a  section  is  made,  then  the  stresses  ia 
that  section  are  classified  into  three  sets  (Fig.  432).  (See 
§§  295  and  367)  and  from  §  367  eq.  (3)  we  see  that  the  ver- 
tical intercepts  between  the  rib  and  the  special  equil. 
polygon  being  proportional  to  the  products  Hz  or 
moments  of  the  stress-couples  in  the  corresponding  sec- 
tions form  a  moment  diagram,  on  inspection  of  which  we 

can  trace  the  change    in  this  moment,     Hz  =  £?-  ,      and 

hence  the  variation  of  the  stress  per  square  inch,  p2,  (as 
due  to  stress  couple  alone)  in  the  outermost  fibre  of  any 
section  (tension  or  compression)  at  distance  e  from  the 
gravity  axis  of  the  section),  from  section  to  section  along 
the  rib. 

By  drawing  through  0  lines  On'  and  Of  parallel  re- 
spectively to  the  tangent  and  normal  at  any  point  m  of  the 
rib  axis  [see  Fig.  433]  and  projecting  upon  them,  in  turn, 
the  proper  ray  (R3  in  Fig.  433)  (see  eqs.  1  and  2  of  §  367) 


472 


MECHANICS    OF    ENGINEERING. 


we  obtain  the  values  of  the  thrust  and  shear  for  the  sec- 
tion at  m.  When  found  in  this  way  for  a  number  of  points 
along  the  rib  their  values  may  be  laid  off  as  vertical  lines 
from  a  horizontal  axis,  in  the  verticals  containing  the  re- 
spective points,  and  thus  a  thrust  diagram  and  a  shear  dia- 
gram may  be  formed,  as  constructed  in  Fig.  433.  Notice 
that  where  the  moment  is  a  maximum  or  minimum  the 
shear  changes  sign  (compare  §  240),  either  gradually  or 


FIG.  433. 


suddenly,  according  as  the  max.  or  min.  occurs  between 
two  loads  or  in  passing  a  load ;  see  m',  e.  g. 

Also  it  is  evident,  from  the  geometrical  relations  involv- 
ed, that  at  those  points  of  the  rib  where  the  tangent-line 
is  parallel  to  the  "  proper  ray  "  of  the  force  diagram,  the 
thrust  is  a  maximum  (a  local  maximum)  the  moment  (of 


ARCH   RIBS  478 

stress  couple)  is  either  a  maximum  or  a  minimum  and  the 
shear  is  zero. 

From  the  moment,  Hx  =  &*,  p*  -  — 

e  1 

may  be  computed.   From  the  thrust  =  Fpit  PI  = rus  ,  (F 

F 

=  area  of  cross-section)  may  be  computed.  Hence  the 
greatest  compression  per  sq.  inch  (/>!+#>)  may  be  found  in 
each  section.  A  separate  stress-diagram  might  be  con- 
structed for  this  quantity  (p\-\-p^  Its  max.  value  (after 
adding  the  stress  due  to  change  of  temperature,  or  to  rib- 
shortening,  for  ribs  of  less  than  three  hinges),  wherever  it 
occurs  in  the  rib,  must  be  made  safe  by  proper  designing 
of  the  rib.  The  maximum  shear  Jm  can  be  used  as  in  §256 
to  determine  thickness  of  web,  if  the  section  is  I-shaped, 
or  box-shaped.  See  §  295. 


383.  Temperature  Stresses. — In  an  ordinary  bridge  truss 
and  straight  horizontal  girders,  free  to  expand  or  contract 
longitudinally,  and  in  Classes  A  and  B  of  §  378  of  arch- 
ribs,  there  are  no  stresses  induced  by  change  of  tempera- 
ture ;    for   the  form  of   the   beam   or   truss   is  under   no 
constraint  from  the  manner  of  support ;  but  with  the  arch- 
rib  of  two  hinges  (hinged  ends,  Class  C)  and  of  fixed  ends 
(Class  D)  having  immovable  piers  which  constrain  the  dis- 
tance between  the  two  ends  to  remain  the  same  at  all  tem- 
peratures,  stresses  called  "  temperature  stresses  "  are  in- 
duced in  the  rib  whenever  the  temperature,  t,  is  not  the 
same   as  that,  tot  when  the  rib  was  put  in  place.     These 
may  be  determined,  as  follows,  as  if  they  were  the  only 
ones,  and  then  combined,  algebraically,  with  those  due  to 
the  loading. 

384.  Temperature  Stresses  in  the  Arch-Rib  of  Hinged  Ends. — 
(Class  C,  §  378.)     Fig.  434.     Let  E  and  /be  constant,  with 


474  MECHANICS  OP  ENGINEERING. 

0,  jj  other  postulates  as  in  §  379. 

Let  t0  =  temperature  of 
erection,  and  I  =  any  other 
temperature  ;  also  let  I  = 
length  of  span  =  OB  (in- 
Flo  m  variable)  and  rt—  co-efficient 

of  linear  expansion  of  the 

material  of  the  curved  beam  or  rib  (see  §  199).  At  tempera- 
ture t  there  must  be  a  horizontal  reaction  H  at  each  hinge 
to  prevent  expansion  into  the  form  O'B  (dotted  curve), 
which  is  the  form  natural  to  the  rib  for  temperature  t  and 
without  constraint.  We  may  /.  consider  the  actual  form 
OB  as  having  resulted  from  the  unstrained  form  O'B  by 
displacing  0'  to  0,  ie.,  producing  a  horizontal  displace- 
ment O'O  =l(t-Qrj. 

But  O'O  =  Ax  (see  §§  373  and  374)  ;  (N.B.  E'a  tangent 
has  moved,  but  this  does  not  affect  Ax,  if  the  axis  X  is 
horizontal,  as  here,  coinciding  with  the  span  ;)  and  the 
ordinate  y  of  any  point  m  of  the  rib  is  identical  with  its 
z  or  intercept  between  it  and  the  spec,  equil.  polygon, 
which  here  consists  of  one  segment  only,  viz.  :  OB.  Its 
force  diagram  consists  of  a  single  ray  Oi  n'  \  see  Fig.  434 
Now  (§  373) 

4x  =  —  CMyds  ;  and  M=Hz  =  in  this  case,  Hy 
EIJo 

If.     .^     _  H    /»B2  7      j  hence  for  graphics,  and 
..it-t--as'  , 


-t0)y=H  As  J»  f     .     .     .     .      (1) 

From  eq.  (1)  we  determine  Ht  having  divided  the  rib  -curve 
into  from  twelve  to  twenty  equal  parts  each  called  As  . 

For  instance,  for  wrought  iron,  t  and  t09  being  expressed 
in  Fahrenheit  degrees,  //  =  0.0000066.  If  E  is  expressed 
in  Ibs.  per  square  inch,  all  linear  quantities  should  be  in 
inches  and  H  will  be  obtained  in  pounds. 

-To2/2  may  be  obtained  by  §  375,  or  may  be  computed.  H 
being  known,  we  find  the  moment  of  stress-couple  =  Hyt 


ARCH-RIBS. 


475 


at  any  section,  while  the  thrust  and  shear  at  that  section 
are  the  projections  of  H,  i.e.,  of  OX  upon  the  tangent  and 
normal.  The  stresses  due  to  these  may  then  be  determined 
in  any  section,  as  already  so  frequently  explained,  and 
then  combined  with  those  due  to  loading. 

385.  Temperature  Stresses  in  the  Arch-Ribs  with  Fixed  Ends.— 
See  Fig.  435.  (Same  postulates  as  to  symmetry,  E  and  1 
constant,  etc.,  as  in  §  380.)  t  and  t0  have  the  same  meaning 
as  in  §  384. 

Here,  as  before,  we 
consider  the  rib  to 
have  reached  its  ac- 
tual form  under  tern- 
perature  t  by  having 
had  its  span  forcibly 
shortened  from  the  _ 
length  natural  to 
temp,  t,  viz.:  O'B', 

to  the  actual  length  OB,  which  the  immovable  piers  compel 
it  to  assume.  But  here,  since  the  tangents  at  0  and  B  are 
to  be  the  same  in  direction  under  constraint  as  before,  the  two 
forces  Hy  representing  the  action  of  the  piers  on  the  rib, 
must  be  considered  as  acting  on  imaginary  rigid  prolonga- 
tions at  an  unknown  distance  d  above  the  span.  To  find 
H  and  d  we  need  two  equations. 

From  §  373  we  have,  since  M=Hz=H  (y—d\ 


FI0.435. 


or,  graphically,  with  equal 


•  (2) 


(3) 


Also,  since  there  has  been  no  change  in  the  angle  between 
end-tangents,  we  must  have,  from  §  374, 


476  MECHANICS   OF   ENGINEERING. 

or   for  graphics,  with  equal  A>  's,  I*y  —  nd     .     .          .     (4) 

in  which  n  denotes  the  number  of  Js's.  From  (4)  we 
determine  d,  and  then  from  (3)  can  compute  H.  Drawing 
the  horizontal  F  G,  it  is  the  special  equilibrium  polygon 
(of  but  one  segment)  and  the  moment  of  the  stress-couple 
at  any  section  =  Hz,  while  the  thrust  and  sheaT  are  the 
projections  of  H=  O^n'  on  the  tangent  and  normal  respect- 
ively of  any  point  m  of  rib. 

For  example,  in  one  span,  of  550  feet,  of  the  St.  Louis 
Bridge,  having  a  rise  of  55  feet  and  fixed  at  the  ends,  the 
force  H  of  Fig.  435  is  =  108  tons,  when  the  temperature  is 
80°  Fahr.  higher  than  the  temp,  of  erection,  and  the  en- 
forced span  is  3*^  inches  shorter  than  the  span  natural  to 
that  higher  temperature.  Evidently,  :.f  the  actual  temp- 
erature t  is  lower  than  that  tn,  of  erection,  ZTmust  act  in  a 
direction  opposite  to  that  of  Figs.  435  and  434,  and  the 
"thrust  "  in  any  section  will  be  negative,  i.e.,  a  pull. 

386.  Stresses  Due  to  Rib-Shortening  —In  §  369,  Fig.  407,  the 
shortening  of  the  element  AE  to  a  length  A'E,  due  to  the 
uniformly  distributed  thrust,  p\F,  was  neglected  as  pro- 
ducing indirectly  a  change  of  curvature  and  form  in  the 
rib  axis  ;  but  such  will  be  the  case  if  the  rib  has  less  than 
three  hinges.  This  change  in  the  length  of  the  different 
portions  of  the  rib  curve,  may  be  treated  as  if  it  were  due 
to  a  change  of  temperature.  For  example,  from  §  199  we 
see  that  a  thrust  of  50  tons  coming  upon  a  sectional  area 
of  F  =  10  sq.  inches  in  an  iron  rib,  whose  material  has  a 
modulus  of  elasticity  =  E  =  30,000,000  Ibs.  per  sq.  inch, 
and  a  coefficient  of  expansion  #  =  .0000066  per  degree 
Fahrenheit,  produces  a  shortening  equal  to  that  due  to  a 
fall  of  temperature  (t0—t)  derived  as  follows :  (See  §  199) 
(units,  inch  and  pound) 

(t  -t}=     P  100,000  g0o 

FE?i    10  x  30,000,000  x .  0000066" 
Fahrenheit. 

Practically,  then,  since  most  metal  arch  bridges  of 
glasses  C  and  D  are  rather  flat  in  curvature,  and  the  thrusts, 


AKCH-IIIBS.  477 

due  to  ordinary  modes  of  loading  do  not  vary  more  than  20 
or  80  per  cent,  from  each  other  along  the  rib,  an  imagin- 
ary fall  of  temperature  corresponding  to  an  average  thrust 
in  any  case  of  loading  may  be  made  the  basis  of  a  con- 
struction similar  to  that  in  §  384  or  §  385  (according  as  the 
ends  are  hinged,  or  fixed)  from  which  new  thrusts,  shears, 
and  stress-couple  moments,  may  be  derived  to  be  combin- 
ed with  those  previously  obtained  for  loading  and  for 
change  of  temperature. 

387.  Resume  — It  is  now  seen  how  the  stresses  per  square 
inch,  both  shearing  and  compression  (or  tension)  may  be 
obtained  in  all  parts  of  any  section  of  a  solid  arch-rib  or 
curved  beam  of  the  kinds  described,  by  combining  the  re- 
sults due  to  the  three  separate  causes,  viz.:  the  load, 
change  of  temperature,  and  rib-shortening  caused  by  the 
thrusts  due  to  the  load  (the  latter  agencies,  however,  com- 
ing into  consideration  only  in  classes  C  and  D,  see  §  378). 
That  is,  in  any  cross-section,  the  stress  in  the  outer  fibre 
is,  [letting  TV,  5Ph",  ZV",  denote  the  thrusts  due  to  the 
three  causes,  respectively,  above  mentioned ;  (Hz)',  (Hz)'\ 
(Hz)'"1,  the  moments] 

T7'  4-  T7  "      T7  '"      *>  r  -i 

=:£*_:  ±^^±e^(Hzy±(B»y'±(H*y»l    .    .     .     (1) 

Le.,  Ibs.  per  sq.  inch  compression  (if  those  units  are  used). 
The  double  signs  provide  for  the  cases 
where  the  stresses  in  the  outer  fibre,  due 
to  a  single  agency,  may  be  tensile.  Fig. 
436  shows  the  meaning  of  e  (the  same 
used  heretofore)  /is  the  moment  of  in- 
ertia of  the  section  about  the  gravity 
axis  (horizontal)  C.  F  =  area  of  cross- 
section.  [BI  =  e  ;  cross  section  symmet- 
rical about  C].  For  a  given  loading  we 
may  find  the  maximum  stress  in  a  given  rib,  or  design  the 
rib  so  that  this  maximum  stress  shall  be  safe  for  the  ma- 
terial employed.  Similarly,  the  resultant  shear  (total,  not 


478 


MECHANICS  OF  ENGINEERING. 


per  sq.  inch)  =  J'  ±  J"  ±  J'"  is  obta-ned  for  any  section 
to  compute  a  proper  thickness  of  web,  spacing  of  rivets, 
etc. 

388  The  Arch-Truss,  or  braced  arch.  An  open-work 
truss,  if  of  homogeneous  design  from  end  to  end,  may  be 
treated  as  a  beam  of  constant  section  and  constant  moment 
of  inertia,  and  if  curved,  like  the  St.  Louis  Bridge  and  the 
Coblenz  Bridge  (see  §  378,  Class  D),  may  be  treated  as  an 
arch-rib.*  The  moment  of  inertia  may  be  taken  as 


where  F^  is  the  sectional  area  of  one  of  the  pieces  I  to  the 
curved  axis  midway  between  them,  Fig.  437,  and  h  =  dis- 
tance between  them. 


FIG.  438. 


FIG.  437. 


Treating  this  curved  axis  as  an  arch-rib,  in  the  usual 
way  (see  preceding  articles),  we  obtain  the  spec,  equil.  pol. 
and  its  force  diagram  for  given  loading.  Any  plane  "J  to 
the  rib -axis,  where  it  crosses  the  middle  m  of  a  "  web- 
member,"  cuts  three  pieces,  A,  B  and  C,  the  total  com- 

*The  St  Louis  Bridge  is  not  strictly  of  constant  moment  of  inertia,  being  somewhat 
strengthened  near  each  pier 


ARCH-RIBS. 


479 


pressioas  (or  tensions)  in  which  are  thus  found  :  For  the 
point  m,  of  rib-axis,  there  is  a  certain  moment  =  Hz,  a 
thrust  =  Th9  and  a  shear  =  J,  obtained  as  previously  ex- 
plained. We  may  then  write  Psin/3  =  J  .  .  .  .  (1) 
and  thus  determine  whether  P  is  a  tension  or  compres- 
sion ;  then  putting  P'+P"  ±  P  cos  ft  -  T]} 2 

(in  which  P  is  taken  with  a  plus  sign  if  a  compression,  and 
minus  if  tension),  and 


(3) 


we  compute  P  and  P",  which  are  assumed  to  be  both  com- 
pressions here.  ft  is  the  angle  between  the  web  member 
and  the  tangent  to  rib-axis  at  ra,  the  middle  of  the  piece. 
See  Fig.  406,  as  an  explanation  of  the  method  just 
adopted. 


HORIZONTAL,  STRAIGHT  GIRDERS. 

389.  Ends  Free  to  Turn. — This  corresponds  to  an  arch- 
rib  with  hinged  ends,  but  it  must  be  understood  that  there 
is  no  hindrance  to  horizontal  motion.  (Fig.  439.)  In 


Fio   439. 


treating  a  straight  beam,  slightly  bent  under  vertical  forces 
only  (as  in  this  case  with  no  horizontal  constraint),  as  a 


480  MECHANICS   OF   ENGINEERING. 

particular  case  of  an  arch-rib,  it  is  evident  that  since  the 
pole  distance  must  be  zero,  the  special  equil.  polygon  will 
have  all  its  segments  vertical,  and  the  corresponding  fore, 
diagram  reduces  to  a  single  vertical  line  (the  load  line,. 
The  first  and  last  segments  must  pass  through  A  and  .&. 
(points  of  no  moment)  respectively,  but  being  vertical  will 
not  intersect  PA  and  P2 ;  i.e.,  the  remainder  of  the  special 
equilibrium  polygon  lies  at  an  infinite  distance  above  the 
span  AB.  Hence  the  actual  spec,  equil.  pol.  is  useless. 

However,  knowing  that  the  shear,  «7,  and  the  moment 
-M"(of  stress  couple)  are  the  only  quantities  pertaining  to 
any  section  m  (Fig.  439)  which  we  wish  to  determine  (since 
there  is  no  thrust  along  the  beam),  and  knowing  that  an 
imaginary  force  H'1 ',  applied  horizontally  at  each  end  of  the 
beam,  would  have  no  influence  in  determining  the  shear 
and  moment  at  m  as  due  to  the  new  system  of  forces,  we 
may  therefore  obtain  the  shears  and  moments  graphically 
from  this  new  system  (viz.:  the  loads  Pl}  etc.,  the  vertical 
reactions  V  and  VM  and  the  two  equal  and  opposite  H"'&). 
[Evidently,  since  H'  has  no  moment  about  the  neutral 
axis  (or  gravity  axis  here),  of  m,  the  moment  at  m  will  be 
unaffected  by  it ;  and  since  H"  has  no  component  "|  to  the 
beam  at  m,  the  shear  at  m  is  the  same  in  the  new  system 
of  forces,  as  in  the  old,  before  the  introduction  of  the 
IT*] 

Hence,  lay  off  the  load-line  1  .  .  2.  .  3,  Fig.  439,  and  con- 
struct an  equil.  polyg.  which  shall  pass  through  A  and  B 
and  have  any  convenient  arbitrary  H"  (force)  as  a  pole 
distance.  This  is  dona  by  first  determining  ri  on  the  load- 
line,  using  the  auxiliary  polygon  A'a'B,  to  a  pole  0'  (arbi- 
trary"), and  drawing  O'n'  ||  to  A'B'.  Taking  O1'  on  a  hori- 
zontal through  ri,  making  0''n'=H",  we  complete  the 
force  diagram,  and  equil.  pol.  AaB.  Then,  z  being  the  ver- 
tical intercept  between  m  and  the  equil.  polygon,  we  have: 
Moiueat  at  m—M=H"z  (or=H'zf  also),  and  shear  at  m,  or 
•/,~2  .  .  n'9  i.e.,  =  projection  of  the  proper  ray  R2,  or 
0"  .  .  2,  upon  the  vertical  through  m.  Similarly  we  ob- 
tain M  and  J  at  any  other  sect:  rn  for  the  given  load.  (See 


AECH-RIBS ;   SPECIAL   CASE  ;   STRAIGHT. 


481 


§§  329,  337  and  367).     The  moment  of  inertia  need  not  be 
constant  in  this  case. 

390,  Straight  Horizontal  Prismatic  Girder  of  Fixed  Ends  at  Same 
Level. — No  horizontal  constraint,  hence  no  thrust.  I  con- 
stant. Ends  at  same  level,  with  end-tangents  horizontal. 
We  may  consider  the  whole  beam  free  (cutting  close  to  thet 
walls)  putting  in  the  unknown  upward  shears  J0  and  «/m 


and  the  two  stress  couples  of  unknown  moments  Mn  and 
Mn  at  these  end  sections.  Also,  as  in  §  388,  an  arbitrary 
H"  horizontal  and  in  line  of  beam  at  each  extremity.  Now 
(See  Fig.  33)  the  couple  at  0  and  the  force  H"  are  equiv- 
alent to  a  single  horizontal  H"  at  an  unknown  vertical  dis- 
tance c  below  0 ;  similarly  at  the  right  hand  end.  The 
special  polygon  FG  is  to  be  determined  for  this  new  sys- 
tem, since  the  moment  and  shear  will  be  the  same  at  any 
section  under  this  new  system  as  under  the  real  system. 
The  conditions  for  determining  it  are  as  follows  :  Since 
the  end-tangents  are  fixed,  -M4s=0  .\  ™z^s=Q  and  since 


482  MECHANICS   OF   ENGINEERING. 

O's  displacement  relatively  to  .B's  tangent  is  zero  we  have 
SMx Js  =0  .-.  2H"sx48f=;Q  >'•  2™As  =0.  See  §  374.  Hence 
for  Equal  Js's,  2'(*)=0  and  2(xz)=Q.  Now  for  any  pole  0'" 
draw  an  equil.  poL  F'"G'"  and  in  it  (by  §  377;  see  Note) 
locate  t;'"m'"  so  as  to  make  J(s'")  =  0  and  2(x»'")=Q. 
Draw  verticals  through  the  intersections  E'"  and  .Z/",  to 
determine  E  and  Z  on  the  beam,  these  are  the  points  of 
inflection  (i.e.,  of  zero  moment),  and  are  points  in  the  re- 
quired special  polygon  FG. 

Draw  0"V'  1  toV'm'"  to  fix  n".  Take  a  pole  0"  on 
the  horizontal  through  n",  making  tXV=lf"  (arbitrary), 
draw  the  force  diagram  0"  1234  and  a  corresponding 
equilibrium  polygon  beginning  at  E.  It  should  cut  L, 
and  will  fulfil  the  two  requirments  2*(z)=Q  and  J™(a?«)=0, 
with  reference  to  the  axis  of  the  beam  O'B'.  The  moment  of 
the  stress-couple  at  any  section  m  will  be  M—H"z,  and  the 
shear  J  =  the  projection  of  the  "  proper  ray  "  of  the  force 
diagram  0"  .  .  1,  2,  etc.,  upon  the  vertical  (not  in  the  trial 
diagram  0'".  .  1,  2,  etc.).  As  far  as  the  moment  is  concern- 
ed the  trial  polygon  F'n '(?'"  will  serve  as  well  as  the  special 
polygon  FG  ;  i.e.,  M=H"'z'"  as  well  as  H"z,  H"f  being  the 
pole-distance  of  0'"  ;  but  for  the  shear  we  must  use  the 
rays  of  the  final  and  not  the  trial  diagram. 

The  peculiarity  of  this  treatment  of  straight  beams, 
considered  as  a  particular  case  of  curved  beams,  consists 
in  the  substitution  of  an  imaginary  system  of  forces  in- 
volving the  two  equal  and  opposite,  and  arbitrary  ZTs,  for 
the  real  system  in  which  there  is  no  horizontal  force  and 
consequently  no  "  special  equilibrium  polygon,"  and  thus 
determining  all  that  is  desired,  i.e.,  the  moment  and  shear 
at  any  section. 

In  the  polygon  FG  the  student  will  recognize  the  "  mo- 
ment-diagram "  of  the  problems  in  Chaps.  Ill  and  IV. 

He  will  also  see  why  the  shear  is  proportional  to  the 

slope  — —  of  the  moment  curve   in  those  chapters.     For 
dx 

example,  the  "  slope  "  of  the  second  segment  of  the  poly- 
gon FG,  that  segment  being  ||  to  0"  2,  is 


ARCH-BIBS;   SPECIAL   CASE;    STRAIGHT.  483 

tang,  of  angle  20'V'=^'^07V7=shear  -*-  H" 

and  similarly  for  any  other  segment ;  i.e.,  the  tangent  oi 
the  inclination  of  the  "  moment  curve,"  or  line,  is  propor-. 
tional  to  the  shear. 

It  is  also  interesting  to  notice  with  the  present  problem 
of  a  straight  beam,  that  in  the  conditions 

J(«Js)=0  and  2(z4s)x=0, 

t>r  locating  the  polygon  FG,  each  ds  is  T  to  its  «,  and 

^hat  consequently  each  zAs  is  the  area  of  a  small  vertical 
strip  of  area  between  the  beam  and  the  polygon,  and 
(z4s)x  is  the  "  moment"  of  this  strip  of  area,  about  0'  the 
origin  of  x.  Hence  these  conditions  imply  ;  first,  that  the 
area  EWL  between  the  polygon  and  the  axis  of  the  beam 
on  one  side  is  equal  to  that  (O'FE+LB'G)  on  the  other 
Side,  and,  secondly,  that  the  centre  of  gravity  of  EWL  lies 
in  the  same  vertical  as  that  of  O'FE  and  LB'  G  combined. 
Another  way  of  stating  the  same  thing  is  that,  if  we  join 
FG,  the  area  of  the  trapezoid  FO'B'  G  is  equal  to  that  of  the 
figure  FEWLG,  and  their  centres  8f  gravity  lie  in  the  same 
vertical.  A  corresponding  statement  may  be  made  (if  we  join 
F'"G'"]  for  the  trapezoid  F"'v'"m'"G'"  and  figure 


484 


MECHANICS   Oi    ENGINEERING. 


CHAPTER  XII. 


OF  CONTINUOUS  GIRDERS. 


[MAINLY  DUE  TO  PROF.  MOHR.  OF  AIX-LA-CHAPELLE  ] 

391.  The  Elastic  Curve  of  a  Horizontal  Loaded  Beam,  Homoge- 
neons  and  Originally  Straight  and  Prismatic,  is  an  Equilibrium 
Polygon,  whose  "load-line"  it  vertical 
and  consists  of  the  successive  products 
Mdx  [treated  as  if  they  were  loads 
each  applied  through  the  middle  of 
its  proper  dx~],  and  whose  "pole  dis- 
tance" is  EL  Fig  441  (exaggerated). 

Let  AO  and  00  be  any  two  con- 
secutive equal  elements  of  a  very 
flat  elastic  curve  (as  above  described). 
Prolong  AO  to  cut  NO.  Then  from 
§  231,  eq.  (7),  we  have 


dx 


(1) 


where  cP^ZJtfjand,  hence,  if  a  triangle  (Fig.  442)  O'D'C', 
be  formed  with  O'D'  \  to  OD,  O'C'  \\  to  00,  and  D'&  ver- 
tical, while  its  (horizontal)  altitude  O'n  is  made  equal,  by 
scale,  to  El  of  the  beam,  then  from  the  similarity  of  the 
triangle  OZ>(7and  O'D'C'  and  the  proportion  ineq.  (l)we 
see  that  D'C'  must  represent  the  product  Mdx  on  the  same 
scale  by  which  O'n  represents  EL  M  is  the  moment  of 


CONTINUOUS  GIKDER  BY  GRAPHICS.  483 

the  stress-couple  at  the  section  whose  neutral  axis  ib  pro- 
jected in  0. 

Similarly,  if  M'  is  the  moment  of  the  stress -couple  at 
0,  and  we  draw  O'F',  II  to  CF,  G'F'  must  represent  M'di 
(on  same  scale).  It  is  therefore  apparent  that  the  line 
AOCF  bears  to  the  figure  O'D'C'F'  the  same  relation 
which  an  equilibrium  polygon  (for  vertical  forces)  does  to 
its  force  diagram,  the  "  loads  "  of  the  force-diagram  being 
the  successive  values  of  Mdx  laid  off  to  scale,  while  its 
"  pole-distance  "  is  El  laid  off  on  the  same  scale.  [As  if 
Mdx  and  M'dx  were  loads  suspended  at  0  and  C  respec- 
tively.] 

Practically,  since  any  actual  elastic  curve  is  very  flat, 
and  since  a  change  of  pole-distance  will  change  all  verti- 
cal dimensions  of  the  equilibrium  polygon  in  an  inverse 
equal  ratio,  we  may  exaggerate  the  vertical  dimensions  of 
the  elastic  curve  by  choosing  a  pole  distance  smaller  than 
El  in  any  convenient  ratio,  n.  Any  deflection  in  the  elas- 
tic curve  thus  obtained  will  be  greater  than  its  true  value 
in  the  same  ratio  n. 

Graphically,  in  order  to  draw  exaggerated  elastic  curves 
according  to  this  principle,  we  obtain  approximate  results 
by  dividing  the  length  of  the  beam  into  a  number  of  equal 
Jsc's,  draw  verticals  through  the  middles  of  the  J#'s  as 
"  force-verticals, "and  lay  off  as  a  "  load-line  "  to  any  con- 
venient scale  the  corresponding  values  of  MAx  in  their 
proper  order. 

The  quality  of  the  product  M  Ax  is  evidently  (length)2  x 
force,  and  with  the  foot  and  pound  as  units  such  a  product 
may  be  called  so  many  (sq.  ft.)  (Ibs.).  It  will  be  noticed 
that  these  products  (MAx)  are  proportional  to,  and  maybe 
represented  by,  the  areas  of  the  corresponding  vertical 
strips  of  the  "  moment-diagram  "  proper  to  the  case  in 
hand,  These  strips  together  make  up  the  " moment-area" 
as  it  may  be  called,  lying  between  the  moment  curve  and 
its  horizontal  axis  (which  is  the  axis  of  the  beam  itself, 
according  to  §§  389  and  390). 


486  MECHANICS    OF    ENGINEERING. 

392.  Mohr's  Theorem. — The  principle  of  the  previous 
paragraph  may  therefore  be  enunciated  as  follows  :  That 
just  as  the  moment  curve  (of  a  straight  prismatic  horizontal 
beam)  between  tivo  consecutive  supports  is  an  equilibrium  poly- 
gon/or the  loading  between  those  supports,  so  also  is  the  elastic 
curve  itself  an  equilibrium  polygon  for  the  "  moment-area  "  con- 
sidered as  a  loading. 

In  dealing  with  the  moment-curve  of  a  single  span  the 
pole  distance  is  arbitrary  (§§  389  and  390),  but  the  position 
of  the  pole  relatively  to  the  load  line  in  other  respects,  and 
the  location  of  the  moment-curve  (equil.-pol.)  relatively  to 
the  beam  (considered  to  be  still  straight  for  this  purpose), 
depend  on  whether  the  beam  simply  rests  on  the  two  sup- 
ports, without  projecting  beyond ;  or  is  built  in,  and  at 
what  angles  ;  or  as  with  a  continuous  girder,  on  the  inclina- 
tion of  the  tangent-lines  at  the  supports,  as  influenced  by 
the  presence  of  loads  on  all  the  spans,  and  on  whether  all 
supports  are  on  the  same  level  or  not. 

For  example,  in  §  389,  for  a  single  span,  the  ends  of 
beam  being  simply  supported  without  overhanging,  the 
pole  0"  must  be  on  a  horizontal  through  n',  and  the  mo- 
ment curve  must  pass  through  the  extremities  A  and  B  of 
the  beam,  thus  giving  a  "  moment-area  "  lying  entirely  on 
one  side  of  the  beam  (or  axis  from  which  the  moment  or- 
dinates,  z,  are  to  be  measured)  ;  whereas,  in  §  390,  also  a 
single  span,  where  the  ends  of  the  beam  are  built  in  hori- 
zontally and  at  the  same  level,  the  pole  must  be  taken  on 
the  horizontal  through  n"9  and  the  moment-curve  FEWLG 
must  intersect  the  beam  in  the  points  E  and  L  (E,  L,  and 
n"  being  found  as  prescribed  in  that  problem),  and  thus 
lies  partly  above  and  partly  below  the  beam.  It  will  be 
necessary,  later,  to  distinguish  the  upper  and  lower  parts 
of  the  moment-area  as  positive  and  negative. 

In  drawing  the  equilibrium  polygon  which  constitutes 
the  actual  elastic  curve,  however,  and  hence  making  use  of 
the  successive  small  ^vertical  strips  of  the  moment-area, 
(when  found)  as  if  they  were  loads,  to  form  a  load-line  ac- 
cording to  a  convenient  scale,  the  pole  distance  is  not  ar- 


CONTINUOUS  GIKDEE,  BY  GRAPHICS. 


487 


bitrary  but  must  be  =  El  on  the  same  scale.  Still,  since 
for  convenience  we  must  always  greatly  exaggerate  the 
vertical  scale  of  the  elastic  curve,  we  may  make  the  pole 
distance  =  El-r-n  and  thus  obtain  an  elastic  curve  whose 
vertical  dimensions  are  n  times  as  large  as  those  of  the  real 
curve ;  while  the  position  of  the  pole  will  depend  on  the 
direction  of  the  tangent  lines  at  the  extremities  of  the  span. 
An  example  will  now  be  given. 

* 

393  Example  of  an  Elastic  Curve  (Beam  Prismatic)  Drawn  as 
an  Equilibrium  Polygon  Supporting  the  Moment- Area  as  Loading. 
—Let  the  beam  be  simply  supported  at  its  extremities  (at 
the  same  level),  and  bear  a  single  eccentric  load  P,  Fig. 
443,  its  own  weight  being  neglected.  The  moment-area 


consists  of  a  triangle  AGE  [see  first  part  of  §260,  or  use 
the  graphic  method  of  §  389,  thus  utilizing  a  force  diagram 
012.],  its  altitude  being  the  moment  represented  by  the 

Pll 
ordinate  CD  and  having  a  value  —  — .     Hence  the  total 


moment-area  =  }&  base  AB  X  mom.  CD 


i.e.,  = 


488  MECHANICS   OF   ENGINEERING. 

Divide  AB  into  (say)  eight  equal  Jx's  (eight  are  rather 
few  in  practice ;  sixteen  or  twenty  would  be  better)  and 
draw  a  vertical  through  the  middle  of  each.  Note  the 
portion  of  each  of  these  vertical  intercepts  between  the 
axis  of  the  beam  and  the  moment-curve  ACS.  The  pro- 
ducts M  Ax  for  the  different  subdivisions  are  proportional 
to  these  intercepts,  since  all  the  ^ce's  are  equal,  and  are  the 
respective  moment-areas  of  the  ^to's. 

Treating  these  products  as  if  they  were  loads,  we  lay  off 
the  corresponding  intercepts  (or  their  halves,  or  quarters, 
or  other  convenient  fractional  part  or  multiple),  from  E 
downwards  to  form  a  vertical  "  load-line,"  beginning  with 
fche  left-hand  intercept  and  continuing  in  proper  order. 

As  to  what  scale  this  implies,  we  determine  by  dividing 
the  total  moment-area  thus  laid  off,  viz.  :  y2  Plh,  say  in 
(sq.  in.)  (Ibs.),  by  the  length  of  EF  in  inches,  thus  obtain- 
ing the  number  of  (sq.  in.)  (Ibs.)  which  each  linear  inch  of 
paper  represents. 

On  this  scale  the  number  of  inches  of  paper  required  to 
represent  the  El  of  the  beam  is  so  enormous,  that  in  its 
stead  we  use  the  nth  portion,  n  being  an  arbitrary  abstract 
number  of  such  magnitude  as  to  make  El  -f-  n  a  con- 
venient pole-distance,  TS. 

The  proper  position  of  the  pole  0'  on  the  vertical  TWt 
is  fixed  by  the  fact  that  the  elastic  curve,  beginning  at  A, 
must  terminate  in  B,  at  the  same  level  as  A.  Hence, 
assuming  any  trial  pole  as  0",  and  drawing  rays  in  the 
usual  manner  (except  that,  as  henceforth,  the  pole  is  taken 
on  the  right  of  the  "  load-line,"  instead  of  on  the  left,  so 
as  to  make  the  resulting  equilibrium  polygon  correspond 
in  direction  of  curvature  to  the  actual  elastic  curve),  we 
draw  the  corresponding  equilibrium  polygon  A"B".  De- 
termining n'  by  drawing  through  0"  a  line  II  to  the  right 
line  A"B",  we  draw  a  horizontal  through  n'  to  intersect 
TWiu  0',  the  required  pole. 

With  0'  as  pole  a  new  equilibrium  polygon  begun  at  A* 
will  terminate  in  B'  and  its  vertical  ordinates  will  be  n 
times  as  great  as  those  of  corresponding  points  on  the 


CONTINUOUS  GrKDER  BY  GRAPHICS.  489 

actual  elastic  curve  AB.  The  same  relation  holds  between 
the  tangents  of  the  angle  of  inclination  to  the  horizontal 
at  corresponding  points  (i.e.,  those  in  same  vertical)of  the 
two  curves. 

394,  Numerical  Case  of  Foregoing  Example, — With  the  inch 
and  pound  as  units,  let  P  =  120  Ibs.,  ^  =  40  in.,  12  =  80  in. 
while  the  prismatic  beam  is  of  timber  having  a  modulus 
of  elasticity  E  =  2,000,000  Ibs.  per  sq.  inch,  d&id  is  rectan- 
gular in  section,  being  2  in.  wide  and  4  in.  high,  so  that 
(its  width  being  placed  horizontally)  the  moment  of  iner- 
tia of  the  section  is  /  =  1/Mbhz  =  yi2x2x64  =  10^  bi- 
quadratic inches  (§  90.)  Eequired  the  maximum  deflection. 
Adopting  1:20  as  the  scale  for  distances  (i.e.,  one  linear 
inch  of  paper  to  twenty  inches  of  actual  distance)  we  make 
the  horizontal  AB  6  in.  long,  Fig.  443,  and  AD  2  in.,  tak- 
ing a  point  G  at  convenience  in  the  vertical  through  Z>, 
and  joining  AC  and  CB,  thus  determining  the  moment- 
diagram  for  this  case.  [As  to  what  pole  distance,  H,  is 
implied  in  this  selection  of  (7,  is  immaterial  in  this  simple 
case  of  a  single  load ;  hence  we  do  not  draw  the  corre- 
sponding force-diagram  at  all.]  We  divide  AB  into  eight 
equal  parts  and  draw  a  vertical  through  the  middle  of 
each.  The  intercepts,  in  these  verticals,  between  AB  and 
the  broken  liuQ'ACB  we  lay  off  from  .#  toward  F  as  pre- 
scribed in  §  393.  (By  taking  DC  small  enough  the  line  EF 
will  not  be  inconveniently  long.) 

Suppose  this  length  EF  measures  6.4  inches  on  the 
paper  (as  in  the  actual  draft  by  the  writer).  Since  it  rep- 
resents a  moment-area  of 

y2Pl^=  1^x120x40x80=192,000  (sq.  in.)  (Ibs.),  the  scale 
of  our  "moment-area-diagram,"  as  we  may  call  it,  must  be 
192,000-J-6.4=30,000  (sq.  in.)  (Ibs.)  per  linear  inch  of  paper. 

Now  #7=21,333,333  (sq.  in.)  (Ibs.),  which  on  the  above 
scale  would  be  represented  by  711  linear  inches  of  paper. 
With  w=100,  however,  we  lay  off  ST=EI+n=7.1l  inches 
of  paper  as  a  pole  distance,  and  with  a  trial  pole  0"  in 


490  MECHANICS   OF    ENGINEERING. 

the  vertical  TW  draw  the  trial  equilibrium  polygon  or 
elastic  curve  A"B",  and  with  it  determine  n',  then  the  final 
polygon  A'B'  as  already  prescribed.  In  A'B'  we  find  the 
greatest  ordinate,  NK,  to  measure  0.88  inches  of  paper, 
which  represents  an  actual  distance  of  0.88  x  20  =  17.6 
inches  But  the  vertical  dimensions  of  the  exaggerated 
elastic  curve  A'B'  are  n=WO  times  as  great  as  those  of  the 
actual,  hence  the  actual  max.  deflection  is  <i=17.6-^n=:0.176 
in.  [This  maximum  deflection  could  also  be  obtained  from 
the  oblique  polygon  A"B"  whose  vertical  dimensions  are 
equal  to  those  of  A'B'.  By  the  formula  of  §  235  we  ob- 
tain d= 0.174  inches.] 

395.  Direction  of  End-Tangents  of  Elastic  Curve  in  the  Foregoing 
Problem. — As  an  illustration  bearing  on  subsequent  work 
let  us  suppose  that  the  only  result  required  in  §  394  is 
tan  «0,  i.e.,  the  tangent  of  the  angle  B'A'T',  which  the 
t.-ingent-line  A '  T'  to  the  elastic  curve  at  the  extremity  A', 
Jb'ig.  443,  makes  with  the  horizontal  line  A'B',  (tan.  aQ  is 
called  the  "slope"  at  A.)  Let  £'£'  be  the  tangent-line  at 
B'.  These  two  "  end-tangents  "  are  parallel  respectively  to 
EO'  and  FO',  and  intersect  at  some  point  R.  Now  since 
A' KB'  is  an  equilibrium  polygon  sustaining  an  imaginary 
set  of  loads  represented  by  the  successive  vertical  strips  of 
the  moment-area  ACB,  the  intersection  R  must  lie  in  the 
vertical  containing  the  centre  of  gravity,  U,  of  that  mo- 
ment-area [§  336|. 

Hence,  if  the  vertical  containing  U  is  known  in  advance, 
or,  as  in  the  present  case,  is  easily  constructed  without 
making  the  strip-subdivision  of  §  394,  we  may  determine 
the  end-tangents  very  briefly  by  considering  the  whole 
moment-area,  M.A.,  (considered  as  a  load)  applied  in  the 
vertical  through  U,  as  follows  : 

Since  AGB  is  a  triangle,  we  find  U  by  bisecting  AB  in 
X,  joining  CX,  and  making  XU  =•  %  XC,  and  then  draw  a 
vertical  through  U.  Laying  off  EF=6A  inches  [so  as  to 
represent  a  moment-area  of  192,000  (sq.  in.)  (Ibs.)  on  a 
scale  of  30,000  (sq.  in.)  (Ibs.)  per  linear  inch  of  paper], 


CONTINUOUS    GrlKDER  BY   GRAPHICS.  491 

and  making  ST=7.ll  inches  as  before,  we  assume  a  trial 
pole  0"  on  TW,  draw  the  two  rays  0"E  and  Q"F,  construct 
the  corresponding  trial  polygon  of  two  segments  A"R"B", 
for  the  purpose  of  finding  n'.  With  a  pole  0'  on  TW&nd 
on  a  level  with  n'  we  draw  the  two  rays  O'E  and  O'F,  and 
the  corresponding  segments  A'R%  and  RB'.  (JB'  should  be 
on  a  level  with  A',  as  a  check.)  These  two  segments  are 
the  end-tangents  required. 
We  have,  therefore, 


tan  «= 


In  the  present  numerical  problem  we  find  B'T'  to  mea- 
sure 3  in.  of  paper,  i.e»,  60  in.  of  actual  distance  for  the 
exagg.  elastic  curve,  and  therefore  0.60  in.  in  the  real  elas- 
tic curve  (with  n  =  100) 


o 

120  in. 

It  is  now  evident  that  the  position  and  direction  of  the 
end-tangents  of  the  elastic  curve  lying  between  any  two  sup- 
ports are  independent  of  the  mode  of  distribution  of  the 
moment-area  so  long  as  the  amount  of  that  moment-area  and 
the  position  of  its  centre  of  gravity  remain  unchanged.  This 
relation  is  to  be  of  great  service. 

396.  Re-Arrangement  of  the  Moment-  Area.  —  As  another  illus- 
tration conducing  to  clearness  in  later  constructions,  let 
us  determine  by  still  another  method  the  end-tangents  of 
the  beam  of  §§  394  and  395.  See  Fig.  444.  As  already 
seen,  their  location  is  independent  of  the  arrangement  of 
the  moment-area  between.  Let  us  re-arrange  this  moment- 
area,  viz.,  the  triangle  AGB,  in  the  following  manner  : 

By  drawing  AX  parallel  to  BC,  and  prolonging  BCio  V 
in  the  vertical  through  A,  we  may  consider  the  original 
moment-area  ACS  to  be  compounded  of  the  positive  mom.- 
area  VBXA,  a  parallelogram,  with  its  gravity-vertical 
passing  through  /),  the  middle  of  the  span  ;  of  the  negative 
mom.  -area  VCA,  a  triangle  whose  gravity-vertical  passes 


492 


MECHANICS    OF    ENGINEERING. 


through  Z>!  making  ADL=%  AD  \  and  of  another  negative 
mom. -area,  the  triangle  ABX,  whose  gravity-vertical  passes 
through  D3  at  one-third  the  span  from  B.  That  is,  the 
(ideal)  positive  load  ACE  is  the  resultant  of  the  positive 
load  (M.A.)2  and  the  two  negative  loads  (or  upward  pulls) 
(M.A.\  and  (M.A.)&  and  may  therefore  be  replaced  by  them 
without  affecting  the  location  of  the  end-tangents,  at  A 


and  B,  of  the  elastic  curve  AB.  These  three  moment- 
areas  are  represented  by  arrows,  properly  directed,  in  the 
figure,  but  must  not  be  confused  with  the  actual  loads  on 
the  beam  (of  which,  here,  there  is  but  one,  viz.,  P). 

From  the  given  shapes  and  dimensions,  since  ACE  — 
192,000  (sq.  in.)  (Ibs.),  we  easily  derive  by  geometrical 
principles  : 

(KA.\=  +576,000  (sq.  in.)  (Ibs.) 
(M.A.\=  -  96,000 
(M.A.)Z  =  -288,000 

Hence,  with  a  pole  distance  El  -4-  n  =  7.11  in.  as  before, 

and  a  "moment-load-line"  formed  of  1'2'  =  (M.A.)l9  (on 
scale  of  30,000  (sq.  in.  Ibs.)  to  one  inch)  2'3'  =  (M.A.),,  and 
3'4'  —  (M.A.\t  first  with  a  trial  pole  0",  construct  the  trial 


CONTINUOUS  GIRDER  BY  GRAPHICS.  498 

polygon  A"B",  and  find  n'  in  usual  way  (§  337) ;  then  take 
a  pole  0'  on  the  horizontal  through  n'  and  the  vertical 
TW,  and  draw  the  new  polygon  A'  123  J5'.  It  should 
pass  through  B'  on  a  level  with  A',  and  A'\  and  B'3  are  the 
required  end-tangents  (of  the  exagg.  elastic  curve). 

[NOTE — If  B'  were  not  at  the  same  level  as  A' ',  but 
(say)  0.40  in.  below  it,  Bf  should  be  placed  at  m,  a  distance 

'   x =  2  inches  (on  the  paper)  below  its  present  posi- 

^0 

tion,  (since  the  distance  scale  is  1:20  and  n  =  100,  in  this 
case)  and  the  "  abutment-line  "  of  final  polygon  would  be 
A'm\. 

Of  course,  this  special  re-arrangement  of  the  moment- 
area  is  quite  superfluous  in  the  present  problem  of  a  dis- 
continuous girder  (not  built  in),  but  considerations  of  this 
kind  will  be  found  indispensable  with  the  successive  spans 
of  a  continuous  girder. 

397.  Positive  and  Negative  Moment- Areas  in  Each  Span  of  a 
Continuous  Girder  (Prismatic).— In  the  foregoing  problem  of  a 
discontinuous  girder  (covering  one  span  only)  not  built  in  at 
the  ends  (otherwise  it  would  be  classed  among  continuous 
girders),  the  moment-curve,  or  equilibrium  polygon  of 
arbitrary  H,  is  easily  found  by  §  389  without  the  aid  of  the 
elastic  curve,  and  the  end  moments  are  both  zero ;  (i.e.,  the 
moment-curve  meets  the  beam  in  the  end- verticals)  but  in 
each  span  of  a  continuous  girder  the  end-moments  are  not 
zero  (necessarily),  and  the  points  in  the  end-verticals  where 
the  moment-curve  must  terminate  (for  an  assumed  H)  can 
not  be  found  without  the  use  of  the  elastic  curve  (or  of 
some  of  its  tangents)  of  the  whole  beam,  dependent,  as  it  is, 
upon  the  loading  on  all  the  spans,  and  the  heights  of  the 
supports. 

Let  Fig.  445  show,  in  general,  any  one  span  of  a  pris- 
matic continuous  girder  (prismatic  ;  hence  /  is  constant), 
between  two  consecutive  supports  A0  and  BQ.  Pb  P2>  etc., 
are  the  loads  on  the  span. 

[If  the  displacement  of  An  relatively  to  the  end-tangent 
at  B»,  and  the  angle  between  the  end-tangents  (of  elastic 


494 


MECHANICS   OF   ENGINEERING. 


curve)  were  known,  the  moment-curve  or  equilibrium 
polygon  FWG,  (AB  being  the  axis  of  beam)  might  be 
found  by  a  process  similar  to  that  in  §_390,  but  the  elastic 
curves  in  successive  spans  are  so  inter-dependent  that  the 
above  elements  can  not  be  found  directly.] 


Ithli 


Fio.  445. 

We  now  suppose,  for  the  sake  of  discussion,  that  the 
whole  girder  has  been  investigated  (by  a  process  to  be 
presented)  for  the  given  loads,  spans,  positions  of  supports, 
etc.,  and  then  the  moment-curve  FEWLG  found,  with  the 
corresponding  force-diagram,  for  the  span  in  the  figure 
and  some  arbitrary  H.  The  horizontal  line  AB  represents 
the  axis  of  the  beam  (for  this  purpose  considered  straight 
and  horizontal)  as  an  axis  from  which  to  measure  the 
moment  ordinates. 

Thus,  the  moment  (of  the  stress-couple)  at  A  is  =  H  X 
AF\  at  B,  H  x  BG  ;at  E  and  L,  zero  (points  of  inflection). 

Now,  according  to  the  usual  conceptions  of  analytical 
geometry,  we  may  consider  the  portion  EWL,  of  the  mo- 
ment-area, above  AB  as  positive,  and  those  below,  AEF 
and  LB  G,  as  negative ;  but  since  not  one  of  these  three 


CONTINUOUS    GIRDEK    BY    GRAPHICS.  495 

areas,  nor  the  position  of  its  gravity-vertical,  is  known  in 
advance,  since  they  are  not  independent  of  the  other 
spans,  a  more  advantageous  re -arrangement  of  the  moment- 
area  may  be  made  thus : 

Join  FG  and  FB,  and  we  may  consider  the  original 
nioment-£,rea  replaced  by  the  following  three  component 
areas :  the  positive  moment-area  FEWLGF  (shaded  by  ver- 
tical lines) ;  the  negative  triangular  moment-area  AFB ; 
and  the  negative  triangular  moment-area  BFG  (the  nega- 
tive moment-areas  being  shaded  by  horizontal  lines).  (In 
subsequent  paragraphs,  by  positive  and  negative  moment- 
areas  will  be  implied  those  just  mentioned.) 

These  three  moment-areas,  treated  as  loads,  each  applied 
in  its  own  gravity -vertical,  and  considered  in  any  order, 
may  be  used  instead  of  the  real  distributed  moment- area, 
as  far  as  determining,  or  dealing  with,  the  end-tangents  of  the 
dastic-curve  at  A0  and  B0  is  concerned  (§  396),  and  the  fol- 
lowing advantages  will  have  been  gained  : 

(1.)  The  amount  of  the  positive  moment-area,  (M.A.\  in 
Fig.  445,  (depending  on  the  area  lying  between  the  polygon 
FEWG  and  the  abutment-line  F  G  of  the  latter),  and  the 
position  of  its  gravity -vertical,  are  independent  of  other  spans, 
and  can  be  easily  found  in  advance,  since  this  moment-area 
and  gravity -vertical  are  the  same  as  if  the  part  of  the  beam 
covering  this  span  were  discontinuous  and  simply  rested  on  the 
supports  A0  and  Bu,  as  in  §  389. 

(2.)  The  gravity-vertical  of  the  left-hand  negative  mo- 
ment-area, (M.A.)i,  is  always  one-third  the  span  from  the 
left  end-vertical,  A0A';  that  of  the  other,  (M.A.\,  an  equal 
distance  from  the  right  end-vertical,  B0B'. 

(3.)  The  two  (right  and  left)  negative  moment-areas 
are  triangular,  each  having  the  whole  span  I  for  its  alti- 
tude, and  for  its  base  the  intercept  AF  (or  BG)  on  which 
the  end-moment  depends.  Hence,  if  the  amounts  of  these 
negative  moment-areas  have  been  found  in  any  span,  we 
may  compute  the  values  which  AF  and  BG  must  have  for 
a  given  H,  and  thus  determine  the  terminal  points  F  and 
G  of  the  moment-curve  of  that  span  (for  that  value  of  H). 


496  MECHANICS  OF   ENGINEERING. 

For  example,  if  (M.A.\  has  been  found  (by  a  process  not 
yet  given)  to  be  160,000  (sq.  in.)  (Ibs.)  while  AB  =  I  =  160 
in.,  then  the  moment  M^  which  AF  represents,  is  computed 
from  the  relation 


(M.A.\  =  y2  AB  x 
or  Jf=  -2000,  in.  Ibs. 


If  #has  been  chosen  =  100  Ibs.  we  put  HxAF  =  2000 
and  obtain  AF  =  20  inches  of  actual  distance,  so  that  with 
a  scale  of  1:20  for  distances  AF  would  be  one  linear  inch 
of  paper.  (Of  course,  in  computing  BG  the  same  value  of 
H  must  be  used.)  "With  H=  100  Ibs.,  then,  and  F  and  G 
as  known  points  of  the  equilibrium  polygon  FEW  G,  it 
is  easily  drawn  by  the  principles  of  §  341. 

We  thus  notice  that  the  amounts  of  the  two  negative 
moment-areas  are  the  only  elements  affected  by  the  con- 
tinuity of  the  girder,  in  this  re-arrangement  of  the  actual 
moment-areas. 

In  the  lower  part  of  Fig.  445  A'  and  B'  represent  the 
extremities  of  the  (exagg.)  elastic  curve,  the  vertical  dis- 
tance B'B'",  of  B9  from  the  horizontal  through  A'  (in  case 
the  two  supports  A0  and  B0  are  not  at  same  level,  as  we 
here  suppose  for  illustration)  being  laid  off  in  accordance 
with  the  principles  of  the  note  in  §  396. 

NOTE.  —  It  is  now  evident  that  if  the  "false  polygon"  (as 
it  will  be  called)  ^4'123^'  has  been  obtained  (and  means  for 
doing  this  will  be  given  later)  in  which  the  first  and  last 
segments  are  the  end  tangents  of  the  (exagg.)  elastic  curve, 
and  which  bears  the  same  relation  to  the  three  moment 
areas  just  mentioned,  as  that  illustrated  in  Fig.  444,  we 
may  proceed  further  to  determine  the  amounts  of  (M.A.\ 
and  (M.A.\  as  follows,  by  completing  the  moment-area  dia- 
gram : 

Having  laid  off  the  known  (M.A.\  (or  positive  moment- 
area)=2'3',  and  ST=EI-±-n,  a  line  parallel  to  12  drawn 
through  2',  determines  the  pole  0',  through  which  paral- 


CONTINUOUS  GIRDER  BY  GRAPHICS. 


497 


lels  to  A'l  and  B'3  will  fix  1'  and  4'  on  the  vertioal  8V> 
and  thus  determine  l"2'=(M.A.\  and  S'4f=(M.A.^  Their 
numerical  values  are  then  computed  in  accordance  with 
the  scale  of  the  moment-area  diagram. 

The  polygon  A'V&B'  will  be  called  the  "false  poh/gon  " 
of  the  span  in  question,  its  end-segments  being  the  end- 
tangents  of  the  elastic  curve. 

398.  Values  of  the  Positive  Moment-Area  in  Special  Cases.— 
For  several  special  cases  these  are  easily  computed,  and 
as  an  illustration,  Fig.  446  shows  a  continuous  girder,  AF 


FIG.  446. 

of  five  spans,  all  six  supports  on  a  level,  and  the  weight  of 
the  beam  neglected.  At  the  extremities  A  and  F,  as  at 
the  other  supports,  the  beam  is  not  built  in,  but  simply 
touches  each  support  in  one  point ;  hence  the  moments  at 
A  and  F  are  zero,  i.e.,  the  moment  curve  must  pass  through 
A  and  F,  so  that  in  the  first  span  the  left  negative  mo- 
ment-area, and  in  the  fifth  span  the  right  negative  mo- 
ment-area, are  zero.  The  positive  moment-area*  are 
shaded. 

On  the  first  span  is  placed  a  uniformly  distributed  load 
W  over  the  whole  span  I'.  .*.  the  positive  moment-area 
for  that  span  is  the  same  as  in  the  case  of  Fig.  235  [soo 
(1)  §  397]  and  being  represented  by  a  parabolic  segment 
whose  area  is  two-thirds  that  of  the  circumscribing  rec- 
tangle, its  value  is 


498  MECHANICS  OF  ENGINEERING. 


[se«  uq.  (2)  §  242],  while  its  gravity-  vertical  bisects  the 
span. 

The  only  load  on  the  second  span  is  a  concentrated  one, 
P"t  at  distances  Z/'  and  1%  from  the  extremities  of  the  span; 
hence  the  positive  moment-area  is  triangular  and  has  a 
value 

(M.AX'=#P'W        •        •         -      (2) 

as  in  §  393.  Its  gravity  vertical  may  easily  be  constructed 
as  in  Fig.  443  [see  (1)  §  397], 

The  third  span  carries  no  load  ;  hence  its  positive  mo- 
ment-area is  zero,  and  the  actual  moment-area  is  composed 
solely  of  the  two  triangular  negative  moment-areas  GDH 
and  DHI,  the  moment-curve  consisting  of  the  single 
straight  line  HI. 

The  fourth  span  carries  a  uniform  load  ?Fnr=f^viIV,  and 
.*.  has  a  positive  moment-area 

(M.Aff^/vW"(VV       ...    (3) 

as  in  eq.  (1),  acting  through  the  middle  of  the  span  (grav- 
ity-vertical). 

Since  the  fifth  and  last  span  carries  no  load,  its  positive 
moment-area  is  .zero,  the  moment  curve  being  the  straight 
line  JFj  so  that  the  actual  moment-area  is  composed  of 
the  left-hand  negative  moment-area. 

At  F  it  is  noticeable  that  the  reaction  or  pressure  of  the 
support  must  be  from  above  downward  to  prevent  the 
beam  from  leaving  the  point  F\  i.e.,  the  beam  must  be 
"  latched  down,"  and  the  reaction  is  negative. 

If  the  beam  were  built  in  at  A  (or  F)  the  moment  at 
that  section  would  not  be  zero,  hence  the  left  (or  right) 
neg.  moment-area  would  not  be  zero  in  that  span,  as  in 
our  present  figure.  But  in  such  a  case  the  tangent  of  the 
elastic  curve  would  have  a  known  direction  at  A  (or  F)  and 
the  problem  would  still  be  determinate  as  will  be  seen. 


CONTINUOUS   GIRDER  BY   GRAPHICS.  499 

A'  .  .  .  F!  gives  an  approximate  idea  (exaggerated)  of 
the  form  of  the  elastic  curve  of  the  entire  girder.  A  change 
in  the  loading  on  any  span  would  affect  the  form  of  this 
curve  throughout  its  whole  length  as  well  as  of  all  the 
moment  curves. 

NOTE.  —  It  is  important  to  remark  that  any  two  of  the 
triangular  negative  moment-areas  which  have  a  common 
base  (hence  lying  in  adjacent  spans)  are  proportional  to 
their  altitudes,  i.e.,  to  the  lengths  of  the  spans  in  which  they 
occur  ;  thus  the  neg.  mom.-areas  (Fig.  446)  GCH&ndDCH 
have  a  common  base  OH, 

_V  ... 

'-~- 


(The  notation  explains  itself  ;  see  figure.)  It  also  follows, 
that  the  resultant  of  these  two  neg.  moment-areas  (if  re- 
quired in  any  construction  ;  see  §  400)  acts  in  a  vertical 
which  divides  the  horizontal  distance  between  their  gravity  ver- 
ticals in  the  inverse  ratio  of  the  spans  to  which  they  belong 
[§21  and  eq.  (4)  above]. 

Hence,  since  this  horizontal  distance  is  X^"~f"K^'"  their 
resultant  must  act  in  a  vertical  Y'"9  whose  distance  from 
the  gravity-vertical  of  GCH  is  %l"ft  and  from  that  of 
CUD,  yd". 

399.  Amount  and  Gravity-Vertical  of  the  Positive  Moment 
Area  of  One  Span  as  Due  to  Any  Loading.  —  Since  we  can  not 
deal  directly  with  a  continuous  load  by  graphics,  but  must 
subdivide  it  into  a  number  of  detached  loads  sufficiently 
numerous  to  give  a  close  approximation,  let  us  suppose 
that  this  has  already  been  done  if  necessary,  and  that  P19 
P2,  etc.,  are  the  detached  loads  resting  in  the  span  AS  in 
question  ;  see  Fig.  447. 

Since  [by  (1),  §  397]  the  positive  moment-area  is  the 
same  as  the  total  moment-area  would  be  if  this  portion  of 
the  beam  simply  rested  on  the  extremities  of  the  span,  not 
extending  beyond  them,  we  may  use  the  construction  in  § 
389  for  finding  it,  remembering  that  in  that  paragraph  the 


500 


MECHANICS   OF   ENGINEERING. 


oblique  polygon  in  the  lower  part  of  Fig.  439  will  serve 
as  well  as  the  (upper)  one  whose  abutment -line  is  the 
beam  itself,  as  far  as  moments  are  concerned. 

Hence,  Fig.  447,  lay  off  the  load-line  LL'y  take  any  pole 
0,  with  any  convenient  pole-distance  H,  and  draw  the 
equilibrium  polygon  FWG.  After  joining  FG,  FWGf1 
will  be  the  positive  moment-area  required. 

To  find  its  gravity  vertical,  divide  the  span  AB,  or  FG't 
into  from  ten  to  twenty  equal  parts  (each  =  As]  and  draw  a 


PIG.  447. 


vertical  through  the  middle  of  each.  The  lengths  xi9  z^ 
etc.,  on  these  verticals,  intercepted  in  the  moment-area, 
are  proportional  to  the  corresponding  strips  of  moment- 
area,  each  of  width =As,  and  of  an  amount  —HzAs. 

Form  a  load  line,  SK,  of  the  successive  z's,  and  with 
any  pole  0',  draw  the  equilibrium  polygon  A'B'  (for  the 
^-verticals).  The  intersection,  R,  of  the  extreme  segments* 
is  a  point  in  the  required  gravity-vertical  (§  336). 

The  amount  of  the  moment-area  is  (M.A.\=  2[Hzds\ 


CONTIGUOUS  GIRDER  BY  GRAPHICS.  501 

For  example,  with  the  span  =Z=120  in.,  subdivided  into 
twelve  equal  ^s's,  we  have  Js— 10  inches  (of  actual  distance). 
If  H=  4  inches  of  paper  and  SK=2(z)=W.Z  inches  of  paper, 
the  force-scale  being  80  Ibs.  to  the  inch,  and  the  distance** 
scale  15  inches  to  the  inch  (1:15),  we  have 

( M.A.)2= [4 x 80]  x  10  x  [10.2x15]  =489600.  (sq.  in.)(lbs.) 

400.  'Construction  of  the  "False-Polygons"  For  All  the  Spans 
of  a  Given  (Prismatic)  Continuous  Girder,  Tinder  Given  Loading, 
and  With  Given  Heights  of  Supports. — [See  note  in  §  397  for 
meaning  of  "false-polygon  ".]  Let  us  suppose  that  the 
given  girder  covers  three  unequal  spans,  Fig.  448,  with 
supports  at  unequal  heights,  and  that  both  extremities  A 
and  D  are  built-in,  or  "  fixed,"  horizontally.  To  clear  the 
ground  for  the  present  construction,  we  suppose  that,  from 
the  given  loading  in  each  span,  the  positive  moment-area 
of  each  span  has  been  obtained  in  numerical  form  [so 
many  (sq.  in.)  (Ibs.)  or  (sq.  in.)  (tons)]  and  its  gravity-ver- 
tical determined  by  §  398  or  §  399  ;  that  the  horizontal 
distances  (i.e.,  the  spans  Z',  I",  and  I'"  and  the  distance  be- 
tween the  above  gravity -verticals  and  the  supports)  have 
been  laid  off  on  some  convenient  scale ;  that  El  has  been 
computed  from  the  material  and  shape  of  section  of  the 
girder  and  expressed  in  the  same  units  as  the  above  mo- 
ment-areas ;  that  a  convenient  value  for  n  has  been  se- 
lected (since  EI-^-n  is  to  be  the  pole  distance  of  all  the 
moment-area-diagrams),  and  that  the  vertical  distances  of 
B,  C,  and  D,  from  the  horizontal  through  A,  have  been 
laid  off  accordingly  (see  note  in  §  396). 

In  the  figure  (448)  verticals  are  drawn  through  the 
points  of  support ;  also  verticals  dividing  each  span  into 
thirds,  since  the  unknown  negative  moment-areas  (sub- 
scripts 1  and  3)  act  in  the  latter  (§  397) ;  and  the  gravity- 
verticals  of  the  known  positive  moment-areas.  The  ver- 
ticals Y',  Y",  V'\  and  V"',  are  to  be  constructed  later. 

The  problem  may  now  be  stated  as  follows : 

the  positions  of  the  supports,  the  value  of  El  and  nt 


502 


MECHANICS   Ol     ENGINEERING. 


the  fact  that  the  girder  is  fixed  horizontally  at  A  and  D,  tM 
heights  of  supports,  the  location  of  the  gravity -verticals  of  all 
the  positive  and  negative  moment-areas,  and  the  amounts  of  ikt 
positive  moment-area^  ;  it  is  required  to  find  graphically  the 
"false-polygon"  in  each  span. 

The  "  false -polygons,"  viz.:  A 123  B  for  the  first  span  (on 
the  left),  B 123  G  for  the  second,  etc.,  are  drawn  in  the  figure 


Fie.  448. 

lor  the  purpose  of  discussing  their  properties  at  the  out- 
set. Since  3B  and  B\  are  both  tangent  to  the  elastic  curve 
at  B,  they  form  a  single  straight  line ;  similarly  (71  is  but 
the  prolongation  of  3(7.  Also  Al  and  3D  must  be  hori- 
zontal since  the  beam  is  built  in  horizontally  at  its  ex- 
tremities A  and  D. 

That  is,  the  three  false  polygons  form  a  continuous 
equilibrium  polygon  A  .  . .  D,  in  equilibrium  under  the 
"loads" 

(M.A.y,  (M.A.\'t  (KA.y,  etc., 

ao  that  we  might  use  a  single  mom. -area-diagram  in  con- 
nection with  it,  but  for  convenience  the  latter  will  b& 


CONTINUOUS    GIRDER  BY   GRAPHICS.  503 

drawn  in  portions,  one  under  each  span,  with  a  pole  dis- 
tance =  — . 
» 

Of  this  polygon  A  ...  D,  we  have  the  two  segments  AL 

and  3-D  already  drawn,  and  know  that  it  passes  through 
the  points  B  and  C ;  we  shall  next  determine  by  construc- 
tion other  points  (called  "  fixed  points  ")  pQ't  p",  pQ",  p'", 
and  pQ'"  in  (the  prolongations  of)  certain  other  segments. 

To  find  the  "fixed  point "  prf,  where  the  segment  23  in  the 
first  span  cuts  the  vertical  F',  the  gravity -vertical  of 
(M.A.y.  The  vertex  p',  or  1,  is  already  known,  being  the 
intersection  of  Al  with  V.  Lay  off2'  3'  =  (M.A.y  which  is 
known,  and  take  a  trial  pole  0/with  a  pole-distance^/-^-7i; 
join  0t'  2'  and  0t'  3'.  Draw  120  II  to  2'0t  to  determine  2,, 
on  the  vertical  (M.  A.)2',  then  through  20  a  line  ||  to  0/3'  to 
cut  V  in  PQ.  The  unknown  segment  23  must  cut  V:  in 
the  same  point ;  since  all  positions  of  0'  on  the  vertical 
T'U'  will  result  in  placing  pQ'  in  this  same  point,  and  one 
of  these  positions  must  be  the  real  pole  0'  (unknown). 
[This  is  easily  proved  in  detail  by  two  pairs  of  similar 
triangles]. 

To  determine  the  "fixed-point  "  p"9  in  the  prolongation 
of  segment  12  of  second  span.  The  prolongations 
of  the  segments  23  (of  first  span)  and  12  (of  second  span) 
must  meet  in  a  point  k'  in  the  (vertical)  line  of  action  of  the 
resultant  of  (M.  A.y  and  (M.  A.\"  (§  336).  Although  the 
amounts  of  (M.  A.)$  and  (M.  A.\"  are  unknown,  still  the 
Vertical  line  of  action  of  their  resultant  (by  §  398,  Note)  is 

I" 

known  to  be  F',  a  horizontal  distance  —   to    the    right    of 

o 

(M.A.\' ;  hence  Yf  is  easily  drawn.  Therefore,  the  unknown 
triangle  &'31  has  its  three  vertices  on  three  known  verti- 
cals, the  side  k'3  passes  through  the  known  point  pQ'y  and 
the  side  13  through  the  known  point  B.  Now  by  pro- 
longing PQ  20  (or  any  line  through  p0')  to  cut  (M.  A.\'  and 
Yf  in  30  and  &/,  respectively,  joining  30B  and  prolonging 
this  line  to  cut  (M.  A.\"  in  some  point  10,  and  then  joining 
fc0'  10,  we  have  a  triangle  &0'3010  °f  which  we  can  make  a 


504  MECHANICS   OF   ENGINEERING. 

statement  precisely  the  same  as  that  just  made  for  k*  3  1 

But  if  two  triangles  [as  &'31  and  kQ'  30  10]  have  theii 
vertices  on  three  parallel  lines  (or  on  three  lines  which 
meet  in  a  point)  the  three  intersections  of  their  correspond- 
ing sides  must  lie  on  the  same  straight  line  [see  reference 
to  Chauvenet,  §  378  a].  Of  these  intersections  we  have 
two,^>0'  and  B  ;  hence  the  third  must  lie  at  the  intersection 
of  the  line  pj  B  (prolonged)  with  &0'  10,  and  in  this  way  the 
"  fixed  point "  p"9  a  point  in  &'l  and  .'.  in  the  segment  12 
(of  second  span;  prolonged,  is  found.  Draw  a  vertical 
through  it  and  call  it  V". 

The  fixed  point  p0"  (in  prolongation  of  segment  23  of  sec- 
ond span)  lies  in  the  vertical  V"  and  is  found  from  p"  and 
the  known  value  of  (M.A.\"  precisely  as^>0'  was  found  from 
p'.  That  is,  we  lay  off  vertically  2"  3"  =  (M.A.\",  and  join 
2"  and  3"  to  Ot",  which  is  any  point  at  distance  El  —  n 
to  the  right  of  2"3".  Through  p"  draw  a  line  ||  to  2"  Ot" 
to  cut  ( M.  A.)2"  in  20,  then  20_p0"  II  to  Ot"  3"  to  determine 
p0"  on  the  vertical  V". 

The  fixed  points  p'"  and^></"  in  the  third  span  lie  in  the 
prolongations  of  the  segments  12  and  23,  respectively,  of 
that  span,  p'"  being  found  from  the  points  pQ"  and  G  and 
the  verticals  (M.A.\"t  Y",  and  (M.A.\'"9  in  the  same  manner 
as  p"  was  determined  with  similar  data,  while  p0'",  in  the 
same  vertical  V"  as  p'",  depends  on  (M.  A.)2'"  and  its 
gravity  vertical  as  already  illustrated ;  hence  the  detail 
need  not  be  given  ;  see  figure. 

In  this  way  ior  any  number  of  spans  we  proceed  from 
span  to  span  toward  the  right  and  determine  the  succes- 
sive fixed  points,  until  the  points  p  and  p0  of  the  last  span 
have  been  constructed,  which  are  p'"  and  p0"'  in  our 
present  problem.  Since  p0'"  is  a  point  in  the  segment  23 
(prolonged)  of  the  last  span,  we  have  only  to  join  it  with 
3  in  that  span,  a  point  already  known,  and  the  segment  23 
is  determined.  Joining  the  intersection  2  with  p'"  we 
determine  the  next  segment  21  and  of  coarse  the  vertex  1, 
which  is  then  joined  with  C  and  prolonged  to  intersect 
M.  A.\"  to  fix  the  segment  103  and  the  point  3.  Join 


CONTINUOUS  GIKDER  BY  GRAPHICS.  505 

3  PQ>  and  proceed  in  a  similar  manner  toward  the  left, 
until  the  whole  equilibrium  polygon  (or  series  of  "  false 
polygons  ")  is  finally  constructed  ;  the  last  step  being  the 
joining  of  2  with  p'. 

401.  Treatment  of  Special  Features  of  the  Last  Problem. —  (1.) 
If  the  beam  is  simply  supported  at  A,  Fig.  448,  instead  of 
built-in,  (M.  A.)'  becomes  zero,  and  the  two  segments  A\ 
and  12  of  that  span  form  a  single  segment  of  unknown 
direction.  Hence,  the  point  A  will  take  the  place  of  p', 
and  the  vertical  FA  that  of  V. 

(2.)  Similarly,  if  D,  in  the  last  span,  is  a  simple  support 
(beam  not  built  in)  (M.  A.)B'"  becomes  zero,  and  the  seg- 
ments D3  and  32  form  a  single  segment  of  unknown 
direction,  so  that  after  pj'  has  been  found,  we  join  pf 
and  D  to  determine  the  segment  Z>2  ;  i.e.,  in  this  last  span, 
D  takes  the  place  of  3  of  the  previous  article. 

(3.)  If  the  first  span  carries  no  load  (M.  A.)2f  is  zero,  and 
the  segments  12  and  23  will  form  a  single  segment  23. 
Hence  if  the  beam  is  built  in  at  A,  pd  will  coincide  with 
the  known  point  p'  (i.e.,  1),  while  if  A  is  a  simple  support 
p0'  and_p'  coincide  with  A,  since,  then,  (M,  A.)L'  is  zero  and 
A  123  is  a  single  segment. 

(4.)  If  the  last  span  is  unloaded  (third  span  in  Fig.  448), 
(M.  A.)2'"  is  zero,  123  becomes  a  single  segment,  and  hence 
PQ"  will  coincide  with  p'"  ;  so  that  after  p'"  has  been  con- 
structed it  is  to  be  directly  joined  to  3,  if  the  beam  is 
built  in  at  D,  and  will  thus  determine  the  segment  13  ;  or 
to  D,  if  D  is  a  simple  support,  (for  then  (M.  A.)z'"  is  zero 
and  the  three  segments  12,  23,  and  3  D  form  a  single  seg- 
ment.) 

(5.)  If  an  intermediate  span  is  unloaded  (say  the  second 
span,  Fig.  448)  the  positive  mom. -area,  (M.  A.\"t  is  zero, 
123  becomes  a  straight  line,  i.e.  a  single  segment,  and 
therefore  p"  coincides  with p"  ;  hence,  when  p"  has  been 
found  we  proceed  as  if  it  were  pQ". 

402.  To  Find  the  Negative  Mom. -Areas,  the  Mom. -Curves,  Shears, 
and  Reactions  of  the  Supports. — (1.)  Having  constructed  the 


506  MECHANICS   OF   ENGINEERING. 

false  polygons  according  to  the  last  two  articles,  the  negto* 
tive  moment-areas  of  each^  span  are  tken  to  be  found  by  the 
note  in  §  397,  Fig.  445,  and  expressed  in  numerical  form. 

[If  the  positive  niom.-area  of  the  span  is  zero  the  points 
2'  and  3'  will  coincide,  Fig.  445,  and  in  the  case  mentioned 
in  (3),  (or  (4)),  of  §  401,  if  A  (or  D)  were  a  simple  support, 
in  Fig.  448,  the  mom. -area-diagram  of  Fig.  445  would  have 
but  two  rays.] 

(2.)  The  moments  at  the  supports  (or  "end-moments  "  of 
the  respecL\e  spawns)  depending,  as  they  do,  directly  on 
the  negative  mom. -areas,  can  now  be  computed  as  illustrated 
in  (3)  §  397.  The  fact  that  each  "  end-moment  "  may  be 
obtained  from  two  negative  mom.-areas,  separately,  one  in 
each  adjacent  span  (except,  of  course  at  the  extremities  of 
the  girder)  forms  a  check  on  the  accuracy  of  the  work. 
The  two  values  should  agree  within  one  or  two  per  cent. 

(3.)  The  "  moment-curve  "  of  each  span  or  equilibrium 
polygon  formed  from  a  force-diagram  whose  load-line  con- 
sists of  the  actual  loads  on  the  span  laid  off  in  proper 
order,  can  now  be  drawn,  a  convenient  value  for  H  having 
been  selected  (the  same  Hfor  aU  the  spans,  that  the  moment- 
curves  of  successive  spans  may  form  a  continuous  line  for 
the  whole  girder);  since  we  may  easily  compute  the  proper 
moment  ordinate  at  each  support  to  represent  the  actual 
moment,  then,  for  the  H  adopted,  by  (3)  §  397.  The 
moment-curve  of  each  span,  since  we  know  its  two  extreme 
points  and  its  pole-distance  H,  is  then  constructed  by 
§  341. 

(4.)  The  shear.  Since  the  last  construction  involves 
drawing  the  special  force-diagram  for  each  span,  with  a 
ray  corresponding  to  each  part  of  the  span  between  two 
consecutive  loads,  the  shear  at  any  section  of  the  beam  is 
easily  found  as  being  the  length  of  the  vertical  projection 
of  the  "  proper  ray/'  interpreted  by  the  force-scale  of  the 
force-diagram,  as  in  §§  389  and  390.  With  the  shears  as 


CONTINUOUS  GIRDER  BY  GRAPHICS.  507 

__  ax  _____  A  ordinates   a   shear  -diagram   may 

now  be  constructed,  if  desired,  for 
each  span.  The  directions  of  the 
shears  should  be  carefully  noted. 
(5.)  Reactions  of  supports.  Let 
us  consider  "  free  "  the  small  por- 
tion of  the  girder,  at  each  point 
of  support,  included  between  two 
sections,  one  close  to  the  support 
on  each  side,  Fig.  449.  Suppose 
it  is  the  support  (7,  and  call  the 
reaction,  or  pressure  at  that  sup- 
port  Bc.  Then,  for  vertical  equi- 
librium, i.e.,  27  =  0  (§  36),  we  have 


(5) 


and,  in  general,  the  reaction  at  a  support  equals  the 
(algebraic)  sum  of  the  two  shears,  one  close  to  the  support 
ou  the  right,  the  other  on  the  left.  The  meaning  of  the 
subscripts  is  evident.  In  applying  this  rule,  however,  a 
free  body  like  that  in  Fig.  449  should  always  be  drawn,  or 
conceived  ;  for  the  two  shears  are  not  always  in  the  same 
direction  ;  hence  the  phrase  "  algebraic  sum." 

At  a  terminal  support,  as  A  or  F,  Fig.  446,  if  the  beam 
is  not  built  in,  the  reaction  is  simply  equal  to  the  shear 
(since  the  beam  does  not  overhang)  just  as  in  §§  241  and 
243.  Fig.  446  presents  the  peculiarity  that  the  reaction 
of  the  support  F  is  negative,  (as  compared  with  J?c  in  Fig. 
449);  i.e.,  the  support  at  F  must  be  placed  above  the  beam 
to  prevent  its  rising  (this  might  also  be  the  case  at  (7,  01 
D,  in  Fig.  446,  for  certain  relations  between  the  loads). 

403.  Numerical  Example  of  Preceding  Methods.  —  As  illustrat- 
ing the  constructions  just  given,  it  is  required  to  investi- 
gate the  case  of  a  rolled  wrought-iron  "  I-beam,"  [a  15- 
inch  heavy  beam  of  the  N.  J.  Steel  and  Iron  Co.,]  extend- 
ing over  four  supports  at  the  same  level,  covering  three 


508  MECHANICS  OF   ENGINEERING. 

spans  of  16,  20,  and  14  feet  respectively,  and  bearing  a 
single  load  in  each  of  the  extreme  spans,  but  a  uniform 
load  over  the  entire  central  span.  As  indicated  thus  : 


6ft. 


W" 

40  tons  32  tons 

[This  is  a  practical  case  where  W"  is  the  weight  of  a 
biick  wall,  and  P'  and  P"r  are  loads  transmitted  by  col- 
umns from  the  upper  floors  of  the  building ;  A  and  D  are 
simple  supports,  and  the  weight  of  the  girder  is  neglected.] 

The  beam  has  a  moment  of  inertia  /  =  707  biquad. 
inches,  and  the  modulus  of  elasticity  of  the  iron  is  E  = 
25,000,000  Ibs.  per  sq.  in.,  =  12,500  tons  per  sq.  in. 

Although  with  a  prismatic  continuous  girder  under 
given  loading,  with  supports  at  the  same  level,  it  may  easily 
be  shown  that  the  moments,  shears,  and  reactions,  to  be 
obtained  graphically,  are  the  same  for  all  values  of  /,  so 
long  as  the  elastic  limit  is  not  surpassed,  still,  on  account 
of  the  necessity  of  its  use  in  other  problems  (supports 
not  on  a  level\  we  shall  proceed  as  if  the  value  of  /  wore 
essential  in  th-is  one. 

Selecting  the  inch  and  ton  as  units  for  numerical  work, 
we  have 

El=  12500  x  707.  =  8,837,500  (sq.  in.)  (tons)  while  the 
respective  positive  mom. -areas,  from  eqs.  (2)  and  (3)  of 
§  398,  are : 

(M.  A.\'  =  y2  x  30  x  84  x  108  =  136,080  (sq.  in.)  (tons) 
(M.  A.\"  =  yu  x  40  x  2402        =  192,000    "      " 
(M.  A.}j"  =  */2  x  32  x  96  x72  =  110,592    «      " 

Adopting  a  scale  of  60,000  (sq.  in.)  (tons)  to  the  lineal  inch 
of  paper,  for  mom.-area  diagrams,  we  have  for  the  above 
mom. -areas  2.27  in.,  3.20  in.,  and  1.84  in.,  respectively,  OB 
the  paper,  for  use  in  Fig.  448. 


CONTINUOUS   GIRDER   BY   GRAPHICS.  509 

Having  laid  off  the  three  spans  on  a  scale  of  60  inches 
to  the  inch  of  paper,  with  A,  B,  C,  and  D  in  the  same  hori- 
zontal line,  we  find  by  the  construction  of  Fig.  443,  that  the 
gravity- vertical  of  (M.  A.\'  lies  4.0  in.  to  the  left  of  the 
middle  in  the  first  span,  that  of  (M.  A.)J"  4.0  in.  to  the 
right  of  the  middle  of  the  third  span ;  while  that  of 
(M.  A.).2",  of  course,  bisects  the  central  span.  Hence  we 
draw  these  verticals ;  and  also  those  of  the  unknown  neg- 
ative mom.-areas  through  the  one-third  points  ;  remember- 
ing [§  401,  (1)  and  (2)]  that  (M.  A.\'  and  (M.  A.)f"  are 
both  zero  in  this  case. 

Since  El  =  8,837,500  (sq.  in.)  (tons),  it  would  require 
147.29  in.  to  represent  it,  as  pole-distance,  on  a  scale  of 
60,000  (sq.  in.)  (tons)  to  the  inch  ;  hence  let  us  take  n  =  50 
for  the  degree  of  (vertical)  exaggeration  of  the  false  poly- 

TjlT 

gons,  since  the  corresponding  pole-distance  -  -  =  2.94  in. 

of  paper  is  a  convenient  length  for  use  with  the  values  of 
(M.  A.)2't  (M.  A.\",  etc.,  above  given. 

Following  the  construction  of  Fig.  448,  except  that  pf  is 
at  A,  and  p§"  is  to  be  joined  to  D  (§  401),  (the  student  will 
do  well  to  draft  the  problem  for  himself,  using  the  pre- 
scribed scales,)  and  thus  determining  the  false-polygons, 
we  then  construct  and  compute  the  neg.  mom.-areas 
according  to  §  402  (1),  and  the  note  in  §  397,  obtaining  th© 
following  results : 


((  €( 

(I  tf 


(M.  A.\',   1.43  in.  of  pap.,  =    85,800  (sq.  in.)  (tons) 

(M.  A.y,  1.77  «    «     "      =  106,200 

(J£  A.y,  1.68  "    "     "      =  100,800 

(M.  A.y',l.\l"    "     "      =    70,200    "     «        « 

The  remaining  results  are  best  indicated  by  the  aid  ol 
Fig.  450.     Following  the  items  of  §  402,  we  find  [(3)  §  397] 
^  that  the  moment  at  B,  using  (M.  A.y,  is 


=  885  inch-tons. 


510  MECHANICS   OF   ENGINEERING. 

[or,  using  (M.  A.),',  'JfB  -  2x^800=893  in. 

Similarly,  Mc  =  2x^800  =  840  in.  tons, 
[or,  using  (M.  A.\'",  Mc  =  835.7  in.  tons.] 
Hence,  taking  means,  we  have,  finally, 

JfA=0 ;  JfB=889  in.  tons  ;  3fc=837.8  ;  JfD=0. 

Fig.  450  shows  the  actual   mom. -areas  and  shear-dia 
grams,  which  are  now  to  be  constructed. 


Fie.  450. 

Selecting  a  value  H  =  20  tons  for  the  pole-distance  of 
the  successive  force-diagrams,  (the  scale  of  distances  being 
5  ft.  (60  in.)  to  the  inch  we  have  [(3)  §  397] 

20  x  ~BO  =  MR  =  889in.-tons  .-.  WG  =  44.4  in.  of  actual 
distance,  or  0.74  in.  of  paper ;  also  20  x  GK  =  MG  =  837.8 
in. -tons  .-.  CK  —  41.89  in.,  or  0.698  in.  of  paper. 

Having  thus  found  G  and  K,  and  divided  BG  into  ten 
equal  parts,  applying  four  tons  in  the  middle  of  each,  we 
construct  by  §  341  an  equilibrium  polygon  which  shall 
pass  through  G  and  TTand  have  20  tons  as  a  pole-distance. 
(We  take  a  force-scale  of  10  tons  to  the  inch.)  It  will 
form  a  (succession  of  short  tangents  to  a)  parabola,  and  is 
the  moment  curve  for  span  SO.  Similarly,  for  the  single 


CONTINUOUS    GlilDEli  BY   GRAPHICS.  5H 

loads  P'  and  P'"  in  the  other  two  spans,  we  draw  the 
equilibrium  polygons  AN'G  and  KZ'"D,  for  the  same  H 
as  before,  and  passing  through  A  and  6r,  and  K  and  D,  re- 
spectively. 

Scaling  the  moment  -ordinates  NN'9  QQ",  and  ZZ'", 
reducing  to  actual  distance  and  multiplying  by  H,  we  have 
for  these  local  moment  maxima,  M$  =  1008,  M^  =  336,  and 
Mz  =  936,  in.  tons. 

Evidently  the  greatest  moment  is  Mv  and  .'.  the  stress 
in  the  outer  fibre  at  JV  will  be  (§  239) 

tons  per  sq.  inch  which  is  much 


too  large.  If  we  employ  a  20-inch  heavy  beam,  with  J= 
1650  biquad.  in.,  the  preceding  moments  will  still  be  the 
same  (supports  all  at  same  level)  and  we  have 


or  nearly   12,000  Ibs.   per   sq.  in.,  and  is  therefore  safe 
(§  183). 

If  three  discontinuous  beams  were  to  be  used,  the  20- 
inch  size  of  beam  (heavy)  would  be  much  too  weak,  in 
each  of  the  three  spans,  as  may  be  easily  shown  ;  hence  the 
economy  of  the  continuous  girder  in  such  a  case  is  readily  per- 
ceived. It  will  be  seen,  however,  that  the  cases  of  conti- 
nuity and  of  discontinuity  do  not  differ  so  much  in  the 
shear-diagrams  as  in  the  moment  curves.  By  scaling  the 
vertical  projection  of  the  proper  rays  in  the  special  force 
diagrams  (as  in  §§  389  and  390)  we  obtain  the  shear  for 
any  section  on  AN,  as  J^R  (see  Fig.  449  for  notation)  = 
12.3  tons  ;  on  NB,  JBL  =  17.7  tons  ;  from  B  to  G  it  varies 
uniformly  from  J"BR  =  20.3  tons,  through  zero  at  Q,  to  JCIi 
=  19.7  tons  of  opposite  sign.  Also,  for  CZt  JCR  =  18.6 
tons  ;  and  for  ZD,  e7DL  =  13.4  tons.  Hence,  the  reactions 
of  the  supports  are  as  follows  : 

tons  ;  £B=JBL+Jj,R=S8.0  tons. 
38.3  tons  ;  .RD=e7DL=13.4  tons. 


512  MECHANICS    OF    ENGINEERING. 

[In  the  shear-diagram,  the  shear-ordinates  are  laid  oft 
below  the  axis  when  the  shear  points  down,  the  "free  body  " 
extending  to  the  right  of  the  section  considered,  (as  e7CL  in  Fig. 
449)  ;  and  above,  when  the  shear  points  upward  for  the 
same  position  of  the  free  body.] 

If  we  divide  the  max.  shear,  20.3  tons  by  the  area  of  the 
web,  13.75  sq.  in.,  of  the  20-inch  heavy  beam,  (§256),  we 
obtain  1.5  tons  or  3000  Ibs.  per  sq.  in.,  which  is  <  4000 
(§  183).  Notice  the  points  of  inflection,  i',  i/',  etc.,  where 
Mis  zero. 

Sufficient  bearing  surface  should  be  provided  at  the 
bupports. 

A  swing-bridge  offers  an  interesting  case  of  a  continu- 
ous girder, 

404.  Continuous  Girder  of  Variable  Mom.  of  Inertia.  —  If  /  is 
variable  and  IQ  denote  the  mom.  of  inertia  of  some  con- 
venient standard  section,  then  we  may  write  /  =  I0  -f-  m, 
when  ra  denotes  the  number  of  times  IQ  contains  /.  In  a 
non-prismatic  beam,  ra  is  different  for  different  sections 
but  is  easily  found,  and  will  be  considered  given  at  each 
section. 

In  eq.  (1)  of  §  391,  then,  we  must  put  IQ  -*•  m  in  place  of 
/  and  thus  write 

d*y  __  [mMdx]  /-ft 

~ 


and  (pursuing  the  same  reasoning  as  there  given)  may 
therefore  say  that  in  a  girder  of  variable  section  if  each 
small  vertical  strip  (Mdx)  of  the  moment-area  be  multiplied  by 
the  value  of  m  proper  to  that  section,  and  these  products  (or  "vir- 
tual mom.-area  strips)  considered  as  loads,  the  elastic  curve  is  an 
equilibrium  polygon  for  these  loads  with  a,  pole  distance  =  EI0. 

In  modifying  §  400  for  a  girder  of  variable  section,  then, 
besides  taking  E!Q  -f-  n  as  pole  distance,  proceed  as 
follows  : 

Construct  the  positive  mom.-area  for  each  span  accord- 
ing to  §  399  ;  for  each  z  of  Fig.  447,  substitute  mz  (each  2 


CONTINUOUS    GIRDER    BY    GRAPHICS.  51'J 

having  in  general  its  own  m),  and  thus  obtain  the  "  virtual 
positive  mom. -area,"  and  its  gravity  vertical. 

Similarly,  there  will  be  an  unknown  "  virtual  neg.  mom.' 
area"  not  triangular,  replacing  each  neg.  mom.-area  of 
§  400.  Though  it  is  not  triangular,  each  of  its  ordi- 
nates  equals  the  corresponding  ordinate  of  the  unknown 
triangular  neg.  mom. -area  multiplied  by  the  proper  m,  and 
its  gravity -vertical  (which  is  independent  of  the  amount  of  the 
unknown  neg.  mom.-area)  is  found  in  advance  by  the  process 
of  Fig.  447,  using,  for  B'S,  a  set  of  ordinates  obtained  thus : 
Draw  any  two  straight  lines  AB  and  FB9  Fig.  445,  (for  a 
left-hand  trial  neg.  mom.-area;  or  FB  and  GFio?  a  right- 
hand  one)  meeting  in  the  end-vertical  of  the  span,  divide 
the  span  into  ten  or  twenty  equal  spaces,  draw  a  vertical 
through  the  middle  of  each,  noting  their  intercepts  between 
AB  and  FB.  Add  these  intercepts  and  call  the  sum  S. 
Multiply  each  intercept  by  the  proper  m,  and  with  these 
new  values  as  a's  construct  their  gravity  vertical  as  in  Fig. 
447.  Add  these  new  intercepts,  call  the  sum  $T,  and  denote 
the  quotient  S  -r-  Sv  by  /?. 

"We  substitute  the  three  verticals  mentioned,  therefore, 
for  the  mom.-area  verticals  of  §  400,  and  the  "  virtual  pos* 
rnom.-area  "  for  the  pos.  mom.-area,  in  each  span ;  pro- 
ceed in  other  respects  to  construct  the  "  false  polygons  " 
according  to  §  400.  Then  the  result  of  applying  the  con- 
struction in  the  note  §  397  will  be  the  "  virtual  neg.  mom.- 
areas,"  each  of  which  is  to  be  multiplied  by  the  proper 
/3  to  obtain  the  corresponding  triangular  neg.  mom.-area, 
with  which  we  then  proceed,  without  further  modifica- 
tions in  the  process,  according  to  (2),  (3),  etc.  of  §  402. 

[The  conception  of  these  "  virtual  mom. -areas  "  is  due 
to  Prof.  Eddy  ;  see  p.  36  of  his  "  Kesearches  in  Graphical 
Statics,"  referred  to  in  the  preface  of  this  work.] 

405.  Remarks, — It  must  be  remembered  that  any  unequal 
settling  of  the  supports  after  the  girder  has  been  put  in 
place,  may  cause  considerable  changes  in  the  values  of  the 
moments,  shears,  etc.,  and  thus  cause  the  actual  stresses  to 
be  quite  different  from  those  computed  without  taking 


514  MECHANICS  OF   ENGINEERING. 

into  account  a  possible  change  in  the  heights  of  the  sup- 
ports. See  §  271. 

For  example,  if  some  of  the  supports  are  of  masonry, 
while  others  are  the  upper  extremities  of  high  iron  or 
steel  columns,  the  fluctuations  of  length  in  the  latter  due 
to  changes  of  temperature  will  produce  results  of  the 
nature  indicated  above. 

If  an  open-work  truss  of  homogeneous  design  from  end 
to  end  (treated  as  a  girder  of  constant  moment  of  inertia^ 
whose  value  may  be  formulated  as  in  §  388,)  is  used  as  a 
continuous  girder  under  moving  loads,  it  will  be  subject 
to  "  reversal  of  stress  "  in  some  of  its  upper  and  lower  hor- 
izontal members,  i.e.,  the  latter  must  be  of  a  proper  de- 
sign to  sustain  both  tension  and  compression,  (according 
to  the  position  of  the  moving  loads,)  and  this  may  disturb 
the  assumption  of  homogeneity  of  design.  Still,  if  /  is 
variable,  §  404  can  be  used ;  but  since  the  weight  of  the 
truss  must  be  considered  as  part  of  the  loading,  several 
assumptions  and  approximations  may  be  necessary  before 
establishing  satisfactory  dimensions. 

Prof.  Cotterill's  Applied  Mechanics  contains  an  admirable 
semi-graphic  demonstration  of  Clapeyron's  Theorem  of 
Three  Moments  (already  mentioned  on  p.  332  of  this  work). 


PART  IV. 

HYDRAULICS. 


CHAPTEE  I. 
DEFINITIONS— FLUID  PRESSURE— HYDROSTATICS  BEGUN. 

406.  A  Perfect  Fluid  is  a  substance  the  particles  of  which 
are  capable  of  moving  upon  each  other  with  the  greatest  free 
dom,  absolutely  without  friction,  and  are  destitute  of  mutual 
attraction.     In  other  words,  the  stress  between  any  two  con- 
tiguous portions  of  a  perfect  fluid  is  always  one  of  compression 
and  normal  to  the  dividing  surface  at  every  point ;  i.e.,  no 
shear  or  tangential  action  can  exist  on  any  imaginary  cutting 
plane. 

Hence  if  a  perfect  fluid  is  contained  in  a  vessel  of  rigid  ma- 
terial  the  pressure  experienced  by  the  walls  of  the  vessel  is 
normal  to  the  surface  of  contact  at  all  points. 

For  the  practical  purposes  of  Engineering,  water,  alcohol, 
mercury,  air,  steam,  and  all  gases  may  be  treated  as  perfect 
fluids  within  certain  limits  of  temperature. 

407.  Liquids  and  Gases. — A  fluid  a  definite  mass  of  which 
occupies  a  definite  volume  at  a  given  temperature,  and  is  in- 
capable both  of  expanding  into  a  larger  volume  and  of  being 
compressed  into  a  smaller  volume  at  that  temperature,  is  called 
a  Liquid,  of  which  water,  mercury,  etc.,  are  common  examples; 
whereas  a  Gas  is  a  fluid  a  mass  of  which  is  capable  of  almost 
indefinite  expansion  or  compression,  according  as  the  space 
within  the  confining  vessel  is  made  larger  or  smaller,  and  al- 
ways tends  to  fill  the  vessel,  which  must  therefore  be  closed  in 
every  direction  to  prevent  its  escape. 

515 


516  MECHANICS    OF   ENGINEERING. 

Liquids  are  sometimes  called  inelastic  fluids,  and  gases 
elastic  fluids. 

408.  Remarks. — Though  practically  we  may  treat  all  liquids 
as  incompressible,  experiment  shows  them  to  be  compressible 
to  a  slight  extent.  Thus,  a  cubic  inch  of  water  under  a  pres- 
sure of  15  Ibs.  on  each  of  its  six  faces  loses  only  fifty  millionths 
(0.000050)  of  its  original  volume,  while  remaining  at  the  same 
temperature;  if  the  temperature  be  sufficiently  raised,  how- 
ever, its  bulk  will  remain  unchanged  (provided  the  initial  tem- 
perature is  over  40°  Fahr.).  Conversely,  by  heating  a  liquid  in 
a  rigid  vessel  completely  filled  by  it,  a  great  bursting  pressure 
may  be  produced.  The  slight  cohesion  existing  between  the 
particles  of  most  liquids  is  too  insignificant  to  be  considered  in 
the  present  connection. 

The  property  of  indefinite  expansion,  on  the  part  of  gases, 
by  which  a  confined  mass  of  gas  can  continue  to  fill  a  confined 
space  which  is  progressively  enlarging,  and  exert  pressure 
against  its  walls,  is  satisfactorily  explained  by  the  "  Kinetic 
Theory  of  Gases,"  according  to  which  the  gaseous  particles  are 
perfectly  elastic  and  in  continual  motion,  impinging  against 
each  other  and  the  confining  walls.  Nevertheless,  for  prac- 
tical purposes,  we  may  consider  a  gas  as  a  continuous  sub- 
stance. 

Although  by  the  abstraction  of  heat,  or  the  application  of 
great  pressure,  or  both,  all  known  gases  may  be  reduced  to 
liquids  (some  being  even  solidified);  and  although  by  con- 
verse processes  (imparting  heat  and  diminishing  the  pressure) 
liquids  may  be  transformed  into  gases,  the  range  of  tempera- 
ture and  pressure  in  all  problems  to  be  considered  in  this  work 
is  supposed  kept  within  such  limits  that  no  extreme  changes  of 
state,  of  this  character,  take  place.  A  gas  approaching  the 
point  of  liquefaction  is  called  a  Vapor. 

Between  the  solid  and  the  liquid  state  we  find  all  grades  of 
intermediate  conditions  of  matter.  For  example,  some  sub- 
stances are  described  as  soft  and  plastic  solids,  as  soft  putty, 
moist  earth,  pitch,  fresh  mortar,  etc.;  and  others  as  viscous  and 
sluggish  liquids,  as  molasses  and  glycerine.  In  sufficient  bulk, 


DEFINITIONS  —  FLUID   PRESSURE  —  HYDROSTATICS.    617 


however,  the  latter  may  still  be  considered  as  perfect  fluids. 
Even  water  is  slightly  viscous. 

409.  Heaviness  of  Fluids.  —  The  weight  of  a  cubic  unit  of  a 
homogeneous  fluid  will  be  called  its  heaviness,  or  rate  of 
weight  (see  §  7),  and  is  a  measure  of  its  density.  Denoting  it 
by  y,  and  the  volume  of  a  definite  portion  of  the  fluid  by  V, 
we  have,  for  the  weight  of  that  portion, 


=  Vy. 


This,  like  the  great  majority  of  equations  used  or  derived  in 
this  work,  is  of  homogeneous  form  (§  6),  i.e.,  admits  of  any  sys- 
tem of  units.  E.g.,  in  the  metre-kilogram  second  system,  if  y 
is  given  in  kilos,  per  cubic  metre,  V  must  be  expressed  in 
cubic  metres,  and  G  will  be  obtained  in  kilos.;  and  similarly 
in  any  other  system.  The  quality  of  7,  —  G- -±  V,  is  evidently 
one  dimension  of  force  divided  by  three  dimensions  of  length. 

In  the  following  table,  in  the  case  of  gases,  the  temperature 
and  pressure  are  mentioned  at  which  they  have  the  given 
heaviness,  since  under  other  conditions  the  heaviness  would  be 
different ;  in  the  case  of  liquids,  however,  for  ordinary  pur- 
poses the  effect  of  a  change  of  temperature  may  be  neglected 
(within  certain  limits). 


HEAVINESS  OF  VARIOUS  FLUIDS.* 

[In  ft.  Ib.  sec.  system;  y  =  weight  in  Ibs.  of  a  cubic  foot.] 


Liquids. 


j  At  temp,  of  melting  ice;  and  14.7 
-j     lbg    persq  in    tension- 


Freshwater,  y—  62.5 

Sea  water 64.0 

Mercury 848.7 

Alcohol 49.3 

Crude  Petroleum,  about 55.0 

(N.B. — A  cubic  inch  of  water 
weighs  0  036024  Ibs.;  and  a  cubic 
foot  1000  av.  oz.) 


Atmospheric  Air 0. 08076 

Oxygen 0.0892 

Nitrogen 0.0786 

Hydrogen 0.0056 

Illuminating  )  from 0  0300 

Gas,  fto 0.0400 

Natural  Gas,  about 0.0500 


*  See  Trautwine's  Civ.  Engineer's  Pocket  Book  for  an  extended  table— 
p.  380,  edition  of  1885. 


518 


MECHANICS   OF   ENGINEERING. 


For  use  in  problems  where  needed,  values  for  the  heaviness 
of  pure  fresh  water  are  given  in  the  following  table  (from 
Rossetti)  for  temperatures  ranging  from  freezing  to  boiling ; 
as  also  the  relative  density,  that  at  the  temperature  of  maxi- 
mum density,  39°. 3  Fahr.  being  taken  as  unity.  The  temper- 
atures are  Fahr.,  and  y  is  in  Ibs.  per  cubic  foot. 


Temp. 

Rel. 
Dens. 

•y- 

Temp. 

Rel. 
Dens. 

•v-   - 

Temp. 

Rel. 
Dens. 

Y- 

32° 

.99987 

62.416 

60° 

.99907 

62.366 

140° 

.98338 

61.886 

35° 

.99996 

62.421 

70° 

.99802 

62.300 

150° 

.98043 

61.203 

39°.3 

1.00000 

62.424 

80° 

.99669 

62.217 

160° 

.97729 

61.006 

40° 

.99999 

62.423 

90° 

.99510 

62.118 

170° 

.97397 

60.799 

43° 

.99997 

62.422 

100° 

.99318 

61.998 

180° 

.97056 

60.586 

45° 

.99992 

62.419 

110° 

.99105 

61.865 

190° 

.96701 

60.365 

50° 

.99975 

62.408 

120° 

.98870 

61.719 

200° 

.96333 

60.135 

55° 

.99946 

62.390 

130° 

.98608 

61.555 

212° 

.95865 

59.843 

From  D.  K.  Clark's  j  for  temp.  = 
"Manual."     \      Y  — 

230° 
59  4 

250° 

58.7 

270° 

58.2 

290° 
57.6 

298° 
57.3 

338° 
56.1 

366° 
55.3 

390° 
54.5 

EXAMPLE  1.  What  is  the  heaviness  of  a  gas,  432  cub.  in.  of 
which  weigh  0.368  ounces?     Use  ft.-lb.-sec.  system. 

432  cub.  in.  =     cub.  ft.  and  0.368  oz,  =  0.028  Ibs. 


f1 


=T=  -T 


=  0.092  Ibs.  per  cub.  foot.  • 


EXAMPLE  2.  Required  the  weight  of  a  right  prism  of  mer- 
cury of  1  sq.  inch  section  and  30  inches  altitude. 

V  =30  X  1  =  30  cub.  in.  =  -||_-  cub.  feet  ;  while  from  the 
table,  y  for  mercury  =  848.7  Ibs.  per  cub.  ft. 

OA 

.-.  its  weight  =  O  =  Vy  =  -^--  X  848.7  =  14.73  Ibs. 


410.  Definitions.  —  By  Hydraulics  we  understand  the  me- 
chanics of  fluids  as  utilized  in  Engineering.  It  may  be  divided 
into 

Hydrostatics,  treating  of  fluids  at  rest  'and 

Hydrodynamics  (ov  Hydrokinetics),  which  deals  with  fluids 
in  motion.  (The  name  Pneumatics  is  sometimes  used  to  cover 
both  the  statics  and  dynamics  of  gaseous  fluids.) 


DEFINITIONS  —  FLUID   PRESS  UKE—  HYDKOSTATIOS.    619 

[Rankine's  nomenclature  has  been  adopted  in  the  present 
work.  Some  recent  writers  use  the  term  Hydromechanics  for 
mechanics  of  fluids,  subdividing  it  into  Hydrostatics  and 
Hydrokinetics,  as  above  ;  they  also  use  the  term  Dynamics  to 
embrace  both  of  the  two  divisions  called  Statics  and  Dynamics 
by  Rankine,  which  by  them  are  called  Statics  and  Kinetics  re- 
spectively. Though  unusual,  perhaps,  the  term  Hydraulics  is 
here  used  to  cover  the  applied  Mechanics  of  Gases  as  well  as 
of  Liquids.] 

Before  treating  separately  of  liquids  and  gases,  a  few  para- 
graphs will  be  presented  applicable  to  both  kinds  of  fluids. 

411.  Pressure  per  Unit  Area,  or  Intensity  of  Pressure.  —  As  in 

§  180  in  dealing  with  solids,  so  here  with  fluids  we  indicate  the 
pressure  per  unit  area  between  two  contiguous  portions  of 
fluid,  or  between  a  fluid  and  the  wall  of  the  containing  vessel, 
by  p,  so  that  if  dP  is  the  total  pressure  on  a  small  area  dF, 
we  have 

dP 

.......     (1) 


as  the  pressure  per  unit  area,  or  intensity  of  pressure  (often, 
though  ambiguously,  called  the  tension  in  speaking  of  a  gas) 
on  the  small  surface  dF.  If  pressure  of  the  same  intensity 
exists  over  a  finite  plane  surface  of  area  =  F^  the  total  pres- 
sure on  that  surface  is 


or 


P  =  fpdF=pfdF=  Fp,  } 

P  [     ....    (2) 


(N.B.—  For  brevity  the  single  word  "  pressure"  will  some- 
times be  used,  instead  of  intensity  of  pressure,  where  no  am- 
biguity can  arise.)  Thus,  it  is  found  that,  under  ordinary  con- 
ditions at  the  sea  level,  the  atmosphere  exerts  a  normal  pressure 
(normal,  because  fluid  pressure)  on  all  surfaces,  of  an  intensity 
of  about  p  —  14*7  Ibs.  per  sq.  inch  (=  2116.  Ibs.  per  sq.  ft.). 
This  intensity  of  pressure  is  called  one  atmosphere.  For  ex- 


520 


MECHANICS    OF   ENGINEERING. 


ample,  the  total  atmospheric  pressure  on  a  surface  of  100  sq. 
in.  is  [inch,  lb.,  sec.] 

P  =  Fp  =  100  X  14.7  =.1470  Ibs.     (=  0.735  tons.) 

The  quality  of  p  is  evidently  one  dimension  of  force  divid- 
ed by  two  dimensions  of  length. 

412.  Hydrostatic  Pressure ;  per  Unit  Area,  in  the  Interior  of  a 
Fluid  at  Rest. — In  a  body  of  fluid  of  uniform  heaviness,  at 
rest,  it  is  required  to  find  the  mutual  pressure  per  unit  area  be- 
tween the  portions  of  fluid  on  opposite  sides  of  any  imaginary 
cutting  plane.  As  customary,  we  shall  consider  portions  of 
the  fluid  as  free  bodies,  by  supplying  the  forces  exerted  on 
them  by  all  contiguous  portions  (of  fluid  or  vessel  wall),  also 
those  of  the  earth  (their  weights),  and  then  apply  the  condi- 
tions of  equilibrium. 

First,  cutting  plane  horizontal. — Fig.  451  shows  a  body  of 
homogeneous  fluid  confined  in  a  rigid 
vessel  closed  at  the  top  with  a  small  air- 
tight but  frictionless  piston  (a  horizontal 
disk)  of  weight  =  G  and  exposed  to  at- 
mospheric pressure  (=pa  P^r  unit  area) 
on  its  upper  face.  Let  the  area  of  piston- 
face  be  =  F.  Then  for  the  equilibrium 
of  the  piston  the  total  pressure  between 
its  under  surface  and  the  fluid  at  0  must 


Fia.  451. 


be 


and  hence  the  intensity  of  this  pressure  is 


(1) 


It  is  now  required  to  find  the  intensity,^?,  of  fluid  pressure 
between  the  portions  of  fluid  contiguous  to  the  horizontal  cut- 
ting plane  J?£7at  a  vertical  distance  =  h  vertically  below  the  pis- 
ton 0.  In  Fig.  452  we  have  as  a  free  body  the  right  parallelo- 


FLUID    PRESSURE. 


521 


piped  OBC  of  Fig.  451  with  vertical  sides  (two  ||  to  paper  and 
four  1  to  it).  The  pressures  acting  on  its  six  faces  are  normal 
to  them  respectively,  and  the  weight  of  the  prism  is  =  vol. 
Xy  =  Fhy,  supposing  y  to  have  the  same  value  at  all  parts  of 
the  column  (which  is  practically  true  for  any  height  of  liquid 
and  for  a  small  height  of  gas).  Since  the  f — • — NFpo 

prism  is  in  equilibrium   under  the  forces  oUIll 

shown  in  the  figure,  and  would  still  be  so 
were  it  to  become  rigid,  we  may  put  (§  36) 
2  (vert,  compons.)  =  0  and  hence  obtain 

Fp  _  Fp,  -  Fhy  =  0.  .     .     (2) 

(In  the   figure  the  pressures  on  the  ver- 
tical faces  ||  to  paper  have  no  vertical  com- 
ponents, and  hence  are  not  drawn.)     From  FIG.  458. 
(2)  we  have 


(3) 


(hy,  being  the  weight  of  a  column  of  homogeneous  fluid  of  unity 
cross-section  and  height  A,  would  be  the  total  pressure  on  the 
base  of  such  a  column,  if  at  rest  and  with  no  pressure  on  the 
upper  base,  and  hence  might  be  called  intensity  due  to  weight.) 
Secondly,  cutting  plane  oblique. — Fig.  453.  Consider  free 
an  infinitely  small  right  triangular  prism  ~bcd,  whose  bases  are 

||  to  the  paper,  while  the  three  side 
faces  (rectangles),  having  areas  =  dF, 
dF, ,  and  dF^ ,  are  respectively  hori- 
zontal, vertical,  and  oblique  ;  let  angle 
cbd  =  of.  The  surface  be  is  a  portion 
of  the  plane  BC  of  Fig.  452.  Given 
p  (—  intensity  of  pressure  on  dF)  and 
<*,  required^,  the  intensity  of  pressure 
on  the  oblique  face  M,  of  area  dF^ 
[N.  B. — The  prism  is  taken  very  small 
in  order  that  the  intensity  of  pressure  may  be  considered  con- 
stant over  any  one  face ;  and  also  that  the  weight  of  the  prism 
may  be  neglected,  since  it  involves  the  volume  (three  dimen- 


FIG.  453. 


522  MECHANICS    OF  ENGINEERING. 

eions)  of  the  prism,  while  the  total  face  pressures  involve  only 
two,  and  is  hence  a  differential  of  a  higher  order.] 
From  2  (vert,  compons.)  =  0  we  shall  have 

pjLF,,  cos  a  —  pdF—  0  ;  but  dF~-  dF.,  —  cos  a  ; 


which  is  independent  of  the  angle  a. 

Hence,  the  intensity  of  fluid  pressure  at  a  given  point  is 
the  same  on  all  imaginary  cutting  planes  containing  the 
point.  This  is  the  most  important  property  of  a  fluid,  and  is 
true  whether  the  liquid  is  at  rest  or  has  any  kind  of  motion  ; 
for,  in  case  of  rectilinear  accelerated  motion,  e.g.,  although  the 
sum  of  the  force-components  in  the  direction  of  the  accelera- 
tion does  not  in  general  =  0,  but  =  mass  X  ace.,  still,  the 
mass  of  the  body  in  question  is  =  weight  -r-  <?,  and  therefore 
the  term  mass  X  ace.  is  a  differential  of  a  higher  order  than 
the  other  terms  of  the  equation,  and  hence  the  same  result 
follows  as  when  there  is  no  motion  (or  uniform  rectilinear 
motion). 

413.  The  Intensity  of  Pressure  is  Equal  at  all  Points  of  any 
Horizontal  Plane  in  a  body  of  homogeneous  fluid  at  rest.  If 
we  consider  a  right  prism  of  the  fluid  in  Fig.  451,  of  small 
vertical  thickness,  its  axis  lying  in  any  horizontal  plane  J3C, 
its  bases  will  be  vertical  and  of  equal  area  dF.  The  pressures 
on  its  sides,  being  normal  to  them,  and  hence  to  the  axis,  have 
no  components  ||  to  the  axis.  The  weight  of  the  prism  also 
has  no  horizontal  component.  Hence  from  2  (hor.  comps. 
||  to  axis)  =  0,  we  have,  pl  and  pz  being  the  pressure-intensi- 
ties at  the  two  bases, 

p,dF-p,dF=b\    .:p=p,,    .    .    .    .    (1) 

which  proves  the  statement  at  the  head  of  this  article. 

It  is  now  plain,  from  this  and  the  preceding  article,  that 
the  pressure-intensity  p  at  any  point  in  a  homogeneous  fluid 
at  rest  is  equal  to  that  at  any  higher  point,  plus  the  weight 


FLUID    PRESSURE. 


523 


(hy)  of  a  column  of  the  fluid  of  section  unity  and  of  altitude 
(A)  =  vertical  distance  between  the  points. 


P  =Po 


(2) 


whether  they  are  in  the  same  vertical  or  not,  and  whatever  be 

the  shape   of  the   containing 

vessel  (or  pipes],  provided  the 

fluid  is   continuous   between 

the  two  points;  for,  Fig.  454, 

by   considering    a    series    of 

small  prisms,  alternately  ver- 

tical and  horizontal,  obcde,  we 


know  that 


FiG.454. 


;       Pc=pb', 

Pd=pc  —  hjs  ;  and  pe—p&\ 
hence,  finally,  by  addition  we  have 


(in  which  h  =  h^  —  A2). 

If,  therefore,  upon  a  small  piston  at  o,  of  area  =  7^,  a  force 
P0  be  exerted,  and  an  inelastic  fluid  (liquid)  completely  fills  the 
vessel,  then,  for  equilibrium,  the  force  to  be  exerted  upon  the  pis- 
ton at  0,  viz.,  Pe  ,  is  thus  computed  :  For  equilibrium  of  fluid 
Pe  =Po  +  ^r  ;  and  for  equil.  of  piston  0,  p0  =  P0  ~  J?Q  ;  also, 


(3) 


From  (3)  we  learn  that  if  the  pistons  are  at  the  same  level 
(h  =  0)  the  total  pressures  on  their  inner  faces  are  directly 
proportional  to  their  areas. 

If  the  fluid  is  gaseous  (2)  and  (3)  are  practically  correct  if 
h  is  not  >  100  feet  (for,  gas  being  compressible,  the  lower 
strata  are  generally  more  dense  than  the  upper),  hut  in  (3)  the 
pistons  must  be  fixed^and  Pe  and  P0  refer  solely  to  the  in- 
terior pressures* 


524  MECHANICS    OF   ENGINEERING. 

Again,  if  h  is  small  or  p0  very  great,  the  term  hy  may  be 
omitted  altogether  in  eqs.  (2)  and  (3)  (especially  with  gases, 
since  for  them  y  (heaviness)  is  usually  small),  and  we  then 
have,  from  (2), 

P=P*',    ........     (4) 


being  the  algebraic  form  of  the  statement  :  A  body  o 
at  rest  transmits  pressure  with  equal  intensity  in  every  direc- 
tion and  to  all  of  its  parts.  [Principle  of  "  Equal  Transmis- 
sion of  Pressure."] 

414.  Moving  Pistons.  —  If  the  fluid  In  Fig.  454  is  inelastic 
and  the  vessel  walls  rigid,  the  motion  of  one  piston  (o)  through 
a  distance  s0  causes  the  other  to  move  through  a  distance  se  de- 
termined by  the  relation  F^Q  =  Fese  (since  the  volumes  de- 
scribed by  them  must  be  equal,  as  liquids  are  incompressible); 
but  on  account  of  the  inertia  of  the  liquid,  and  friction  on  the 
vessel  walls,  equations  (2)  and  (3)  no  longer  hold  exactly,  still 
are  approximately  true  if  the  motion  is  very  slow  and  the 
vessel  short,  as  with  the  cylinder  of  a  water-pressure  engine. 

But  if  the  fluid  is  compressible  and  elastic  (gases  and  vapors  ; 
steam,  or  air)  and  hence  of  small  density,  the  effect  of  inertia 
and  friction  is  not  appreciable  in  short  wide  vessels  like  the 
cylinders  of  steam-  and  air-engines,  and  those  of  air-compres- 
sors ;  and  eqs.  (2)  and  (3)  still  hold,  practically,  even  with  high 
piston-speeds.  For  exam  pie,  in  the  space  AB, 
^S-  455,  between  the  piston  and  cylinder-head 
of  a  steam-engine  (piston  moving  toward  the 
right)  the  intensity  of  pressure,  p,  of  the 
steam  against  the  moving  piston  B  is  prac- 


FIO.  455.          tically  equal  to  that  against  the  cylinder-head 
A  at  the  same  instant. 

415.  An  Important  Distinction  between  gases  and  liquids 
(i.e.,  between  elastic  and  inelastic  fluids)  consists  in  this : 

A  liquid  can  exert  pressure  against  the  walls  of  the  contain- 
ing vessel  only  by  its  weight,  or  (when  confined  on  all  sides) 
by  transmitted  pressure  coming  from  without  (due  to  piston 
pressure,  atmospheric  pressure,  etc.);  whereas — 


FLUID   PRESSURE.  525 

A  gas,  confined,  as  it  must  be,  on  all  sides  to  prevent  dif- 
fusion, exerts  pressure  on  the  vessel  not  only  by  its  weight, 
but  by  its  elasticity  or  tendency  to  expand.  If  pressure  from 
without  is  also  applied,  the  gas  is  compressed  and  exerts  a  still 
greater  pressure  on  the  vessel  walls. 

416.  Component,  of  Pressure,  in  a  Given  Direction.  —  Let 
ABCD,  whose  area  =  dF,  be  a  small  element  of  a  surface, 
plane  or  curved,  and  p  the  intensity  of 
fluid  pressure  upon  this  element,  then 
the  total  pressure  upon  it  \$>pdF,  and  is 
of  course  normal  to  it.  Let  A  '  B'  CD  be 
the  projection  of  the  element  dF  upon 
a  plane  CDM  making  an  angle  a  with 
the  element,  and  let  it  be  required  to 
find  the  value  of  the  component  of  pdF 
in  a  direction  normal  to  this  last  plane  (the  other  component 
being  understood  to  be  ||  to  the  same  plane).  We  shall  have 


Compon.  ofpdF  ~]  to  CDM  =  pdFcos  a  =p(dF.coB  a).  (I) 

But  dF  .  cos  a  =  area  A'B'CD,  the  projection  of  dF  upon 
the  plane  CDM. 

.-.  Compon.  "1  to  plane  CDM.  =p  X  (project,  of  dF  on  CDM)\ 

i.e.,  the  component  of  fluid  pressure  (on  an  element  of  a  sur- 
face) in  a  given  direction  (the  other  component  being  "|  to 
the  first)  is  found  l>y  multiplying  the  intensity  of  'the  pressure 
by  the  area  of  the  projection  of  the  element  upon  a  plane  ~|  to 
the  give?i  direction. 

It  is  seen,  as  an  example  of  this,  that  if  the  fluid  pressures 
on  the  elements  of  the  inner  surface  of  one  hemisphere  of  a 
hollow  sphere  containing  a  gas  are  resolved  into  components  "I 
and  ||  to  the  plane  of  the  circular  base  of  the  hemisphere,  the 
sum  of  the  former  components  simply  ==  nr*p,  where  r  is  the 
radius  of  the  sphere,  and  p  the  intensity  of  the  fluid  pressure  ; 
for,  from  the  foregoing,  the  sum  of  these  components  is  just 
the  same  as  the  total  pressure  would  be,  having  an  intensity  p, 


526 


MECHANICS    OF   ENGINEERING. 


oil  a  great  circle  of  the  sphere,  the  area,  7r/>a,  of  this  circle  being 
the  sum  of  the  areas  of  the  projections,  upon  this  circle  as  a 
base,  of  all  the  elements  of  the  hemispherical  surface.  (Weight 
of  fluid  neglected.) 

A  similar  statement  may  be  made  as  to  the  pressures  on 
the  inner  curved  surface  of  a  right  cylinder. 

417.  Non-planar  Pistons. — From  the  foregoing  it  follows  that 
the  sum  of  the  components  ||  to  the  piston-rod,  of  the  fluid 
pressures  upon  the  piston  at  A,  Fig.  457,  is  just  the  same  as  at 
B^  if  the  cylinders  are  of  equal  size  and  the  steam,  or  air,  is  at 
the  same  tension.  For  the  sum  of  the  projections  of  all  the 
elements  of  the  curved  surface  of  A  upon  a  plane  ~|  to  the 
piston-rod  is  always  =  TIT*  =  area  of  section  of  cylinder-bore. 


FIG.  457. 

If  the  surface  of  A  is  symmetrical  about  the  axis  of  the  cylin- 
der the  other  components  (i.e.,  those  ~|  to  the  piston-rod)  will 
neutralize  each  other.  If  the  line  of  intersection  of  that  sur- 
face with  the  surface  of  the  cylinder  is  not  symmetrical  about 
the  axis  of  the  cylinder,  the  piston  may  be  pressed  laterally 
against  the  cylinder- wall,  but  the  thrust  along  the  rod  or 
" working  force'*  (§  128)  is  the  same  (except  for  friction  in- 
duced by  the  lateral  pressure),  in  all  instances,  as  if  the  surface 
were  plane  and  ~1  to  piston-rod. 

418.  Bramah,  or  Hydraulic,  Press. — This  is  a  familiar  instance 
of  the  principle  of  transmission  of  fluid  pressure.  Fig.  458. 
Let  the  small  piston  at  0  have  a  diameter  d  =  1  inch  =  -fa  ft., 
while  the  plunger  E,  or  large  piston,  has  a  diameter  df  =  AB 
=  CD  =  15  in.  =  f  ft.  The  lever  MJV  weighs  #,  =  3  Ibs., 
and  a  weight  O  =  40  Ibs.  is  hung  at  M.  The  lever-arms  of 
these  forces  about  the  fulcrum  N  are  given  in  the  figure. 
The  apparatus  being  full  of  water  (oil  is  otten  used),  the  fluid 
pressure  P9  against  the  small  piston  is  found  by  putting 


FLUID   PRESSURE. 


527 


.  about  ^T)  =  0  for  the  equilibrium  of  the  lever  £ 
whence  [ft.,  lb.,  sec.] 

P9  x  i  _  40  X  3  -  3  X  2  =  0.     .-.  P0  =  126  Ibs. 


FIG.  458. 


But,  denoting  atmospheric  pressure  by  pa,  and  that  of  the 
water  against  the  piston  by  p0  (per  unit  area),  we  may  also 
write 

.  -pa\ 


Solving  for  jp9  ,  we  have,  putting  pa  —  14.7  X  144  Ibs.  per 

sq.  ft., 

p.  =  [l26  -f-  1  (-^)3J  +  14.7  X  144  =  25236  Ibs.  per  sq.  ft. 


Hence  at  e  the  press,  per  unit  area,  from  §  409,  and  (2),  §  413,  is 
pe  =p0  +  hy=  25236  +  3  X  62.5  =  25423  Ibs.  per  sq.  ft. 

=  175.6  Ibs.  per  sq.  inch  or  11.9  atmospheres,  and  the  total 
upward  pressure  at  e  on  base  of  plunger  is 

P  =  Fepe  =  n  ~pe  =  J  *(Jff  x  25423  =  31194  Ibs., 

or  almost  16  tons  (of  2000  Ibs.  each).  The  corapressive  force 
upon  the  block  or  bale,  C,  —  P  less  the  weight  of  the  plunger 
and  total  atmos.  pressure  on  a  circle  of  15  in.  diameter. 


52S 


MECHANICS    OF   ENGINEERING. 


419.  The  Dividing  Surface  of  Two  Fluids  (which  do  not  mix)  in 
Contact,  and  at  Rest,  is  a  Horizontal  Plane. — For,  Fig.  459,  sup- 
posing any  two  points  e  and  0  of  this  sur- 
face to  be  at  different  levels  (the  pressure 
at  0  being  j?0,  that  at  epe,  and  the  heavi- 
nesses of  the  two  fluids  yl  and  y\  respec- 
tively), we  would  have,  from  a  considera- 
tion of  the  two  elementary  prisms  eb  and 
10  (vertical  and  horizontal;,  the  relatioa 


FIG.  459. 


while  from  the  prisms  ec  and  cO,  the  relation 


These  equations  are  conflicting,  hence  the  aoove  supposition 
is  absurd.     Therefore  the  proposition  is  true. 

For  stable  equilibrium,  evidently,  the  heavier  fluid  must  oc- 
cupy the  lowest  position  in  the  vessel,  and  if  there  are  several 
fluids  (which  do  not  mix),  they  will  arrange 
themselves  vertically,  in  the  order  of  their  den- 
sities, the  heaviest  at  the  bottom,  Fig.  460.  On 
account  of  the  property  called  diffusion  the  par- 
ticles of  two  gases  placed  in  contact  soon  inter- 
mingle and  form  a  uniform  mixture.  This  fact 
gives  strong  support  to  the  "  Kinetic  Theory  of 
Gases"  (§  408). 


FIG.  460. 


m$ 


420.  Free  Surface  of  a  Liquid  at  Rest, — The  surface  (of  a 
liquid)  not  in  contact  with  the  walls  of  the  containing  vessel 
is  called  a  free  surface,  and  is  necessarily 
horizontal  (from  §  419)  when  the  liquid  is  at 
rest.  Fig.  461.  (A  gas,  from  its  tendency 
to  indefinite  expansion,  is  incapable  of  hav- 
ing a  free  surface.)  This  is  true  even  if  the 
space  above  the  liquid  is  vacuous,  for  if  the 
surface  were  inclined  or  curved,  points  in  the 
body  of  the  liquid  and  in  the  same  horizon- 
tal plane  would  have  different  heights  (or  "  heads")  of  liquid 


TWO   LIQUIDS   IN  BENT  TUBE. 


between  them  and  the  surface,  producing  different  intensities 
of  pressure  in  the  plane,  which  is  contrary  to  §  413. 

When  large  bodies  of  liquid  like  the  ocean  are  considered, 
gravity  can  no  longer  be  regarded  as  acting  in  parallel  lines ; 
consequently  the  free  surface  of  the  liquid  is  curved,  being  ~] 
to  the  direction  of  (apparent)  gravity  at  all  points.  For  ordi- 
nary engineering  purposes  (except  in  Geodesy)  the  free  surface 
of  water  at  rest  is  a  horizontal  plane. 

421.  Two  Liquids  (which  do  not  mix)  at  Rest  in  a  Bent  Tube 
open  at  Both  Ends  to  the  Air,  Fig.  460 ;  water  and  mercury,  for 
instance.  Let  their  heavinesses  be  yL 
and  YI  respectively.  The  pressure  at  e 
may  be  written  (§  413)  either 


or 


JPe  = 


according  as  we  refer  it  to  the  water 
column  or  the  mercury  column  and 
their  respective  free  surfaces  where  the 
pressure^  =Po9  =  Pa  =  atmos.  press. 
«  is  the  surface  of  contact  of  the  two  liquids. 


P 


=Pa 


i.e.,  A, 


Hence  we  have 

r.  :  y,«    •   (3) 


le.,  tike  freights  of  the  free  surfaces  of  the  two  liquids  above  the 
surface  of  contact  are  inversely  proportional  to  their  respec- 
tive heavinesses. 

EXAMPLE.  —  If  the  pressure  at  e  —  2  atmospheres  (§  396)  we 
shall  have  from  (2)  (inch-lb.-sec.  system  of  units) 

A,x,  =  JP,  —  pa  =  2  X  14.7  -  14.T  =  14.7  Ibs.  per  sq.  inch. 

.-.  AB  must  =  14.7  -5-  [848.7  -5-  1728]  =  30  inches 

*- 

(since,  for  mercury,  y9  =  848.7  Ibs.  per  cub.  ft.).  Hence, 
from  (3), 


,         ,. 

A,  =          = 


30 


=  408  inches  =  34  feet. 


530  MECHANICS   OF   ENGINEERING. 

i.e.,  for  equilibrium,  and  thatjye  may  =  2  atmospheres,  A2  and 
hl  (of  mercury  and  water)  must  be  30  in.  and  34  feet  respec- 
tively. 

422.  City  Water-pipes. — If  h  =  vertical  distance  of  a  point 
B  of  a  water-pipe  below  the  free  surface  of  reservoir,  and  the 
water  be  at  rest,  the  pressure  on  the  inner  surface  of  the  pipe 
at  B  (per  unit  of  area)  is 

p  =p0  -\-  hy  ;   and  here  p0  =pa  =  atmos.  press. 
EXAMPLE. — If  h  =  200  ft.  (using  the  inch,  lb.,  and  second) 
p  =  14.7  +  [200  X  12] [62.5  ^-  1728]  =  101.5  Ibs.  per  sq.  in. 

The  term  hy.  alone,  =  86.8  Ibs.  per  sq.  inch,  is  spoken  of  as  the 
hydrostatic  pressure  due  to  200  feet  height,  or  "Head,"  of 
water.  (See  Trautwine's  Pocket  Book  for  a  table  of  hydro- 
static pressures  for  various  depths.) 

If,  however,  the  water  is  flowing  through  the  pipe,  the  pres- 
sure against  the  interior  wall  becomes  less  (a  problem  of  Hy- 
drodynamics to  be  treated  subsequently),  while  if  that  motion 
is  suddenly  checked,  the  pressure  becomes  momentarily  much 
greater  than  the  hydrostatic.  This  shock  is  called  "  water- 
ram"  and  "  water-hammer,"  and  may  be  as  great  as  200  to  300 
Ibs.  per  sq.  inch. 

423.  Barometers  and  Manometers  for  Fluid  Pressure. — If  a 
tube,  closed  at  one  end,  is  filled  with  water,  and  the  other  ex- 
tremity is  temporarily  stopped  and  afterwards 
opened  under  water,  the  closed  end  being  then 
a  (vertical)  height  =  h  above  the  surface  of 
the  water,  it  is  required  to  find  the  intensity, 
p0 ,  of  fluid  pressure  at  the  top  of  the  tube,  sup- 
posing it  to  remain   filled  with  water.     Fig. 
463.     At  E  inside  the   tube  the  pressure  is 
14.7  Ibs.  per  sq.  inch,  the  same  as  that  outside 
at  the  same  level  (§  413) ;  hence,  from  pE  —  pn 

P*=pE-hy (1)' 


BAROMETERS.  531 

EXAMPLE. — Let  h  =  10  feet  (with  inch-lb.-sec.  system) ;  then 
p.  =  14.7  -  120  X  [62.5  H-  1728]  =  10.4  Ibs.  per  sq.  inch, 

or  about  f  of  an  atmosphere.  If  now  we  inquire  the  value 
of  h  to  make  p0  =  zero,  we  put /^  —  hy  =  0  and  obtain  h  = 
408  inches,  =  34  ft.,  which  is  called  the  height  of  the  water- 
barometer.  Hence,  Fig.  463&,  ordinary  atmospheric  pressure 
will  not  sustain  a  column  of  water  higher  than  34  feet.  If 
mercury  is  used  instead  of  water  the  height  supported  by  one 
atmosphere  is 

I  =  14.7  -r-  [848.7  -5-  1728]  =  30  inches, 

=  76  centime,  (about),  and  the  tube  is  of  more  manageable 
proportions  than  with  water,  aside  from  the  ad- 
vantage that  no  vapor  of  mercury  forms  above 
the  liquid  at  ordinary  temperatures.  [In  fact,  the 
water-barometer  height  b  =  34  feet  has  only  a 
theoretical  existence  since  at  ordinary  tempera- 
tures (40°  to  80°  Fahr.)  vapor  of  water  would 
form  above  the  column  and  depress  it  by  from 
0.30  to  1.09  ft.]  Such  an  apparatus  is  called  a 
Barometer,  and  is  used  not  only  for  measuring 
the  varying  tension  of  the  atmosphere  (from  14.5 
to  15  Ibs.  per  sq.  inch,  according  to  the  weather  and  height 
above  sea-level),  but  also  that  of  any  body  of  gas.  Thus,  Fig. 
464,  the  gas  in  ZHsput  in  communication  with 
the  space  above  the  mercury  in  the  cistern  at 
(7;  and  we  have  p  —  hy,  where  y  —  heav.  of 
mercury,  and  JP  is  the  pressure  on  the  liquid  in 
the  cistern.  For  delicate  measurements  an  at- 
tached thermometer  is  also  used,  as  the  heavi- 
ness y  varies  slightly  with  the  temperature. 

If  the  vertical  distance  CD  is  small,  the  ten- 
sion in  C  is  considered  the  same  as  in  D. 

For  gas-tensions  greater  than  one  atmosphere, 
the  tube  may  be  left  open  at  the  top,  forming  an  open  ma- 


FIG.  463a. 


FIG.  464. 


532 


MECHANICS    OF    ENGINEERING. 


nometer,  Fig.  465.     In  this  case,  the  tension  of  the  gag  above 
the  mercury  in  tilt  cistern  is 


FIG.  465. 


in  which  b  is  the  height  of  mercury  (about  30 
in.)  to  which  the  tension  of  the  atmosphere  above 
the  mercury  column  is  equivalent. 


D        EXAMPLE. — If   h  =  51    inches,    Fig.   465,   we 
have  (ft.,  lb.,  sec.) 


p  —  [4.25  ft.  +  2.5  ft.]  848.7  =  5728  Ibs.  per  sq.  foot 
=  39.7  Ibs.  per  sq.  inch  —  2.7  atmospheres. 

Another  form  of  the  open  manometer  consists  of  a  U  tube, 
Fig.  464,  the  atmosphere  having  access  to  one  branch,  the  gas 
to   be  examined,  to  the  other,  while  the 
mercury  lies  in  the  curve.     As  before,  we 
have 

where pa  =  atmos.  tension,  and  b  as  above. 
The  tension  of  a  gas  is  sometimes  spoken 
of  as  measured  by  so  many  inches  of  mer? 
cury.  For  example,  a  tension  of  22.05 
Ibs.  per  sq.  inch  (Ij-  atmos.)  is  measured  by  45  inches  of  mer- 
cury in  a  vacuum  manometer  (i.e.,  a  common  barometer), 
Fig.  464.  With  the  open  manometer  this  tension  (1£  atmos.) 
would  be  indicated  by  15  inches  of  actual  mercury,  Figs.  465 
and  466.  An  ordinary  steam-gauge  indicates  the  excess  of 
tension  over  one  atmosphere ;  thus  "  40  Ibs.  of  steam"  implies 
a  tension  of  40  +  14.7  =  54.7  Ibs.  per  sq.  in. 

The  Bourdon  steam-gauge  in  common  use  consists  of  a 
curved  elastic  metal  tube  of  flattened  or  elliptical  section 
(^with  the  long  axis  ~|  to  the  plane  of  the  tube),  and  has  one 
end  fixed.  The  movement  of  the  other  end,  which  is  free  and 


FIG.  466. 


TENSION   OF   GASES. 


533 


closed,  by  proper  mechanical  connection  gives  motion  to  the 
pointer  of  a  dial.  This  movement  is  caused  by  any  change  of 
tension  in  the  steam  or  gas  admitted,  through  the  fixed  end,  to 
the  interior  of  the  tube.  As  the  tension  increases  the  ellip< 
tical  section  becomes  less  flat,  i.e.,  more  nearly  circular,  caus- 
ing the  two  ends  of  the  tube  to  separate  more  widely,  i.e.,  the 
free  end  moves  away  from  the  fixed  end  ;  and  vice  versa. 

Such  gauges,  however,  are  not  always  reliable.  They  are 
graduated  by  comparison  with  mercury  manometers ;  and 
should  be  tested  from  time  to  time  in  the  same  way.* 

424.  Tension  of  Illuminating  Gas. — This  is  often  spoken  of  as 
measured  by  inches  of  water  (from  1  to  3  inches  usually). 
Strictly  it  should  be  stated  that  this 
water-height  measures  the  excess  of 
its  tension  over  that  of  the  atmos- 
phere. Thus,  in  Fig.  466,  water 
being  used  instead  of  mercury,  h  = 
say  2  inches,  while  b  =  408  inches. 

This  difference  of  tension  may  be 
largely  affected  by  a  change  in  the 
barometer  due  to  the  weather,  or  by 
a  difference  in  altitude,  as  the  follow- 
ing example  will  illustrate: 

EXAMPLE. — Supposing  the  gas  at  rest,  and  the  tension  at  the 
gasometer  A.,  Fig.  467,  to  be  "two  inches  of  water,"  required 
the  water-column  h"  (in  open  tube)  that  the  gas  will  support 
in  the  pipe  at  B,  120  feet  (vertically)  above  the  gasometer. 
Let  the  temperature  be  freezing  (nearly),  and  the  outside  air  at 
a  tension  of  14.7  Ibs.  per  sq.  inch ;  the  heaviness  of  the  gas  at 
this  temperature  being  0.036  Ibs.  per  cubic  foot.  For  the 
small  difference  of  120  ft.  we  may  treat  both  the  atmosphere 
and  the  gas  as  liquids,  that  is,  of  constant  density  throughout 
the  vertical  column,  and  therefore  apply  the  principles  of 
§  413  ;  with  the  following  result : 

The  tension  of  the  outside  air  at  .B,  supposed  to  be  at  the 
same  temperature  as  at  A,  will  sustain  a  water-column  less 
than  the  408  inches  at  A  by  an  amount  corresponding  to  the 

*  Of  late  years  gauges  have  come  into  use  constructed  of  boxes  with  cor- 
rugated sides  of  thin  metal  like  the  aneroid  barometer.  Motion  of  the  sides, 
under  varying  internal  fluid  pressure,  causes  movement  of  a  pointer  on  a  dial. 


FIG.  467. 


534  MECHANICS    OF    ENGINEERING. 

120  feet  of  air  between,  of  the  heaviness  .0807  Ibs.  per  cub. 
ft.  120  feet  of  air  weighing  .0807  Ibs.  per  cub.  ft.  will  balance 
0.154  ft.  of  water  weighing  62.5  Ibs.  per  cubic  ft.,  i.e.,  1.85 
inches  of  water.  Now  the  tension  of  the  gas  at  B  is  also  less 
than  its  tension  at  A,  but  the  difference  is  not  so  great  as  with 
the  outside  air,  for  the  120  ft.  of  gas  is  lighter  than  the  120  ft. 
of  air.  Since  120  ft.  of  gas  weighing  0.036  Ibs.  per  cubic  ft. 
will  balance  0.0691  ft.,  or  0.83  inches,  of  water,  therefore  the 
difference  between  the  tensions  of  the  two  fluids  at  B  is  greater 
than  at  A  by  (1.85  -  0.83  =  )  1.02  inches;  or,  at  B  the  total 
difference  is  2.00  +  1.02  =  3.02  inches. 

,  Hence  if  a  small  aperture  is  made  in  the  pipe  at  B  the  gas 
will  flow  out  with  greater  velocity  than  at  A.  At  Ithaca, 
N.  Y.,  where  the  University  buildings  are  400  ft.  above  the 
gas-works,  this  phenomenon  is  very  marked. 

When  the  difference  of  level  is  great  the  decrease  of  tension 
as  we  proceed  upward  in  the  atmosphere,  even  with  constant 
temperature,  does  not  follow  the  simple  law  of  §413;  see 
§4-77. 

For  velocity  of  flow  of  gases  through  orifices,  see  §  548,  etc. 

425.  Safety-valves. — Fig.  468.  Eequired  the  proper  weight 
G  to  be  hung  at  the  extremity  of  the  horizontal  lever  AB, 

with  fulcrum  at  B,  that  the  flat 

« a Jt         disk- val ve  E  shall  not  be  forced 

upward  by  the  steam  pressure, £/, 
until  the  latter  reaches  a  given 
value  =  p.  Let  the  weight  of 
the  arm  be  Gl ,  its  centre  of  grav- 
ity being  at  C,  a  distance  =  o 
from  B ;  the  other  horizontal  distances  are  marked  in  the 
figure. 

Suppose  the  valve  on  the  point  of  rising;  then  the  forces 
acting  on  the  lever  are  the  fulcrum-reaction  at  B,  the  weights 
G  and  6rl ,  and  the  two  fluid-pressures  on  the  disk,  viz. :  Fp^ 
(atmospheric)  downward,  and  Fp  (steam)  upward.  Hence, 
from  ^(moms.  B)  =  0, 

Gl  +  G,G  +  Fpaa  -  Fpa  =  0.      ...    (1) 


BURSTING   OF  PIPES.  535 

Solving,  we  have 

-A)-0,r '    •    (2) 


EXAMPLE.  —  "With  a  =  2  inches,  J  =  2  feet,  c  .  =  1  foot 
#j  =  4  lbs.,^>  =  6  atmos.,  and  diam.  of  disk  =  1  inch;  with 
the  foot  and  pound, 


G  =  |4.  *  (^)8[6  X  W.7  X  144  -  1  X  14.7  X  144]  -  4:  X*. 
/.  G  =  2.81  Ibs. 


[Notice  the  cancelling  of  the  144;  for  F(p  —  pa)  is  pounds, 
being  one  dimension  of  force,  if  the  pound  is  selected  as  the 
unit  of  force,  whether  the  inch  or  foot  is  used  in  both  fac- 
tors.] Hence  when  the  steam  pressure  has  risen  to  6  atmos. 
(=  88.2  Ibs.  per  square  inch)  (corresponding  to  73.5  Ibs.  per  sq. 
in.  by  steam-gauge)  the  valve  will  open  if  G  =  2.81  Ibs.,  or  be 
on  the  point  of  opening. 

426.  Proper  Thickness  of  Thin  Hollow  Cylinders  (i.e,,  Pipes 
and  Tubes)  to  Resist  Bursting  by  Fluid  Pressure. 

CASE  I.  Stresses  in  the  cross-section  due  to  End  Pressure  / 
Fig.  469.  —  Let  AB  be  the  circular  cap  clos- 
ing the  end  of  a  cylindrical  tube  containing 
fluid  at  a  tension  =  p.  Let  r  =  internal 
radius  of  the  tube  or  pipe.  Then  considering 
the  cap  free,  neglecting  its  weight,  we 
have  three  sets  of  \\  forces  in  equilibrium 
in  the  figure,  viz.  :  the  internal  fluid  pres- 
sure —  nr^p  ;  the  external  fluid  pressure 
=  7tr*pa  ;  while  the  total  stress  (tensile)  on 
the  small  ring,  whose  area  now  exposed  is 
%nrt  (nearly),  is  =  Snrtp^  ,  where  t  is  the  thickness  of  the  pipe, 
and  j!?j  the  tensile  stress  per  unit  area  induced  by  the  end-pres- 
sures (fluid). 


536  MECHANICS    OF    ENGINEERING. 

For  equilibrium,  therefore,  we  may  put  ^(hor.  comps.)  =  0; 

>1==0; 

(1) 


.       _r(p-pa) 
•  •  Jri  ~  o/ 


(Strictly,  the  two  circular  areas  sustaining  the  fluid  pressures 
are  different  in  area,  but  to  consider  them  equal  occasions  but 
a  small  error.) 

Eq.  (1)  also  gives  the  tension  in  the  central  section  of  a  thin 
hollow  sphere,  under  bursting  pressure. 

CASE  II.  /Stresses  in  the  longitudinal  section  of  pipe,  due  to 
radial  fluid  pressures* — Consider  free  the  half  (semi-circular) 

of  any  length  I  of  the  pipe,  be- 
tween two  cross-sections.  Take  an 
axis  X  (as  in  Fig.  470)  ~]  to  the 
longitudinal  section  which  has  been 
made.  Let  p,  denote  the  tensile 
stress  (per  unit  area)  produced  in 
the  narrow  rectangles  exposed  at  A 
and  B  (those  in  the  half -ring  edges, 
having  no  X  components,  are  not 
drawn  in  the  figure).  On  the  in- 
ternal curved  surface  the  fluid  pres- 
sure is  considered  of  equal  intensity 
=.p  at  all  points  (practically  true  even  with  liquids,  if  2r  is 
small  compared  with  the  head  of  water  producing  p).  The 
fluid  pressure  on  any  dF  or  elementary  area  of  the  internal 
curved  surface  is  =  pdF.  Its  X  component  (see  §  416)  is 
obtained  by  multiplying  j?  by  the  projection  of  dF  on  the  ver- 
tical plane  ABC,  and  since  p  is  the  same  for  all  the  dF's  of 
the  curved  surface,  the  sum  of  all  the  ^components  of  the  in- 
ternal fluid  pressures  must  —  p  multiplied  by  the  area  of  rect- 
angle ABCD,  =  %rlp  •  and  similarly  the  X  components  of  the 


FIG.  470. 


*  Analytically  this  problem  is  identical  with  that  of  the  smooth  cord  on 
a  smooth  cylinder,  §  169,  and  is  seen  to  give  the  same  result. 


BURSTING   OF   PIPES.  537 

external  atmos.  pressures  =  '2rlpa  (nearly).  The  tensile  stresses 
(  ||  to  X)  are  equal  to  2%?a  ;  hence  for  equilibrium,  2X=  0 
gives 

2rlpa  =  0  ; 


(2) 


This  tensile  stress,  called  hoop  tension,  p^  opposing  rupture  by 
longitudinal  tearing,  is  seen  to  be  double  the  tensile  stress^ 
induced,  under  the  same  circumstances,  on  the  annular  cross' 
section  in  Case  I.  Hence  eq.  (2),  and  not  eq.  (1),  should  be 
used  to  determine  a  safe  value  for  the  thickness  of  metal,  t,  or 
any  other  one  unknown  quantity  involved  in  the  equation. 

For  safety  against  rupture,  we  must  put  p^  =  T'  ',  a  safe 
tensile  stress  per  unit  area  for  the  material  of  the  pipe  or  tube 
(gee  §§  195  and  203)  ; 


(For  a  thin  hollow  sphere,  t  may  be  computed  from  eq.  (1)  ; 
that  is,  need  be  only  half  as  great  as  with  the  cylinder,  other 
things  being  equal.) 

EXAMPLE.  —  A  pipe  of  twenty  inches  internal  diameter  is  to 
contain  water  at  rest  under  a  head  of  340  feet  ;  required  the 
proper  thickness,  if  of  cast-iron. 

340  feet  of  water  measures  10  atmospheres,  so  that  the  in 
ternal  fluid  pressure  is  11  atmospheres  ;  but  the  external  pres 
sure  j?a  being  one  atmos.,  we  must  write  (inch,  lb.,  sec.) 

(p—Pa)  =  10  X  14.  T  =  147.0  Ibs.  per  sq.  in.,  and  r  =  10  in., 

while  (§  203)  we  may  put  T'  =i  of  9000  =  4500  Ibs.  per  sq, 
in.  ;  whence 


10 
t  =  ^  =  0.326  inches. 


538  MECHANICS    OF   ENGINEERING. 

But  to  insure  safety  in  handling  pipes  and  imperviousness  to 
the  water,  a  somewhat  greater  thickness  is  adopted  in  practice 
than  given  by  the  above  theory. 

Thus,  Weisbach  recommends  (as  proved  experimentally  also) 
for 


Pipes  of  sheet  iron,  t  —  [0.00172  rA  -\-  0.12" 

'0.00476  rA  +  0.34= 


cast  t  = 

copper        t  = 

lead  t  = 


U00296  rA  +  0.16= 
U01014  rA  +  0.211 
zinc  t  =  [0.00484  rA  +  0.16= 


inches ; 


in  which  t  =  thickness  in  inches,  r  =  radius  in  inches,  and  A 
=  excess  of  internal  over  external  fluid  pressure  (i.e.,  p  —  pa) 
expressed  in  atmospheres. 

For  instance,  for  the  example  just  given,  we  should  have 
(cast-iron) 

t  =  .00476  X  10  X  10  +  0.34  =  0.816  inches. 

If  the  pipe  is  subject  to  "  water-ram"  (§  422)  the  strength 
should  be  much  greater.  To  provide  against  "  water-ram," 
Mr.  J.  T.  Fanning,  on  p.  453  of  his  "  Hydraulic  and  Water- 
supply  Engineering,"  advises  adding  230  feet  to  the  static 
head  in  computing  the  thickness  of  cast-iron  pipes. 

For  thick  hollow  cylinders  see  Rankine's  Applied  Mechan- 
ics, p.  290,  and  Cotter-ill's  Applied  Mechanics,  p.  403. 

427.  Collapsing  of  Tubes  under  Fluid  Pressure.  (Cylindrical 
boiler-flues,  for  example.) — If  the  external  exceeds  the  internal 
fluid  pressure,  and  the  thickness  of  metal  is  small  compared 
with  the  diameter,  the  slightest  deformation  of  the  tube  or 
pipe  gives  the  external  pressure  greater  capability  to  produce 
a  further  change  of  form,  and  hence  possibly  a  final  collapse ; 
just  as  with  long  columns  (§  303)  a  slight  bending  gives  great 
advantage  to  the  terminal  forces.  Hence  the  theory  of  §  426 
is  inapplicable.  According  to  Sir  Win.  Fairbairn's  experi- 
ments (1858)  a  thin  wrought-iron  cylindrical  (circular)  tube 
will  not  collapse  until  the  excess  of  external  over  internal 
pressure  is 


COLLAPSE   OF  TUBES.  539 

p(in  Ibs.  per  sq.  in.)  =  9672000  1~.     .     .(!)..  (not  homog.) 

(t,  I,  and  d  must  all  be  expressed  in  the  same  linear  unit.) 
Here  t  ==  thickness  of  the  wall  of  the  tube,  d  its  diameter,  and 
I  its  length  ;  the  ends  being  understood  to  be  so  supported  as 
to  preclude  a  local  collapse. 

EXAMPLE.—  With  Z  =  10  ft.  =  120  inches,  d  =  4  in.,  and  t  = 
^  inch,  we  have 

p  =  9672000  f^-  -T-  (120  X  4)~|  =  201.5  Ibs.  per  sq.  inch. 


For  safety,  -J-  of  this,  viz.  40  Ibs.  per  sq.  inch,  should  not  be 
exceeded  ;  e.g.,  with  14.7  Ibs.  internal  and  54.7  Ibs.  external. 

[NOTE.  —  For  simplicity  the  power  of  the  thickness  used  in  eq.  (1)  above 
has  been  given  as  2.00.  In  the  original  formula  it  is  2.19,  and  then  all 
dimensions  must  be  expressed  in  inches.  A  discussion  of  the  experiments 
of  Mr.  Fairbairn  will  be  found  in  a  paper  read  by  Prof.  Unwin  before  the 
Institute  of  Civ.  Engineers  (Proceedings,  vol.  xlvi.).  See  also  Prof.  Unwin  's 
"  Machine  Design,"  p.  66,  It  is  contended  by  some  that  in  the  actual  con- 
ditions of  service,  boiler-flues  are  subjected  to  such  serious  straining 
actions  due  1o  unequal  expansion  of  the  connecting  parts  as  to  render  the 
above  formula  quite  unreliable,  thus  requiring  a  large  allowance  in  itg 
application.] 


540  MECHANICS   OF   ENGINEERING. 


CHAPTEE  II. 

HYDROSTATICS  (Continued)— PRESSURE  OF  LIQUIDS  IN  TANEi 
AND  RESERVOIRS. 

428.  Body  of  Liquid  in  Motion,  but  in  Relative  Equilibrium.^ 
By  relative  equilibrium  it  is  meant  that  the  particles  are  not 
changing'their  relative  positions,  i.e.,  are  not  moving  among 
each  other.  On  account  of  this  relative  equilibrium  the  fol- 
lowing problems  are  placed  in  the  present  chapter,  instead  of 
under  the  head  of  Hydrodynamics,  where  they  strictly  belong. 
As  relative  equilibrium  is  an  essential  property  of  rigid  bodies, 
we  may  apply  the  equations  of  motion  of  rigid  bodies  to  bodies 
of  liquid  in  relative  equilibrium. 

CASE  I.  All  the  particles  moving  in  parallel  right  lines 
with  equal  velocities  /  at  any  given  instant  (i.e.,  a  motion  of 
translation.) — If  the  common  velocity  is  constant  we  have  a 
uniform  translation,  and  all  the  forces  acting  on  any  one  par- 
ticle are  balanced,  as  if  it  were  not  moving  at  all  (according  to 
iNewton's  Laws,  §  54);  hence  the  relations  of  internal  pressure, 
free  surface,  etc.,  are  the  same  as  if  the  liquid  were  at  rest. 
Thus,  Fig.  471,  if  the  liquid  in  the  moving  tank  is  at  rest  rel- 
v  atively  to  the  tank  at  a  given  instant,  with 
its  free  surface  horizontal,  and  the  motion 
of  the  tank  be  one  of  translation  with  a  uni- 
form velocity,  the  liquid  will  remain  in  this 
condition  of  relative  rest,  as  the  motion 

FIG.  471. 

proceeds. 

But  if  the  velocity  of  the  tank  is  accelerated  with  a  constant 
acceleration  =p  (this  symbol  must  not  be  confused  with  p 
for  pressure),  the  free  surface  will  begin  to  oscillate,  and  finally 
come  to  relative  equilibrium  at  some  angle  a  with  the  horizon- 
tal, which  is  thus  found,  when  the  motion  is  horizontal.  See 
Fig.  472,  in  which  the  position  and  value  of  a  are  the  same, 
whether  the  motion  is  uniformly  accelerated  from  left  to  right 


RELATIVE   EQUILIBRIUM    OF   LIQUIDS. 


Fio.  472. 


or  uniformly  retarded  from  right  to  left.  Let  0  be  the  lowest 
point  of  the  free  surface,  and  Ob  a 
small  prism  of  the  liquid  with  its 
axis  horizontal,  and  of  length  =  x ; 
nb  is  a  vertical  prism  of  length  — 
2,  and  extending  from  the  extremity 
of  Ob  to  the  free  surface.  The 
pressure  at  both  0  and  n  is  pa  = 
atmos.  pres.  Let  the  area  of  cross- 
section  of  both  prisms  be  —  dF. 

Now  since  Ob  is  being  accelerated  in  direction  JT(horizont.), 
the  difference  between  the  forces  on  its  two  ends,  i.e.,  its  2X, 
must  =  its  mass  X  accel.  (§  109). 

(y  =  heaviness  of  liquid ;  pb  —  press,  at  b) ;  and  since  the  ver- 
tical prism  nb  has  no  vertical  acceleration,  the  ^(vert.  com- 
pons.)  for  it  must  =  0. 

.\pbdF-padF-zdF.y=0 (2) 

From  (1)  and  (2), 

?rm-=sr.   ...i=£. 

g  X       Q 

Hence  On  is  a  right  line,  and  therefore 


(3) 


tan 


or-,   =  •£.. 
x         a 


[Another,  and  perhaps  more  direct,  method  of  deriving  this 
result  is  to  consider  free  a  small  particle  of  the  liquid  lying  in 
the  surface.  The  forces  acting  on  this  particle  are  two :  the 
first  its  weight  —  dG  ;  and  the  second  the  resultant  action  of 
its  immediate  neighbor-particles.  Now  this  latter  force  (point- 
ing obliquely  upward)  must  be  normal  to  the  free  surface  of 
the  liquid,  and  therefore  must  make  the  unknown  angle  a  with 
the  vertical.  Since  the  particle  has  at  this  instant  a  rectilinear 
accelerated  motion  in  a  horizontal  direction,  the  resultant  of  the 
two  forces  mentioned  must  be  horizontal  and  have  a  value  = 
mass  X  acceleration.  That  is,  the  diagonal  tormed  on  the  two 


542 


MECHANICS   OF   ENGINEERING. 


forces  must  be  horizontal  and  have  the  value  mentioned,  = 
(dG  -r-  g)p ;  while  from  the  nature  of  the  figure  (let  the  stu- 
dent make  the  diagram  for  himself)  it  must  also  =  dG  tan  a. 


jri  ,  dG  —  p 

.:  dG  tan  a  —  — .p ;  or,  tan  a  =  ^-. 

ff  9 


.    .    Q.  E. 


If  the  translation  were  vertical,  and  the  acceleration  upward 
[i.e.,  if  the  vessel  had  a  uniformly  accelerated  upward  motion 
or  a  uniformly  retarded  downward  motion],  the  free  surface 
would  be  horizontal,  but  the  pressure  at  a  depth  =  h  below  the 
surface  instead  of^>  =pa  -\-hy  would  be  obtained  as  follows: 
Considering  free  a  small  vertical  prism  of  height  =  h  with 
upper  base  in  the  free  surface,  and  putting  J2"(vert.  compons.) 
=  mass  X  acceleration,  we  have 

dF.p  -  dF.pa  -  hdF.  y  =  MF'Y  .pi 


(5) 


If  the  acceleration  is  downward  (not  the  velocity  necessarily) 
we  make  p  negative  in  (5).      If  the  vessel  falls  freely,  p  =  —  g 
and  .'.p  =pa,  in  all  parts  of  the  liquid. 
Query :  Suppose  p  downward  and  >  g. 
CASE  II.    Uniform  Rotation  about  a  VerticalAxis. — If  the 
narrow  vessel  in  Fig.  473,  open  at  top  and  containing  a  liquid, 

be  kept  rotating  at  a  uniform  angu- 
lar velocity  GO  (see  §  110)  about  a 
vertical  axis  Z,  the  liquid  after  some 
oscillations  will  be  brought  (by  fric- 
tion) to  relative  equilibrium  (rotat- 
ing about  Z,  as  if  rigid).  Required 
the  form  of  the  free  surface  (evi- 
dently a  surface  of  revolution)  at 

each    point    of    which    we    know 

^•"^•^ 

FIG.  473.  Let  (9  be  the  intersection  of  the 

axis  Zwith  the  surface,  and  n  any  point  in  the  surface ;  o  being 


UNIFORM   ROTATION   OF   LIQUID  IN  VESSEL.         643 

a  point  vertically  under  n  and  in  same  horizontal  plane  as  0. 
Every  point  of  the  small  right  prism  nb  (of  altitude  =  z  and 
sectional  area  dF)  is  describing  a  horizontal  circle  about  Z,  and 
has  therefore  no  vertical  acceleration.  Hence  for  this  prism, 
free,  we  have  2Z  =  0;  i.e., 

dF.pb  -  dF.pa  -  zdF.-y  =  0  .....     (1) 

Now  the  horizontal  right  prism  Ob  (call  the  direction  0  .  .  .  &, 
X)  is  rotating  uniformly  about  a  vertical  axis  through  one  ex- 
tremity, as  if  it  were  a  rigid  body.  Hence  the  forces  acting 
on  it  must  be  equivalent  to  a  single  horizontal  force,  —  G?aJ/p, 
(§122$,)  coinciding  in  direction  with  X.  [M=  mass  of  prism 
=  its  weight  -r-  ^,  and  p  =  distance  of  its  centre  of  gravity 
from  0  ;  here  p  =  ^x  =  %  length  of  prism].  Hence  the 

of  the  forces  acting  on  the  prism  Ob  must  =  —  <*?  - 


But  the  forces  acting  on  the  two  ends  of  this  prism  are  their 
own  X  components,  while  the  lateral  pressures  and  the  weights 
of  its  particles  have  no  X  compons.  ; 


1.      .    .    (2) 

*y 

From  (1)  and  (2)  we  have 

s=(Jw'=£i> (3) 

where  v  =  cox  =  linear  Telocity  of  the  point  n  in  its  circular 
path. 

[As  in  Case  I,  we  may  obtain  the  same  result  by  considering 
a  single  surface-particle  free,  and  would  derive  for  the  resultant 
force  acting  upon  it  the  value  dG  tan  a  in  a  horizontal  direc- 
tion and  intersecting  the  axis  of  rotation.  But  here  a  is  dif- 
ferent for  particles  at  different  distances  from  the  axis,  tan  a 

being  the  -^-  of  the  curve  On.     As  the  particle  is  moving  uni- 
formly in  a  circle  the  resultant  force  must  point  toward  the 


544  MECHANICS   OF   ENGINEERING. 

centre  of  the  circle,  i.e.,  horizontally,  and  have  a  value  —  -  .  ~  > 

g      x 

where  x  is  the  radius  of  the  circle  [§  74,  eq.  (5)]  ; 


j.  d&  (cox)*  dz        of® 

tan  a  =  -  -  —  '  :  or  tan  a,  =  —  ,  =  —  ; 
g      x  dx         g 


Hence  any  vertical  section  of  the  free  surface  through  the 
axis  of  rotation  Z  is  a  parabola,  with  its  axis  vertical  and  vertex 
at  0;  i.e.,  the  free  surface  is  &  paraboloid  of  revolution,  with 
Z  as  its  axis.  Since  GOX  is  the  linear  velocity  v  of  the  point 
5  in  its  circular  path,  z  =  "height  due  to  velocity"  v  [§  52], 

EXAMPLE.  —  If  the  vessel  in  Fig.  473  makes  100  revol.  per 
minute,  required  the  ordinate  z  at  a  horizontal  distance  of  x  = 
4  inches  from  the  axis  (ft.-lb.-sec.  system).  The  angular  veloc- 
ity GO  =  \Vi7t  100  4-  60]  radians  per  sec.  [N.  B.  —  A  radian  = 
the  angular  space  of  which  3.1415926  .  .  .  make  a  half-revoL* 
or  angle  of  180°].  With  x  =  4  f  t.  and  g  =  32.2, 


and  the  pressure  at  b  (Fig.  471)  is  (now  use  inch,  lb.,  sec.) 

f»0  K 

&  =pa+zy=  14.7+  2±  X  =  14.781  Ibs.  per  sq.  in. 


Prof.  Mendelejeff  of  Eussia  has  recently  utilized  the  fact  an- 
nounced  as  the  result  of  this  problem,  for  forming  perfectly 
true  paraboloidal  surfaces  of  plaster  of  Paris,  to  receive  by 
galvanic  process  a  deposit  of  metal,  and  thus  produce  specula 
of  exact  figure  for  reflecting  telescopes.  The  vessel  contain- 
ing the  liquid  plaster  is  kept  rotating  about  a  vertical  axis 
at  the  proper  uniform  speed,  and  the  plaster  assumes  the  de- 
sired shape  before  solidifying.  A  fusible  alloy,  melted?  may 
also  be  placed  in  the  vessel,  instead  of  liquid  plaster. 


EELATIVE   EQUILIBKIUM. 


545 


REMARK.  —  If  the  vessel  is  quite  full  and  closed  on  top,  ex- 
except  at  0'  where  it  communicates 
by  a  stationary  pipe  with  a  reser- 
voir, Fig.  474,  the  free  surface 
cannot  be  formed,  but  the  pres- 
sure at  any  point  in  the  water  is 
just  the  same  during  uniform  rota- 
tion, as  if  a  free  surface  were  formed 
with  vertex  at  0\ 


See  figure  for  A0  and  0.  (In  subse- 
quent paragraphs  of  this  chapter 
the  liquid  will  be  at  rest.) 


FIG.  474. 


428a.  Pressure  on  the  Bottom  of  a  Vessel  containing  Liquid  at 
Rest.  —  If  the  bottom  of  the  vessel  is  plane  and  horizontal,  the 
intensity  of  pressure  upon  it  is  the  same  at  all  points,  being 


&1 


c 

^l^g^s 

=^="^=5ij 

Fio.  475. 


FIG.  476. 


(Figs.  475  and  476),  and  the  pressures  on  the  ele- 
ments of  the  surface  form  a  set  of  parallel  (vertical)  forces. 
This  is  true  even  if  the  side  of  the  vessel  overhangs,  Fig.  476, 
the  resultant  fluid  pressure  on  the  bottom  in  both  cases  being 


=  Fp-Fpa  =  Fhy. 


(1) 


(Atmospheric  pressure  is  supposed  to  act  under  the  bottom.) 
It  is  further  evident  that  if  the  bottom  is  a  rigid  homogeneous 
plate  and  has  no  support  at  its  edges,  it  may  be  supported  at  a 


546  MECHANICS   OF   ENGINEERING. 

single  point  (Fig.  477),  which  in  this  case  (horizontal  plate) 
is  its  centre  of  gravity.  This  point  is  called 
the  Centre  of  Pressure,  or  the  point  of  appli- 
cation of  the  resultant  of  all  the  fluid  pressures 
acting  on  the  plate.  The  present  case  is  such 
that  these  pressures  reduce  to  a  single  result- 
ant, but  this  is  not  always  practicable. 

EXAMPLE. — In  Fig.  476  (cylindrical  vessel 


FIG.  477.  containing    water),   given    Ji  =  20   ft.,   A,  == 

15  ft.,  ^  =  2  ft.,  7\  =  4  ft.,  required  the  pressure  on  the  bot- 
tom, the  vertical  tension  in  the  cylindrical  wall  CA,  and  the 
hoop  tension  (§  426)  at  G.  (Ft.,  lb.,  sec.)  Press,  on  bottom  = 
Fhy  =  Ttrfhy  =  7rl6  X  20  X  62.5  =  62857  Ibs. ;  while  the 
upward  pull  on  CA  = 

(yrrf  -  7tr?)h,y  =  ar(16  -  4)15  X  62.5  =  35357  Ibs. 

If  the  vertical  wall  is  t  =  -fa  inch  thick  at  C  this  tension  will 
be  borne  by  a  ring-shaped  cross  section  of  area  =  %nr£  (nearly) 
=  2*48  X  TV  =  30.17  sq.  inches,  giving  (35357  -r-  30.17)  = 
about  1200  Ibs.  per  sq.  inch  tensile  stress  (vertical). 
The  hoop  tension  at  C  is  horizontal  and  is 

P"  =  r*(P  ~Pa)  -5-  t  (see  §  426),  where  p  =pa  +  h,y  ; 

=  3125  lb,  per  sq,  in. 
Tfcr 

(using  the  incli  and  pound). 

429.  Centre  of  Pressure. — In  subsequent  work  in  this  chapter, 
sin/36  the  atmosphere  has  access  both  to  the  free  surface  of 
liquid  and  to  the  outside  of  the  vessel  walls,  and  j90  would  canv 
eel  out  in  finding  the  resultant  fluid  pressure  on  any  elemen* 
tary  area  dF  of  those  walls,  we  shall  write  : 

The  resultant  fluid  pressure  on  any  dF  of  the  vessel  wall  is 
normal  to  its  surface  and  is  dP  =pdF=  zydF.  in  which  z 
is  the  vertical  distance  of  the  element  below  the  free  surface 
of  the  liquid  (i.e.,  z  =  the  "head  of  water").  If  the  surface 
pressed  on  is  plane,  these  elementary  pressures  form  a  system 
of  parallel  forces,  and  may  be  replaced  by  a  single  resultant 


CENTRE   OF   PRESSURE. 


547 


(if  the  plate  is  rigid)  which  will  equal  their  sum,  and  whose 
point  of  application,  called  the  Centre  of  Pressure,  may  be 
located  by  the  equations  of  §  22,  put  into  calculus  form. 

If  the  surface  is  curved  the  elementary  pressures  form  a  sys 
tern  of  forces  in  space,  and  hence  (§  38)  cannot  in  general  be 
reduced  to  a  single  resultant,  but  to  two,  the  point  of  applica- 
tion of  one  of  which  is  arbitrary  (viz.,  the  arbitrary  origin, 
§38). 

Of  course,  the  object  of  replacing  a  set  of  fluid  pressures  by 
a  single  resultant  is  for  convenience  in  examining  the  equi- 
librium, or  stability,  of  a  rigid  body  the  forces  acting  on  which 
include  these  fluid  pressures.  As  to  their  effect  in  distorting 
the  rigid  body,  the  fluid  pressures  must  be  considered  in  their 
true  positions  (see  example  in  §  264),  and  cannot  be  replaced 
by  a  resultant. 

430,  Resultant  Liquid  Pressure  on  a  Plane  Surface  forming 
Part  of  a  Vessel  Wall.  Co-ordinates  of  the  Centre  of  Pressure.— 
Fig.  478.  Let  AB  be  a  portion  (of  any  shape)  of  a  plane 
surface  at  any  angle  with  the 
horizontal,  sustaining  liquid 
pressure.  Prolong  the  plane 
of  AB  till  it  intersects  the  free 
surface  of  the  liquid.  Take 
this  intersection  as  an  axis  Y, 
0  being  any  point  on  T.  The 
axis  X,  "|  to  Y,  lies  in  the 
given  plane.  Let  a  =  angle 
between  the  plane  and  the  free 
surface.  Then  x  and  y  are  the 
co-ordinates  of  any  elementary  FIG.  478. 

area  dF of  the  surface,  referred  to  JTand  Y.  &•=.  the  "head 
of  water,"  below  the  free  surface,  of  any  dF.  The  pressures 
are  parallel. 

The  normal  pressure  on  any  dF  =  zydF\  hence  the  sum  of 
these,  =  their  resultant, 


=  P,  =  yfstJF=  Fzy 


(1) 


548  MECHANICS    OF   ENGINEERING. 

in  which  z  =  the  <c  mean  s"  i.e.,  the  s  of  the  centre  of  gravity 
G  of  the  plane  figure  AB^  and  F  '=  total  area  of  AB  \_Fz  = 
fzdF,  from  eq.  (4),  §  23].  y  =  heaviness  of  liquid  (see  §  409). 
That  is,  the  total  liquid  pressure  on  a  plane  figure  is  equal 
to  the  weight  of  an  imaginary  prism  of  the  liquid  having  a 
base  =  area  of  the  given  figure  and  an  altitude  =  vertical 
depth  of  the  centre  of  gravity  of  the  figure  below  the  surface 
of  the  liquid.  For  example,  if  the  figure  is  a  rectangle  with 
one  base  (length  =  £)  in  the  surface,  and  lying  in  a  vertical 
plane, 


Evidently,  if  the  altitude  be  increased,  P  varies  as  its  square. 

From  (1)  it  is  evident  that  the  total  pressure  does  not  de- 
pend on  the  horizontal  extent  of  the  water  in  the  reservoir. 

Now  let  xc  and  yc  denote  the  co-ordinates,  in  plane  YOX, 
of  the  centre  of  pressure,  C,  or  point  of  application  of  the  re- 
sultant pressure  P,  and  apply  the  principle  that  the  sum  of 
the  moments  of  each  of  several  parallel  forces,  about  an  axis  ~| 
to  them,  is  equal  to  the  moment  of  their  resultant  about  the 
same  axis  [§  22].  First  taking  OY  as  an  axis  of  moments, 
and  then  OX,  we  have 

=  f*(zydF)x,    and  Pyc  =  f*(zydF)y.  .    (2) 

But  P  =  Fzy  —  Fx(mi  a)y,  and  the  z  of  any  dF—  a?  sin  a. 
Hence  eqs.  (2)  become  (after  cancelling  the  constant,  y  sin  a) 


___  . 

--  —  —  —  —  ,  ana  yc  — 


in  which  IT  =  the  "  mom.  of  inertia"  of  the  plane  figure  re- 
ferred to  Y  (see  §  85).  [N.  B.  —  The  centre  of  pressure  as 
thus  found  is  identical  with  the  centre  of  oscillation  (§  117) 
and  the  centre  of  percussion  [§  113]  of  a  thin  homogeneous 
plate,  referred  to  axes  Xand  Z,  T  being  the  axis  of  suspen- 
sion.] 

Evidently,  if  the  plane  figure  is  vertical  a  —  90°.  a?  =  z  for 


CENTRE   OF  PRESSURE. 


549 


^all  dF's,  and  x  =  z.    It  is  also  noteworthy  that  the  position 
of  the  centre  of  pressure  is  independent  of  a. 

NOTE. — Since  the  pressures  on  the  equal  dF's  lying  in  any 
horizontal  strip  of  the  plane  figure  form  a  set  of  equal  parallel 
forces  equally  spaced  along  the  strip,  and  are  therefore  equiva- 
lent to  their  sum  applied  in  the  middle  of  the  strip,  it  follows 
that  for  rectangles  and  triangles  with  horizontal  bases,  the 
centre  of  pressure  must  lie  on  the  straight  line  on  which  the 
middles  of  all  horizontal  strips  are  situated. 

431.  Centre  of  Pressure  of  Rectangles  and  Triangles  with  Bases 
Horizontal. — Since  all  the  dF\  of  one  horizontal  strip  have 
the  same  a?,  we  may  take  the  area  of  the  strip  K 

for  dF  in  the  summation  fx*dF.  Hence  for 
the  rectangle  AB,  Fig  479,  we  have  from  eq. 
(3),  §  430,  with  dF=ldx, 


;(i) 


o 

A 

i, 

! 
' 

'     ft 
j 

Y 

dx 

•  G 
C. 

«—  6—  "*B 

Fio.  479. 

"Va        f*U        O 

while  (see  note,  §  430)  yc  =  %b. 

When  the  upper  base  lies  in  the  surface,  h^  =  0,  and  a?0  = 
4A,  =  %•  of  the  altitude. 

O       3  O       c/ 

For  a  triangle  with  its  base  horizontal  and  vertex  up,  Fig. 
480,  the  length  ^  of  a  horizontal  strip  is  variable  and  dF— 

udx.    From  similar  triangles  u  =  j- -(x  —  A,) ;  therefore 


o 

1 

1*1 

Y 

M 


6— B * 

Flo.  480. 


650  MECHANICS   OF  ENGINEERING. 

Also,  since  the  centre  of  pressure  must  lie  on  the  line  AB  join* 
ing  the  vertex  to  the  middle  of  base  (see  note,  §  430),  we  easily 
determine  its  position. 

Evidently  for  h,  =  0,  i.e.,  when  the  vertex  is  in  the  surface, 

o  t        v     x0  =  f  A3.     Similarly,  for   a   triangle  with 

J*i  base  horizontal  and  vertex  down.  Fig.  481, 

^....0-**  •  —  •»—•»•——— -*t  7  o 

we  find  that 

If  the  base  is  in  the  surface,  hl  =  0  and 
(3)  reduces  to  xc  =  JA2. 

It  is  to  be  noticed  that  in  the  case  of  the  triangle  the  value 
of  o?0  is  the  same  whatever  be  its  shape,  so  long  as  hl  and  Aa 
remain  unchanged  and  the  base  is  horizontal.  If  the  base  is 
not  horizontal,  we  may  easily,  by  one  horizontal  line,  divide 
the  triangle  into  two  triangles  whose  bases  are  horizontal  and 
whose  combined  areas  make  up  the  area  of  the  first.  The  re- 
sultant pressure  on  each  of  the  component  triangles  is  easily 
found  by  the  foregoing  principles,  as  also  its  point  of  applica- 
tion. The  resultant  of  the  two  parallel  forces  so  determined 
will  act  at  some  point  on  the  line  joining  the  centres  of  pres- 
sures of  the  component  triangles,  this  point  being  easily  found 
by  the  method  of  moments,  while  the  amount  of  this  final  re- 
sultant pressure  is  the  sum  of  its  two  components,  since  the 
latter  are  parallel.  An  instance  of  this  procedure  will  be 
given  in  Example  3  of  §  433.  Similarly,  the  rectangle  of  Fig. 
479  may  be  distorted  into  an  oblique  parallelogram  with  hori- 
zontal bases  without  affecting  the  value  of  xc ,  nor  the  amount 
of  resultant  pressure,  so  long  as  hl  and  Aa  remain  unchanged. 

432.  Centre  of  Pressure  of  Circle.— Fig.  482.  It  will  lie  on 
the  vertical  diameter.  Let  r  =  radius.  From  eq.  (3),  §  430, 

7V            IQ  +F  ~x        ±7rr4  +  nr*  ~x 
x*  =  —?= ;   =  -1-1 — 


Fx 
(See  eq.  (4),  §  88,  and  also  §  91.) 


X 


CENTRE   OF   PRESS UEE. 


551 


FIG.  483. 


433.  Examples. — It  will  be  noticed  that  although  the  total 
pressure  on  the  plane  figure  depends  for  its  value  upon  the 
head,  z,  of  the  centre  of  gravity,  its  point  of  application  is  al- 
ways lower  than  the  centre  of  gravity. 

EXAMPLE  1. — If  6  ft.  of  a  vertical  sluice-gate,  4  ft.  wide, 
Fig.  483,  is  below  the  water-surface,  the  total 
water  pressure  against  it  is  (ft.,  lb.,  sec. ;  eq. 
(1),§430) 

P  =  Fzy  =  6  X  4  X  3  X  62.5  =  4500  Ibs., 

and  (so  far  as  the  pressures  on  the  vertical 
posts  on  which  the  gate  slides  are  concerned) 
is  equivalent  to  a  single  horizontal  force  of 
that  value  applied  at  a  distance  XG  =  f  of 
6  =  4  ft.  below  the  surface  (§  431). 

EXAMPLE  2.— To  (begin  to)  lift  the  gate  in  Fig.  483,  the 
gate  itself  weighing  200  Ibs.,  and  the  coefficient  of  friction 
between  the  gate  and  posts  being/*=  0.40  (abstract  numb.)  (see 
§  156),  we  must  employ  an  upward  vertical  force  at  least 

=  P'  =  200  +  0.40  X  4500  =  2000  Ibs. 

EXAMPLE  3. — It  is   required   to  find   the  resultant  hydro- 
static pressure  on  the  trapezoid  in  Fig.  483&  with  the  dimen- 
sions there  given  and  its  bases  horizontal ;  also  its  point  of  ap- 
plication, i.e.,  the  centre  of  pressure  of 

r~r the  plane  figure  in  the  position  there 

A      B  c      D  shown.    From  symmetry  the  C.  of  P.  will 

be  in  the  middle  vertical  of  the  figure, 
as  also  that  of  the  rectangle  B  CFE,  and 
that  of  the  two  triangles  ABE  and 
CDF  taken  together  (conceived  to  be 
shifted  horizontally  so  that  CF  and 
BE  coincide  on  the  middle  vertical, 
thus  forming  a  single  triangle  of  5  ft.  base,  and  .having  the 
same  total  pressure  and  C.  of  P.  as  the  two  actual  triangles 
taken  together).  Let  Pl  =  the  total  pressure,  and  xcf  refer  to 
the  C.  of  P.,  for  the  rectangle ;  P2  and  xe\  for  the  5  ft.  tri- 


f 


AD -10 
EF-    5 


FIG.  483a. 


552 


MECHANICS   OF   ENGINEERING. 


angle;  A1  =  4  ft.  and  A2  =  10  ft.  being  the  same  for  both. 
Then  from  eq.  (1),  §  430,  we  have  (with  the  ft.,  lb.,  and  sec.) 

Pl  =  30  X  1y  =  210>/ ;    and    Pa  =  J-  X  6  X  5  X  67  =  90y; 
while  from  eqs.  (1)  and  (3)  of  §  431  we  have  also  (respectively) 


2    1000-64        2     936 


3  '    100  -  16  "    3 

1    484-80-4-100 


84 


=  7.438  feet; 


228 


8+10 


2X18 


=  6.333  feet. 


The  total  pressure  on  the  trapezoid,  being  the  resultant  of 
Pl  and  Pa  ,  has  an  amount  =  JPl  +  Pt  (since  they  are  parallel), 
and  has  a  lever-arm  xc  about  the  axis  0  Y  to  be  found  by  the 
principle  of  moments,  as  follows  : 

(210  X  7.438  +  90  X  6.33)r 

=  -  -  =  7'09ft- 


"•  = 


The  total  hydrostatic  pressure  on  the  trapezoid  is  (for  fresh 
water) 

P  =  P1  +  P,=  [210  +  90]  62.5  =  18750  Ibs. 

EXAMPLE  4. — Eequired  the  horizontal  force  P',  Fig.  484,  to 
be  applied  at  N  (with  a  leverage  of  a'  =  30  inches  about  the 

fulcrum  M)  necessary  to  (begin 
to)  lift  the  circular  disk  AB  of 
radius  r  =  10  in.,  covering  an 
opening  of  equal  size.  NMAB 
is  a  single  rigid  lever  weighing 
#'  =  210  Ibs.  The  centre  of 
gravity,  G,  of  disk,  being  a  ver- 
tical distance  H  =  O'G  =  40 
inches  from  the  surface,  is  50 
inches  (viz.,  the  sum  of  OM  = 
k  =  20"  and  MG  =  30")  from 
axis  0  Y ;  i.e.,  x  =  50  inches. 
FIG.  484.  The  centre  of  gravity  of  the 

whole  lever  is  a  horizontal  distance  5',  —  12  inches,  from  M. 


EXAMPLES — CENTRE   OF   PRESSURE. 


553 


For  impending  lifting  we  must  have,  for  equilibrium  of  the 
lever, 

-lc);      .     .     .     .    (1) 


where  P  =  total  water  pressure    on  circular  disk,   and  a?0  = 
OC.     From  eq.  (1),  §  430,  (using  inch,  lb.,  and  sec.,) 


P  =  Fzy  =  nr*zy  = 
From  § 432,  xc=  OC  = 


fi9  ^ 

x  40  X  -^  =  454.6  Ibs. 

1 1  Jo 


~\  v*  1 

~  =  50  +  i.         =  50.5 in. 

4   x  4: 


=  JL  [210  X  12  +  454.6  x  30.5]  =  546  Ibs. 

oU 

434.  Example  of  Flood-gate. — Fig.  485.     Supposing  the  rigid 
double  gate  AD,  8  ft.  in  total  width,  to 
have  four  hinges ;  two  at  0,  and  two  at/", 
1  ft.  from  top  and  bottom  of  water  chan-    ~= 
nel ;  required  the  pressures  upon  them,     in-t- 

taking  dimensions  from  the  figure  (ft.,     £E  9\         \ L_ 

lb.,  sec.). 


Wat.  press.  =  P  =  F  zy 

=  72X4^X62.5  =  20250 


Flo.485. 


pounds,  and  its  point  of  application  (cent,  of  press.)  is  a  dis- 
tance xc  =  f  of  9'  =  6'  from  0  (§  431).  Considering  the 
whole  gate  free  and  taking  moments  about  e,  we  shall  have 

(press,  at  f)xT  =  20250  x  5  ;  .-.  press,  at  /=  14464  Ibs. 
(half  on  each  hinge  at/"),  and 

/.  press,  at  e  =  P  —  press,  at/*  =  5875  Ibs. 
(half  coming  on  each  hinge). 


554 


MECHANICS    OF    ENGINEERING. 


If  the  two  gates  do  not  form  a  single  rigid  body,  and  hence 
are  not  in  the  same  plane  when  closed,  a  wedge-like  or  toggle- 
joint  action  is  induced,  producing  much  greater  thrusts  against 
the  hinges,  and  each  of  these  thrusts  is  not  ~]  to  the  plane  of 
the  corresponding  gate.  Such  a  case  forms  a  good  exercise 
for  the  student. 

435.  Stability  of  a  Vertical  Rectangular  Wall  against  Water 
Pressure  on  One  Side.  —  Fig.  486.  All  dimensions  are  shown  in 


«^u^ 


FIG.  486. 


the  figure,  except  I,  which  is  the  length 
of  wall  "1  to  paper.  Supposing  the  wall 
to  be  a  single  rigid  block,  its  weight  G' 
=  h'h'ly'  (y'  being  its  heaviness  (§  7), 
and  I  its  length).  Given  the  water 
depth  =  h,  required  the  proper  width 
V  for  stability.  For  proper  security  : 
First,  the  resultant  of  G'  and  the 
water-pressure  P  must  fall  within  the 

base  BD  (or,  which  amounts  to  the  same  thing),  the  moment 

of  G'  about  D,  the  outer  toe  of  the  wall,  must  be  numerically 

greater  than  that  of  P  ;  and         4 

Secondly,  P  must  be  less  than  the  sliding  friction  fG'  (see 

§  156)  on  the  base  ED. 

Thirdly,  the  maximum  pressure   per  unit  of  area  on  the 

base  must  not  exceed  a  safe  value  (compare  §  348). 

Now  P  =  Fzy  =  hi  —  y  =  —  h*ly  (y  =  heaviness  of  water)  ; 

2          2 

and  x  =    h. 


Hence  for  stability  against  tipping  about  D, 


P\h  must  le  <  G'W  ;    i.e.,  tftly  <  ^fi'l 
while,  as  to  sliding  on  the  base, 

P  mmtle  <fG'  ;    i.e.,  \Ttly  <fVh'lyf. 


(1) 


(2) 


As  for  values  of  the  coefficient  of  friction,/,  on  the  base  of 
wall,  Mr.  Fanning  quotes  the  following  among  others,  from 
various  authorities  : 


STABILITY    OF   WALL. 


555 


For  point-dressed  granite  on  dry  clay,  /  =  0.51 

«  «  "  "  "  moist  clay,  0.33 

"  "  "  "  "  gravel,  0.58 

«  «  "  "  "  smooth  concrete,  0.62 

«  «  "  "  "  similar  granite,  0.70 

For  dressed  hard  limestone  on  like  limestone,  0.38 

"  <•  "  "  brickwork,  0.60 

For  common  bricks  on  common  bricks,  0.64 

To  satisfactorily  investigate  the  third  condition  requires  the 
detail  of  the  next  paragraph. 

436,  Parallelopipedical  Reservoir  Walls.  More  Detailed  and 
Exact  Solution. — If  (1)  in  the  last  paragraph  were  an  exact 
equality,  instead  of  an  inequality, 
the  resultant  R  of  P  and  G' 
would  pass  through  the  corner 
Z>,  tipping  would  be  impending, 
and  the  pressure  per  unit  area  at 
D  would  be  theoretically  infinite. 
To  avoid  this  we  wish  the  wall 
to  be  wide  enough  that  the  re- 
sultant R,  Fig.  487,  may  cut 
BD  in  such  a  point,  E' ,  as  to  cause  the  pressure  per  unit  area, 
pm ,  at  D  to  have  a  definite  safe  value  (for  the  pressure  pm  at 
I),  or  quite  near  Z>,  will  evidently  be  greater  than  elsewhere 
on  BD ;  i.e.,  it  is  the  maximum  pressure  to  be  found  on  BD). 
This  may  be  done  by  the  principles  of  §§  346  and  362. 

First,  assume  that  It  cuts  BD  outside  of  the  middle  third/ 


FIG.  487. 


i.e. 


that 


where  n  denotes  the  ratio  of  the  distance  of  E'  from  the  mid- 
dle of  the  base  to  the  whole  width,  &',  of  base.  Then  the  pres- 
sure (per  unit  area)  on  small  equal  elements  of  the  base  BD 
(see  §  346)  may  be  considered  to  vary  as  the  ordinates  of  a 
triangle  MND  (the  vertex  M  being  within  the  distance  BD\ 
and  E7]}  will  =  jJ/2)  ;  i.e., 


556  MECHANICS   OF  ENGINEERING. 

MD  =  3(i  — n)&'. 
The  mean  pressure  per  unit  area,  on 


and  hence  the  maximum  pressure  (viz.,  at  D\  being  double 
the  mean,  is 

--n)];      .    .    .    .    (0) 


and  if  pmis  to  equal  <7'(see  §§  201  and  203),  a  safe  value  for  the 
crushing  resistance,  per  unit  area,  of  the  material,  we  shall 
have 


To  find  J',  knowing  n,  we  put  the  2(mom&,)  of  the  G'  and  P 
at  E,  about  E',-=  zero  (for  the  only  other  forces  acting  on 
the  wall  are  the  pressures  of  the  foundation  against  it,  along 
MD  ;  and  since  the  resultant  of  these  latter  passes  through  E'> 
the  sum  of  their  moments  about  E'  is  already  zero)  ;  i.e., 

G'nb'  -  P$h  =  0  ;    or,  nbnhfly'=  | 


Having  obtained  Jr,  we  must  also  ascertain  if  P  is  <fG',  the 
friction  ;  i.e.,  if  P  is  <  fl'h'ly'.  If  not,  V  must  be  still  further 
increased.  (Or,  graphically,  the  resultant  of  G'  and  P  must 
not  make  an  angle  >  0,  the  angle  of  friction,  with  the  ver- 
tical. 

If  7i,  computed  from  (1),  should  prove  to  be  <  -J-,  our  first 
assumption  is  wrong,  and  we  therefore  assume  n  <  |,  and  pro- 
ceed thus : 

Secondly,  n  being  <  -J-  (see  §§  346  and  362),  we  have  a 


STABILITY   OF   RESERVOIE   WALLS.  557 

trapezoid  of  pressures,  instead  of  a  triangle,  on  BD.  Let  the 
pressure  per  unit  area  at  D  be  pm  (the  maximum  on  base). 
The  whole  base  now  receives  pressure,  the  mean  pressure  (per 
unit  area)  being  =  G-'  -f-  [b'l]  ;  and  therefore,  from  §  362, 
Case  I,  we  have 

;     .....    (Oa) 


and  since,  here,  G'  =  Vh'ly',  we  may  write 


For  safety  as  to  crushing  resistance  we  put 

(On  +  1)AV  =  C'\  whence  n  =  i  [^  -  l]  .    .    (la) 

Having  found  n  from  eq.  (la),  we  determine  the  proper 
width  of  base  ~b'  from  eq.  (2),  in  case  the  assumption  n  <  £  is 
verified. 

EXAMPLE.—  In  Fig.  486,  let  A'  =  12  ft.,  h  =  10  ft.,  while 
the  masonry  weighs  (yf  —  )  150  Ibs.  per  cub.  ft.  Supposing 
it  desirable  to  bring  no  greater  compressive  stress  than  100  Ibs. 
per  sq.  inch  (=  14400  Ibs.  per  sq.  ft.)  on  the  cement  of  the 
joints,  we  put  C'  =  14400,  using  the  ft.-lb.-sec.  system  of  units. 

Assuming  n  >  -J,  we  use  eq.  (1),  and  obtain 

_  1  _  2    12  X  150  _     5_ 
~2       3*     14400  ~   ^  12' 

which  is  >  %  ;  hence  the  assumption  is  confirmed,  also  the 
propriety  of  using  eq.  (1)  rather  than  (la). 
Passing  to  eq.  (2),  we  have 


-»*-»«f8Sr»T*r~ 


But,  as  regards  frictional  stability,  we  find  that,  with/  =  0.30, 
a  low  value,  and  V  =  3.7  ft.  (ft.,  lb.,  sec.), 


568  MECHANICS    OF    ENGINEERING. 

100  X  62.5 


fVh'lyf       2  X  0.3  X  3.7  X  12  X  150 


which  is  greater  than  unity,  showing  the  friction  to  be  insuf- 
ficient to  prevent  sliding  (with  /  —  0.30)  ;  a  greater  width 
must  therefore  be  chosen,  for  frictional  stability. 

If  we  make  n  =  -J-,  i.e.,  make  R  cut  the  base  at  the  outer 
edge  of  middle  third  (§  362),  we  have,  from  eq.  (2), 


and  the  pressure  at  D  is  now  of  course  well  within  the  safe 
limit  ;  while  as  regards  friction  we  find 

P  +fG'  =  0.92,  <  unity, 

and  therefore  the  wall  is  safe  in  this  respect  also. 

With  a  width  of  base  =  3.7  feet  first  obtained,  the  portion 
MD,  Fig.  487,  of  the  base  which  receives  pressure  [according 
to  Navier's  theory  (§  346)]  would  be  only  0.92  feet  in  length, 
or  about  one  fourth  of  the  base,  the  portion  JBM  tending  to 
open,  and  perhaps  actually  suffering  tension,  if  capable  (i.e.,  if 
cemented  to  a  rock  foundation),  in  which  case  these  tensions 
should  properly  be  taken  into  account,  as  with  beams  (§  295), 
thus  modifying  the  results. 

It  has  been  considered  safe  by-  some  designers  of  high 
masonry  dams,  to  neglect  these  possible  tensile  resistances,  as 
has  just  been  done  in  deriving  ~b'  —  3.7  feet  ;  but  others,  in 
view  of  the  more  or  less  uncertain  and  speculative  character  of 
Navier's  theory,  when  applied  to  the  very  wide  bases  of  such 
structures,  prefer,  in  using  the  theory  (as  the  best  available), 
to  keep  the  resultant  pressure  within  the  middle  third  at  the 
base  (and  also  at  all  horizontal  beds  above  the  base),  and  thus 
avoid  the  chances  of  tensile  stresses. 

This  latter  plan  is  supported  by  Messrs.  Church  and  Fteley, 
as  engineers  of  the  proposed  Quaker  Bridge  Dam  in  connec- 
tion with  the  New  Croton  Aqueduct  of  New  York  City,  in 
their  report  of  1887.  See  §  439. 


RESERVOIR   WALLS. 


559 


437.  Wall  of  Trapezoidal  Profile.  Water-face  Vertical. — 
Economy  of  material  is  favored  by  using  a  trapezoidal  profile. 
Fig.  488.  "With  this  form  the 
stability  may  be  investigated  in 
a  corresponding  manner.  The 
portion  of  wall  above  each 
horizontal  bed  should  be  ex- 
amined similarly.  The  weight 
G-'  acts  through  the  centre  of 
gravity  of  the  whole  mass. 

Detail.— Let  Fig.  488  show 
the  vertical  cross-section  of  a 
trapezoidal  wall,  with  notation 
for  dimensions  as  indicated  ;  the 
portion  considered  having  a  length  =  I,  ~|  to  the  paper.  Let 
y  —  heaviness  of  water,  y'  that  of  the  masonry  (assumed  homo- 
geneous), with  n  as  in  §  436. 

For  a  triangle  of  pressure,  MD,  on  the  base,  i.e.,  with  n  >  -J-, 
or  resultant  falling  outside  the  middle  third  (neglecting  pos- 
sibility of  tensile  stresses  on  left  of  M\  if  the  intensity  of 
pressure^  at  D  is  to  =  C'  (§  201),  we  put,  as  in  §  436, 


whence 


7[i  -  n\  C'  =  f  #',  i.e.,  =  \IK  .  W  +  l"}yf, 


_i_i/ty  v 


For  a  trapezoid  of  pressure,  i.e.  with  n  <  -J-,  or  the  resultant 
of  P  and  G'  falling  within  the  middle  third,  we  have,  as  be- 
fore (§  362,  Case  I), 


or  O',  =  (6»  +  l)£; 


whence 


n  = 


560  MECHANICS   OF   ENGINEERING. 

From  the  geometry  of  the  figure,  having  joined  the  middles 
of  the  two  bases,  we  have 


(§  26,Prob.  6),  and,  by  similar  triangles,  OT  :  KV  ::  gO  :  h',. 
whence 


The  lines  of  action  of  #'  and  P  meet  at  2?,  and  their  result- 
ant cuts  the  base  in  some  point  E'  .  The  sum  of  their  moments 
about  E'  should  be  zero,  i.e.,  P  .  %h  —  G'  .  ON'\  that  is,  (see 
eq.  (a)  above,  and  eq.  (1),  §  430,) 


i.e.,  cancelling, 

W)(1>'  -  V)  +  6nb'(b'  +  b")].     (2)' 


Hence  we  have  two  equations  for  finding  two  unknowns- 
viz.:  (l)'and  (2)'  when  n  >  \  ;  and  (la)'  and  (2)'  when  n  <  -J. 

For  dams  of  small  height  (less  than  40  ft.,  say),  if  we  im- 
mediately put  n  =  -J-,  thus  restricting  the  resultant  pressure  to 
the  edge  of  middle  third,  and  solve  (2)r  for  b',  b'f  being  as- 
sumed of  some  proper  value  for  a  coping,  foot-walk,  or  road- 
way, while  h'  may  be  taken  enough  greater  than  h  to  provide 
against  the  greatest  height  of  waves,  from  2.5  to  6  ft.,  the 
value  of  pm  at  D  will  probably  be  <  C  '  .  In  any  case,  for  a 
value  of  n  —  ,  or  <,  \  we  put  pm  for  C"in  equation  (la)'  and 
solve  for  jt?m,  to  determine  if  it  is  no  greater  than  C'. 

Mr.  Fanning  recommends  the  following  values  for  C'  (in  lbs» 


RESERVOIR   WALLS.  561 

per  sq.foot)  with  coursed  rubble  masonry  laid  in  strong  mor- 
tar: 

For  Limestone.      Sandstone.       Granite.  Brick. 

O       —         50,000         50,000         60,000         35,000 

Av     heaviness    of  / 

the    masonry   in  V   152  132  154  120 

Ibs.  per  cub.  ft.     \ 

As  to  /national  resistance,  P  must  be  </#';  i.e., 

\ltly  <fh!y>W  +  *")  ......     (3)' 


If  the  base  is  cemented  to  a  rock  foundation  with  good 
material  and  workmanship  throughout,  Messrs.  Church  and 
Fteley  (see  §  436)  consider  that  the  wall  may  be  treated  as 
amply  safe  against  sliding  on  the  base  (or  any  horizontal  bed), 
pro\dded  the  other  two  conditions  of  safety  are  already  satis- 
fied. 

438.  Triangular  Wall  with  Vertical  Water-face.—  Making 
l>"  =  0  in  the  preceding  article,  the  trapezoid  becomes  a  right 
triangle,  and  the  equations  reduce  to  the  following  : 


and 

pm  =  Wyr  [6ra  +  1]  for  n  <  -J-      .     .     .     (la)" 

(pm  not  to  exceed  C'  in  any  case)  ;  while  to  determine  the 
breadth  of  base,  I',  after  n  is  computed  [or  assumed,  for  small 
height  of  wall],  we  have  from  eq.  (2)', 


.....      (2)" 
Also,  for  frictional  stability, 

ly  must  be  <  tfh'Vly'  .....     (3)" 


562  MECHANICS   OF   ENGINEERING. 

439.  High  Masonry  Dams. — Although  the  principle  of  the 
arch  may  be  utilized  for  vertical  stone  dikes  of  small  height 

(30  to  50  feet)  and  small  span,  for 
greater  heights  and  spans  the 
formula  for  hoop  tension,  §  426  (or 
rather,  here,  "  hoop  compression"), 
on  the  vertical  radial  joints  of  the 
horizontal  arch  rings,  Fig.  489,  calls 
for  so  great  a  radial  thickness  of 
joint  in  the  lower  courses,  that 
straight  dikes  (or  "gravity  dams") 
are  usually  built  instead,  even 
where  firm  rock  abutments  are  available  laterally. 

For  example,  at  a  depth  of  100  feet,  where  the  hydrostatic 
pressure  is  hy  =  100  X  62.5  =  6250  Ibs.  per  sq.  ft.,  if  we  as- 
sume for  the  voussoirs  a  (radial,  horizontal)  thickness  =  4  ft., 
with  a  (horizontal)  radius  of  curvature  r  —  100  feet,  we  shall 
find  a  compression  between  their  vertical  radial  faces  of  (ft.> 
lb.,  sec.) 

p,,  =  r(p-^  =  100  X6260 
t  4 

or  1085  Ibs.  per  sq.  inch ;  far  too  great  for  safety,  even  if  there 
were  no  danger  of  collapse,  the  dike  being  short.  If  now  the 
thickness  is  increased,  in  order  to  distribute  the  pressure  over 
a  greater  surface,  we  are  met  by  the  fact  that  the  formula  for 
"  hoop  compression"  is  no  longer  strictly  applicable,  the  law  of 
distribution  of  pressure  becoming  very  uncertain ;  and  even 
supposing  a  uniform  distribution  over  the  joint,  the  thickness 
demanded  for  proper  safety  against  crushing  is  greater  than 
for  a  straight  dam  ("  gravity  dam")  at  a  very  moderate  depth 
below  the  water  surface,  unless  the  radius  of  curvature  of  arch 
can  be  made  small.  But  the  smaller  the  radius  the  more  does 
the  dam  encroach  on  the  storage  capacity  of  the  reservoir,  while 
in  no  case,  of  course,  can  it  be  made  smaller  than  half  the  span. 
Another  point  is,  that  as  masonry  is  not  destitute  of  elas- 
ticity, the  longer  the  span  the  more  unlikely  is  it  that  the 
parts  of  the  arch  will  "close  up"  properly,  and  develop  the 


HIGH   MASONRY   DAMS.  563 

abutment  reactions  when  the  water  is  first  admitted  to  the 
reservoir ;  which  should  occur  if  it  is  to  act  as  an  arch  instead 
of  by  gravity  resistance. 

For  these  reasons  the  engineers  of  the  proposed  Quaker 
Bridge  Dam  reported  unfavorably  to  the  plan  of  a  curved  de- 
sign for  that  structure,  and  recommended  that  a  straight  dam 
be  built.  See  reference  in  §  436.  According  to  their  designs 
this  dam  is  to  be  258  feet  in  height  (which  exceeds  by  about  90 
feet  the  height  of  any  dam  previously  built),  about  1400  feet 
in  length  at  the  top,  and  216  feet  in  width  at  the  lowest 
point  of  base,  joining  the  bed-rock. 

More  recently,  however  (1888),  a  board  of  experts,  specially 
appointed  for  the  purpose,  having  examined  a  number  of  dif- 
ferent plans,  have  reported  favorably  to  the  adoption  of  a 
curved  form  for  the  dam,  as  offering  greater  resistance  under 
extraordinary  circumstances  (Impact  of  ice-floes,  earthquakes^ 
etc.),  on  account  of  its  arched  form  (though  resisting  by 
gravity  action  under  usual  conditions)  than  a  straight  struc- 
ture ;  and  also  as  more  pleasing  in  appearance. 

Fig.  490  shows  the  profile  of  a  straight  high  masonry  darn 
as  designed  at  the  present  day.  Assuming  a  width  ~b"  =  from 
6  to  22  feet  at  the  top,  and  a  sufficient  Ji"  (see  figure)  to  ex- 
ceed the  maximum  height  of  waves,  the  up-stream  outline 
A  CM  is  made  nearly  vertical  and  perhaps  somewhat  concave, 
while  the  down-stream  profile  BDN^  by  computation  or 
graphical  trial,  or  both,  is  so  formed  that  when  the  reservoir  is 
full  the  resultant  R,  of  the  weight 
G  of  the  portion  AECD  of  ma-  SS'/7'A'I 
sonry  above  each  horizontal  bed,  as  =  =\ 
CD,  and  the  hydrostatic  pressure  P 
on  the  corresponding  up-stream  face 
A  C,  shall  cut  the  bed  CD  in  such  a 
point  E'  as  not  to  cause  too  great 
compression  pm  at  the  outer  edge  D 
(not  over  85  Ibs.  per  sq.  inch  accord- 
ing to  M.  Krantz  in  "  Keservoir  F10-  49°- 
Walls").  pm  being  computed  by  one  of  the  equations  [(0)  and 
(Da)  of  1 436] 


5(54  MECHANICS    OF    ENGINEERING. 

For  E'  outside  the  middle  third  |  2#          t  _, 

and  neglecting  tension  (   '  T*  ~~  3.^/>./(4_  n\ 


For  £v  inside  middle  third  1  .  .pm  =  (6yl_+  1W  .    (lay// 

C/-/x  •  £ 

where  Z  =  length  of  wall  1  to  paper,  usually  taken  =  one  foot, 
or  one  inch,  according  to  the  unit  of  length  adopted;  f  or  n, 
see  §  436. 

Nor,  when  the  reservoir  is  empty  and  the  water  pressure 
lacking,  must  the  weight  G  resting  on  each  bed,  as  CD,  cut 
the  bed  in  a  point  E"  so  near  the  edge  C  as  to  produce  exces- 
sive pressure  there  (computed  as  above).  The  figure  shows 
the  general  form  of  profile  resulting  from  these  conditions. 
The  masonry  should  be  of  such  a  character,  by  irregular  bond- 
ing in  every  direction,  as  to  make  the  wall  if  possible  a  mono- 
lith. For  more  detail  see  next  paragraph. 

440.  Quaker  Bridge  Dam*  (on  the  New  Croton  Aqueduct).  — 
Attempts,  by  strict  analysis,  to  determine  the  equation  of  the 
curve  BN,  AM  being  assumed  straight,  so  as  to  bring  the 
point  E'  at  the  outer  edge  of  the  middle  third  of  its  joint,  or 
to  make  the  pressure  at  D  constant  below  a  definite  joint,  have 
failed,  up  to  the  present  time  ;  but  approximate  and  tentative 
methods  are  in  use  which  serve  all  practical  purposes.  As  an 
illustration  the  method  set  forth  in  the  report  on  the  Quaker 
Bridge  Dam  will  be  briefly  outlined  ;  this  method  confines  E' 
to  the  middle  third. 

The  width  AB  —  I"  is  taken  =  22'  for  a  roadway,  and  h"  = 
7  ft.  The  profile  is  made  a  vertical  rectangle  from  A  down 
to  a  depth  of  33  ft.  below  the  water  surface  (reservoir  full). 
Combining  the  weight  of  this  rectangle  of  masonry  with  the 
corresponding  water  pressure  (for  a  length  of  wall  =  one  foot), 
we  find  the  resultant  pressure  comes  a  little  within  the  outer 
edge  of  the  middle  third  of  the  base  of  the  rectangle,  while 
pm  is  of  course  small. 

The  rectangular  form  of  profile  might  be  continued  below 
this  horizontal  joint,  as  far  as  complying  with  the  middle 

*  This  dam  was  not  built. 


QUAKER    BRIDGE    DAM.  565 

third  requirement,  and  the  limitation  of  pressure-intensity,  is 
concerned ;  but,  not  to  make  the  widening  of  the  joints  too 
abrupt  in  a  lower  position  where  it  would  be  absolutely  re- 
quired, a  beginning  is  made  at  the  joint  just  mentioned  by 
forming  a  trapezoid  between  it  and  a  joint  11  ft.  farther  down, 
making  the  lower  base  of  the  latter  of  some  trial  width,  which 
can  be  altered  when  the  results  to  which  it  gives  rise  become 
evident.  Having  computed  the  weight  of  this  trapezoid  and 
constructed  its  line  of  action  through  the  centre  of  gravity  of 
the  trapezoid,  the  value  of  the  resultant  G  of  this  weight  and 
that  of  the  rectangle  is  found  (by  principle  of  moments  or  by 
an  equilibrium  polygon)  in  amount  and  position,  and  combined 
with  the  water  pressure  of  the  corresponding  44  ft.  of  water  to 
form  the  force  7?,  whose  point  of  intersection  with  the  new 
joint  or  bed  (lower  base  of  trapezoid)  is  noted  and  the  value  of 
pm  computed.  These  should  both  be  somewhat  nearer  their 
limits  than  in  the  preceding  joint.  If  not,  a  different  width 
should  be  chosen,  and  changed  again,  if  necessary,  until  satis- 
factory. Similarly,  another  layer,  11  ft.  in  height  and  of 
trapezoidal  form,  is  added  below  and  treated  in  the  same  way ; 
and  so  on  until  in  the  joint  at  a  depth  of  66  ft.  from  the 
water  surface  a  width  is  found  where  the  point  E'  is  very 
close  upon  its  limiting  position,  while  pm  is  quite  a  little  under 
the  limit  set  for  the  upper  joints  of  the  dam,  8  tons  per  square 
foot.  For  the  next  three  11  ft.  trapezoidal  layers  the  chief 
governing  element  is  the  middle-third  requirement,  E'  being 
kept  quite  close  to  the  limit,  while  the  increase  of  pm  to  7.95 
tons  per  sq.  ft.  is  unobjectionable;  also,  we  begin  to  move 
the  left-hand  edge  to  the  left  of  the  vertical,  so  that  when  the 
reservoir  is  empty  the  point  E"  shall  not  be  too  near  the  up- 
stream edge  C. 

Down  to  a  depth  of  about  200  ft.  the  value  of  pm  is  allowed 
to  increase  to  10.48  tons  per  sq.  ft.,  while  the  position  of  E' 
gradually  retreats  from  the  edge  of  its  limit.  Beyond  200  ft. 
depth,  to  prevent  a  rapid  increase  of  width  and  consequent 
extreme  flattening  of  the  down-stream  curve,  pm  is  allowed 
to  mount  rapidly  to  16.63  tons  per  sq.  ft.  (=231  Ibs.  per 
sq.  in.),  which  value  it  reaches  at  the  point  N  <A  the  base  of 


MECHANICS   OF   ENGINEERING. 

the  dam,  which  has  a  width  =  216  ft.,  and  is  258  feet  below 
the  water  surface  when  the  reservoir  is  full. 

The  heaviness  of  the  masonry  is  taken  as  y'  —  156.25  Ibs. 
per  cubic  foot,  just  f  of  y  =  62.5  Ibs.  per  cub.  foot,  the  heavi- 
ness taken  for  water. 

When  the  reservoir  is  empty,  we  have  the  weight  G  of  the 
superincumbent  mass  resting  on  any  bed  CD,  and  applied 
through  the  point  E"  ;  the  pressure  per  unit  area  at  G  can 
then  be  computed  by  eq.  (la)"7,  §  439,  n  being  the  quotient  of 
(^CD  —  CE")-±-  VT>  for  this  purpose.  In  the  present  case 
we  find  E"  to  be  within  middle  third  at  all  joints,  and  the 
pressures  at  C  to  be  under  the  limit. 

For  further  details  the  reader  is  referred  to  the  report  itself 
(reprinted  in  Engineering  News,  January,  1888,  p.  20).  The 
graphic  results  were  checked  by  computation,  Wegmann's 
method,  applied  to  each  trapezoid  in  turn. 

441.  Earthwork  Dam,  of  Trapezoidal  Section.  —  Fig.  491.    It  is 
<•.--  £_.  >D  required  to  find  the  conditions  of  sta- 

bility of  the  straight  earthwork  dam 
ABDE,   whose    length  —I,    L    to 
paper,  as  regards  sliding  horizontally 
on  the  plane  AE;  i.e.,  its  frictional 
K  —  c—  •*    stability.     With  the   dimensions   of 
p  the  figure,  y  and  y'  being  the  heavi- 

FIO.  491.  nesses  of  the  water  and  earth  respec- 

tively (see  §  7),  we  have 

Weight  of  dam  =  G,  =  vol.  X  y'  =  lh,\b  +  Jfa  +  c)]/.  (1) 

Resultant  water  press.  =  P  =  Fzy=OAxlX  ±hy.  .  (2) 
Horiz.  comp.  of  P  =  H  =  P  sin  a 

=  [02  sin  a$My  =  \ttly.      ...  (3) 


From  (3)  it  is  evident  that  the  horizontal  component  of  P  is 
just  the  same,  viz.,  =  hi  .  \hy,  as  the  water  pressure  would  be 
on  a  vertical  rectangle  equal  to  the  vertical  projection  of  OA 


EAKTHWORK    DAM.  567 

and  with  its  centre  of  gravity  at  the  same  depth  (JA).     Com- 
pare §416.     Also, 

Yert.  comp.  ofjP=  V  =  P  cos  a 

=  \OA  cos  a]^hly  =  \ahly,        ...     (4) 


and  is  the  same  as  the  water  pressure  on  the  horizontal  projec- 
tion of  OA  if  placed  at  a  depth  —  O'G  =  -JA. 

For  stability  against  sliding,  the  horizontal  component  of  P 
must  be  less  than  the  friction  due  to  the  total  vertical  pressure 
on  the  plane  AE,  viz.,  Gl  -\-  V  \  hence  if  /"is  the  coefficient  of 
friction  on  AE,  we  must  have  H  <f  \_Gl  -\-  "F],  i.e.  (see  above), 


must  be  <  Jlh$  +  i(a,  +  c)]/  +  ^ahly   .  .    (5) 

However,  if  the  water  leak  under  the  dam  on  the  surface  AE^ 
so  as  to  exert  an  upward  hydrostatic  pressure 

V  =  [a,  +  I  +  c\lhy, 
(to  make  an  extreme  supposition,)  the  friction  will  be  only 


and  (5)  will  be  replaced  by 

JZ  </[<?,+  F--  Fa  ......    (6) 


Experiment  shows  ("Weisbach)  that  with  y=0.33  computa- 
tions made  from  (6)  (treated  as  a  bare  equality)  give  satisfactory 
results. 

EXAMPLE.—  (Ft.,  lb.,  sec.)  With  /  =  0.33,  h  =  20  ft,  h,  = 
22  ft,  a  =  24  ft.,  a,  =  26.4  ft.,  and  o  —  30  ft.,  we  have,  mak- 
ing (6)  an  equality,  with  y'  =  %y, 


=f 

/.  i(400)=|[22(5+28.2)2  +^(24X20)  -(26.4+  J+  30)20]; 
whence,  solving  for  J,  the  width  of  top,  b  =  10.3  feet. 


568 


MECHANICS   OF   ENGINEERING. 


442,  Liquid  Pressure  on  Both  Sides  of  a  Gate  or  Rigid  Plate, — 
The  sluice-gate  AB,  for  example,  Fig.  492,  receives  a  pressure. 

Pl ,  from  the  "  head-water"  Jf,  and 
an  opposing  pressure  P3  from  the 
"tail-water"  N.     Since    these  two 
horizontal  forces  are  not  in  the  same 
j,  though  parallel,  their  resultant 
which  =  jP,  —  P^ ,  acts  horizon- 
tally in  the  same  plane,  but  at  a  dis- 
FIG.  492.  tance  below  01  =  u,  which  we  may 

find  by  placing  the  moment  of  R  about  Ot ,  equal  to  the  alge- 
braic sum  of  those  of  P1  and  P2  about  Ov. 


Ru  =  Ppj  -  Pt(a>e"  +  h). 


_ 
u 


_ 

•*  a 


(i) 

(2) 


(7,  and  Q  are  the  respective  centres  of  pressure  of  the  surfaces 
OJZ  and  OJ$,  and  u  =  distance  of  R  from  Ol  ,  while  h  =  dif- 
ference of  level  between  head  and  tail  waters.  If  the  surfaces 
O,-Z?  and  OJ$  are  both  rectangular, 

ac/  =  |A1    and    ve"  =  |A3. 

EXAMPLE.  —  Let  the  dimensions  be  as  in  Fig.  493,  both  sur- 
faces under  pressure  being  rect- 
angular and  8  ft.  wide.  Then  (ft., 
lb.,  sec.)  R  =  P,  -  P,  ,  or  (§  430) 


4]62.5 


FW.  493. 


=  20000  Ibs.  =  10  tons; 
while  f  rom  ex.  (2) 


[12X8X6X8-8X8X 
20000 


That  is,  u  =  6.93  feet,  which  locates  O.    Hence  the  pressure 
of  the  gate  upon  its  hinges  or  other  support  is  the  same  (aside 


WATEK   ON  BOTH   SIDES   OF   GATE. 


569 


from  its  own  weight),  provided  it  is  rigid,  as  if  the  single 
horizontal  force  R  =  10  tons  acted  at  the  point  (7,  2.93  ft.  be- 
low the  level  of  the  tail-water  surface. 

443.  If  the  plate,  or  gate,  is  entirely  below  the  tail-water 
surface,  the  resultant  pressure  is  applied  in  the  centre  of  gravity 
of  the  plate.  —  Proof  as  follows:  Conceive  the  surface  to  be 
divided  into  a  great  number  of  small  equal  areas,  each  =  dF; 
then,  the  head  of  water  of  any  dF  being  =  xl  on  the  head- 
water side,  and  =  #„  on  the  tail-water  side,  the  resultant  pres- 
sure on  the  dF  is  ydF(xl  —  a?a)  =yhdF,  in  which  h  is  the 
difference  of  level  between  head  and  tail  water.  That  is,  the 
resultant  pressures  on  the  equal  dF's  are  equal,  and  hence 
form  a  system  of  equal  parallel  forces  distributed  over  the  plate 
in  the  same  manner  as  the  weights  of  the  corresponding  por- 
tions of  the  plate  ;  therefore  their  single  resultant  acts  through 
the  centre  of  gravity  of  the  plate  ;  Q.  E.  D.  This  single  re- 
sultant =fyhdF  =  yhfdF  =  Fhy. 

EXAMPLE.  —  Fig.  494.  The  resultant  pressure  on  a  circular 
disk  ab  of  radius  =  8  inches,  (in 
the  vertical  partition  OK,)  which 
has  its  centre  of  gravity  3  ft. 
below  the  tail-water  surface,  with 
h  —  2  ft.,  is  (ft.,  lb.,  sec.) 

R  =  Fhy  =  nr*hy 


X 


=  174.6  Ite,   ^ 


and  is  applied  through  the  centre  jj 
of  gravity  of  the   circle. 


Jfyi-  Wf%%^^ 


dently  R  is  the  same  for  any 

depth  Mow  the  tail-water  surface,  so  long  as  h  =  2  ft. 

the  student  find  a  graphic  proof  of  this  statement.] 


[Let 


444.  Liquid  Pressure  on  Curved  Surfaces.  —  If  the  rigid  surface 
is  curved,  the  pressures  on  the  individual  dJ?'B,  or  elements  of 
area,  do  not  form  a  system  of  parallel  forces,  and  the  single  re- 
sultant (if  one  is  obtainable)  is  not  equal  to  their  sum.  In 


570  MECHANICS    OF   ENGINEERING. 

general,  the  system  is  not  equivalent  to  a  single  force,  but  can 
always  be  reduced  to  two  forces  (§  38)  the  point  of  application 
of  one  of  which  is  arbitrary  (the  arbitrary  origin  of  §  38)  and 
its  amount  =  V(2Xy  +  (2  Y)*  +  (2Z)\ 

A  single  Example  will  be  given  ;  that  of  a  thin  rigid  shell 
having  the  shape  of  the  curved  surface  of  a  right  cone,  Fig. 
495,  its  altitude  being  h  and  radius  of  base  =  r.  It  has  no 
bottom,  is  placed  on  a  smooth  horizontal  table,  vertex  up,  and 
is  filled  with  water  through  a  small  hole  in  the  apex  0,  which  is 

left  open  (to  admit  atmospheric 
pressure).  What  load,  besides  its 
own  weight  G')  must  be  placed 
upon  it  to  prevent  the  water  from 
lifting  it  and  escaping  under  the 
edge  A  ?  The  pressure  on  each 
dF  of  the  inner  curved  surface  is 
zydF  &&&  is  normal  to  the  surface. 

FIG.  495. 

Its  vertical  compon.  is  zydFmn  a, 

and  horizontal  compon.  =  zydF  cos  a.  The  dF'&  have  all 
the  same  a,  but  different  z's  (or  heads  of  water).  The  lifting 
tendency  of  the  water  on  the  thin  shell  is  due  to  the  vertical 
components  forming  a  system  of  ||  forces,  while  the  horizon- 
tal components,  radiating  symmetrically  from  the  axis  of  the 
cone,  neutralize  each  other.  Hence  the  resultant  lifting  force 
is 

F=  ^(vert.  comps.)  =  y  sin  afzdF=  y  sin  a  F~z\     (1) 

where  F  =  total  area  of  curved  surface,  and  z  —  the  "head  of 
water"  of  its  centre  of  gravity.  Eq.  (1)  may  also  be  written 
thus: 

V=yFb~z',     .......    (2) 

in  which  Fb  =  F  sin  a  =  area  of  the  circular  base  =  area  of 
the  projection  of  the  curved  surface  upon  a  plane  ~\  to  the 
vertical,  i.e.,  upon  a  horizontal  plane.  Hence  we  may  write 


,    .     ......     (3) 

since  z  =  •§  A,  being  the  z  of  the  centre  of  gravity  of  the  curved 


CONICAL    SHELL.  571 

surface  and  not  that  of  the  base,  y  =  heaviness  of  water.  If 
G'  =  weight  of  the  shell  and  is  <  T7,  an  additional  load  of 
V  —  G'  will  be  needed  to  prevent  the  lifting.  If  the  shell  has 
a  bottom  of  weight  =  G" ,  forming  a  base  for  the  cone  and 
rigidly  attached  to  it,  we  find  that  the  vertical  forces  acting  on 
the  whole  rigid  body,  base  and  all,  are :  V  upward ;  G'  and 
G"  downward;  and  the  liquid  pressure  on  the  base,  viz., 
Vf  =  7tr*hy  (§  428$)  also  downward.  Hence  the  resultant 
vertical  force  to  be  counteracted  by  the  table  is  downward,  and 

-  G'  +  G"  +  V  -  F,  which  =  G'  +  G"  +  \nr*hy  ;    (4) 

i.e.,  the  total  weight  of  the  rigid  vessel  and  the  water  in  it,  as 
we  know,  of  course,  in  advance. 


OHAPTEE  HI. 

EARTH  PRESSURE  AND  RETAINING  WALLS. 

[NOTE. — This  chapter  was  outlined  and  written  mainly  by 
Prof.  0.  L.  Crandall,  and  is  here  incorporated  with  his  permis- 
sion. The  theory  of  earth  pressure  is  arranged  from  Bau- 
meister.] 

445.  Angle  of  Repose. — Granular  materials,  like  dry  sand, 
loose  earth,  soil,  gravel,  pease,  shot,  etc.,  on  account  of  the 
friction  between  the  component  grains,  occupy  an  intermediate 
position  between  liquids  and  large  rigid  bodies.  When  heaped 
up,  the  side  of  the  mass  cannot  be  made  to  stand  at  an  inclina- 
tion with  the  horizontal  greater  than  a  definite  angle  called  the 
angle  of  natural  slope,  or  angle  of  repose,  different  for  each 
material ;  so  that  if  the  side  of  the  mass  is  to  be  retained  per- 
manently at  some  greater  angle,  a  Retaining  Wall  (or  "  Revet- 
ment Wall"  in  military  parlance)  becomes  necessary  to  sup- 
port it.  If  the  material  is  somewhat  moist  it  may  be  made  to 
stand  alone  at  an  inclination  greater  than  that  of  the  natural 
slope,  on  account  of  the  cohesion  thus  produced,  but  only  as 
long  as  the  degree  of  moisture  remains ;  while  if  much  water 
is  present,  it  assumes  the  consistency  of  mud  and  may  require 
a  much  thicker  wall,  if  it  is  to  be  supported  laterally,  than  if 
dry. 

In  dealing  with  earth  to  be  supported  by  a  retaining  wall, 
we  consider  the  former  to  have  lost  any  original  cohesion 
which  may  have  existed  among  its  particles,  or  that  it  will 
eventually  lose  it  through  the  action  of  the  weather ;  and  hence 
treat  it  as  a  granular  material. 

A  few  approximate  values  of  the  angle  of  natural  slope  are 

572 


RETAINING    WALLS. 


573 


given  below,  being  taken  from  Fanning,  p.  345 ;  see  reference 
on  p.  538  of  this  work. 


MATERIAL. 

Angle 
of  Repose. 

Coefficient 
of  Friction. 

Ratio 
of  Slope. 

28° 

.532 

Horiz.  to  vert. 
1.88  to    1 

30° 

.577 

1.73 

45° 

1.000 

1.00 

Wet  clay      «... 

15° 

.268 

3.73 

45° 

1.000 

1.00 

42° 

.900 

1.11 

Gravel            

38° 

.781 

1.28 

36° 

.727 

1.38 

Vegetable  soil                    • 

35° 

.700 

1.43         1 

Peat              

20° 

.364 

2.75         1 

The  angle  of  repose,  or  natural  slope,  is  also,  evidently,  the 
angle  of  friction  between  two  masses  of  the  same  granular 
material. 

446.  Earth  Pressure,  and  Wedge  of  Maximum  Thrust. — Fig. 
496.  Let  AB  be  a  retaining  wall,  having  a  plane  face  AB  in 
contact  with  a  mass  of  earth  ABD,  both  wall  and  earth  being 
of  indefinite  extent  ~|  to  the  paper.  N 

Let  AD  be  the  natural  slope  of  the  earth,  making  an  angle  ft 
with  the  vertical  (ft  is  the  complement  of  the  angle  of  repose ; 
see  preceding  table).  Since  AB,  making  an  angle  a  with 
the  vertical,  is  more  nearly  vertical  than  AD,  the  retaining 
wall  is  necessary,  to  keep  the  mass  ABD  in  the  position 
shown.  The  profile  BCD  may  be  of  any  form  in  this  general 
discussion.  Suppose  the  wall  to  be  on  the  point  of  giving 
way  ;  then  the  following  morions  are  impending  : 

1st.  Sliding  is  impending  between  some  portion  ABCfA  of 
the  mass  of  earth  and  the  remainder  C' AD,  the  surface  of 
rupture  AC'  (Cr  not  shown  in  figure  because  not  found  yet, 
but  lying  somewhere  on  the  profile  BCID]  being  assumed 
plane,  and  making  some  angle  $  (to  be  determined)  with  the 
vertical.  At  this  instant  the  resultant  pressure  Nf  of  AC'D 
on  the  plane  AC'  of  the  mass  ABC'  (a  wedge)  must  make 
an  angle  =  ft  (=  comp.  of  angle  of  friction)  with  AC'  on 
the  upper  side. 


574 


MECHANICS    OF    ENGINEERING. 


2d.  A  downward  sliding  of  the  mass  ABC1  along  the  back 
face  AB  of  the  wall.  That  is,  the  resultant  pressure  P'  of 
the  wall  against  the  mass  BA  C'  at  this  instant  makes  an  angle 


sx-xx     &-V--.S 


FIG.  496. 


complement  of  angle  of  friction  between  the  earth  and 
wall)  with  the  plane  AB  and  on  the  upper  side.  The  weight 
of  the  wedge  of  earth  BAG'  will  be  called  G',  and  we  desire 
to  find  the  pressure  P'  against  the  wall. 

Let  BA  £7 be  a  wedge  (of  the  earth-mass),  in  which  A  C  makes 
any  angle  0  with  A  V,  and  suppose  it  to  be  on  the  point'  of 
moving  down  and  forcing  out  the  wall;  thus  encountering 
friction  both  on  the  plane  A  C  and  the  plane  AB.  Then  the 
forces  acting  on  it  are  three,  acting  in  known  directions  ;  viz. : 
^  G,  its  own  weight,  vertical  ;/W,  the  resultant  pressure  of  the 
earth  below  it,  making  an  angle  /3  with  A  C  on  upper  side  ; 
andxP,  the  resultant  pressure  of  the  wall,  at  angle  6  with  AB 
(see  Fig.  496  for  positions  of  N  and  P).  If  now  we  express 
the  force  P  in  terms  of  0  and  other  quantities,  and  find  that 
value  0',  of  0,  for  which  P  is  a  maximum,  we  thereby  deter- 
mine the  "wedge  of  maximum  thrust"  ABC 'A  ;  while  this 
maximum  thrust,  P' ,  is  the  force  which  the  wall  must  be  de- 
signed to  withstand.  [If  the  wall  is  overturned,  the  earth 
will  sink  with  it  until  this  part  of  its  surface  gradually  as- 
sumes the  natural  slope.] 

Let  G  =  weight  of  prism  of  base  ABC,  and  altitude  —  unity 
"1  to  paper;  then  G  =  y  X  area  ABC,  where  y  =  "  heavi- 
ness" —  wgt.  per  cub.  unit,  of  earth.  JSTow  P,  6r,  and  $ 
balance;  therefore,  in  triangle  aba,  if  a^'and  ac  are  drawn  jj 


RETAINING   WALLS.  575 

and  =  G  and  ^respectively,  be  is  —  and  ||  to  P;  and  from 
Trigonometry  we  have 

p  -  a    sin  W  ~  #1    •  m 

> 


in  which  <?  stands  for  a  +  0,  for  brevity,  being  the  angle 
which  P  makes  with  the  vertical.  N  makes  an  angle  =  ft  —  0 
with  the  vertical. 

The  value,  0',  of  0,  which  makes  P  a  maximum  is  found 

by  placing  — —  =  0.  From  eq.  (1),  remembering  that  G  is  a 
function  of  0,  and  that  ft  and  d  are  constants,  we  have 

:  -  <£)  -  G  cos  O  -  <f»>~|  +  Gsin  (ft  -  </>)  cos  (/3-f  5  -  0) 


For  P  to  be  a  maximum  we  must  put 

numerator  of  above  =  0 


To  find  a  geometrical  equivalent  of  —  ,  denote  AC  by,  Z> 

and  draw  AE,  making  an'  angle  =  d<p  with  AC.  Now  the 
area  ACI  =  AI  X  ±CE=.(L  +  dlftZdQ  =  %L*d<}>  .  .  . 
(neglecting  infinitesimal  of  2d  order).  Now 

dG  =  y  X  area  ACI  X  unity  ;  /.  -=—  -  =  \yl?\  .*.  («)  becomes 


—      ~  sn 


+  G  ein  (/3  -  0)  cos  (/?  +  d  —  0)  =  0  ; 

i.e.,  G  = 

__  jyZ3  ain  (/?  —  0)  sin  (/?  +  tf  —  0)  _ 
sin  (/S  -f  6  —  0)  cos  (/?  —  0)  —  cos  (/?  +  d  —  0)  sin  (ft  —  0) 


576  MECHANICS    OF   ENGINEERING. 

when  P  is  a  maximum  ;  and  hence,  calling  G'  and  0'  and 
the  values  of  G,  0,  and  Z,  for  max.  P,  we  have 


and  therefore  from  (1)  P  max.  itself  is 


447.  Geometric  Interpretation  and  Construction.  —  If  in  Fig. 
496  we  draw  CF,  making  angle  d  with  AD,  G  being  any 
point  on  the  ground  surface  BD,  we  have 

CF-=  L  sin  (ft  ~  <l>\ 
sin  tf 

Drop  a  perpendicular  FH  from  F  to  A  G,  and  we  shall  have 

FH=  CF.  sin  (ft  +  6  -  <A),  =  L  .  «h  (/>-  *)« 

sin  o 

From  this  it  follows  that  the  weight  of  prism  of  fiie^l  CF 
and  unit  height 


sn 


When  ^IC1'  (as  0  varies)  assumes  the  position  and  value 
bounding  the  prism  of  maximum  thrust,  Fig.  497,  L  becomes 
—  Z\  and  0  =  cf)f\  and  eq.  (4)  gives  the  weight  of  the  prism 
AC'F'.  This  weight  is  seen  to  be  equal  to  that  of  the  prism 
(or  wedge)  of  maximum  thrust  ABC',  by  Comparing  eq.  (4) 
with  eq.  (2);  that  is,  AC'  bisects  the  area  ABC'F',  and 
hence  may  be  determined  by  fixing  such  a  point  C ' ',  on  the 
upper  profile  BD,  as  to  make  the  triangular  area  AC'F' 
equal  to  the  sectional  area  of  the  wedge  BC'A\  C'F'  being 
drawn  at  an  angle  =  d  with  AD. 

This  holds  for  any  form   of  ground  surface  BD,  or  any 


RETAINING    WALLS. 


577 


values  of  the  constants  /?,  a,  or  6.     C'  is  best  found  graphic- 
ally by  trial,  in  dealing  with 
an  irregular  profile  BD. 

Having  found  AC'^  = 
L' ,  Pf  can  be  found  from 
(3),  or  graphically  as  fol- 
lows :  (Fig.  497)  With  F' 
as  a  centre  and  radius  = 
C'F')  describe  an  arc  cut- 
ting AD  in  J ',  and  join 
C'J'.  The  weight  of  prism 
with  base  C'J'F'  and  unit  height  will  =  P '.  For  that  prism 
has  a  weight 


FIG.  497. 


'J'  .  Q'H'\ 


but 
and 


Sn          - 


sn 


C'H'  =  L  sin  (ft  -  00  ; 

sina(/5  —  00 . 


weight  of  prism  C'JfF=  \yL  ,  —  ^  . 

sin  o 

[See  eq.  (3).] 

448.  Point  of  Application  of  the  Resultant  Earth  Thrust— 

This  thrust  (called  P'  throughout  this  chapter  except  in  the 
present  paragraph)  is  now  known  in  magnitude  and  direction, 
but  not  in  position ;  i.e.,  we  must  still  determine  its  line  of 
action,  as  follows : 

Divide  AB  into  a  number  of  equal  parts,  ab,  be,  cd,  etc.- 
see  Fig.  498.  Treat  ab  as  a  small  retaining  wall,  and  find  the 
magnitude  P'  of  the  thrust  against  it  by  §  447 ;  treat  ac  simi- 
larly, thus  finding  the  thrust,  P",  against  it ;  then  ad,  ae,  etc., 
the  thrusts  against  them  being  found  to  be  P" ',  PIV,  etc. ;  and 
so  on.  ISTow  the  pressure 

P'  on  ab  is  applied  nearly  at  middle  of  ab, 
P"  __  p'          «  «  «        «     ic 


'678  MECHANICS   OF   EFG1NEEBING. 

and  so  on.  Erect  perpendiculars  at  the  middle  paints  of  ab, 
be,  cd,  etc.,  equal  respectively  to  P' ^ 
P"  -  p^  p>>>  _  p>>,  etc.,  and"  join  the 
ends  of  the  perpendiculars.  The  per- 
pendicular through  the  centre  of  gravity 
of  the  area  so  formed  (Fig.  498)  will 
give,  on  AB,  the  required  point  of  ap- 
plication of  the  thrust  or  earth  pressure 
on  AB,  and  this,  with  the  direction  and 
FIG.  498.  magnitude  already  found  in  §  447,  will 

completely  determine  the  thrust  against  the  wall  AB. 

449.  Special  Law  of  Loading. — If  the  material  to  be  retained 
consists  of  loose  stone,  masses  of  masonry,  buildings,  or  even 
moving  loads,  as  in  the  case  of  a  wharf  or  roadway,  each  can 
be  replaced  by  the  same  weight  of  earth  or  other  material 
which  will  render  the  bank  homogeneous,  situated  on  the  same 
verticals,  and  the  profile  thus  reduced  can  be  treated  by  §§447 
and  448. 

Should  the  solid  mass  extend  below  the  plane  of  rupture, 
AC',  and  the  plane  of  natural  slope, it  will  become  a  retaining 
wall  for  the  material  beyond,  if  strong  enough  to  act  as  such 
(limiting  the  profile  ABCD  of  Fig.  496  to  the  front  of  the 
mass,  or  to  the  front  and  line  of  rupture  for- maximum  thrust 
above  it,  if  it  does  not  reach  the  surface);  if  not  strong  enough, 
or  if  it  does  not  reach  below  the  plane  of  natural  slope,  its 
presence  is  better  ignored,  probably,  except  that  the  increased 
weight  must  be  considered. 

The  spandrel  wall  of  an  arch  may  present  two  of  these 
special  cases ;  i.e.,  the  profile  may  be  enlarged  to  include  a 
moving  load,  while  it  may  be  limited  at  the  back  by  the  other 
spandrel. 

If  the  earth  profile  starts  at  the  front  edge  of  the  top  of 
wall,  instead  of  from  the  back  as  at  B,  Fig.  496,  eq.  (3)  would 
only  apply  to  the  portion  behind  AB  prolonged,  leaving  the 
part  on  the  wall  (top)  to  be  treated  as  a  part  of  the  wall  to  aid 
in  resisting  the  thrust. 

If  the  wall  is  stepped  in  from  the   footings,  or  foundation 


RETAINING   WALLS. 


579 


.courses,  probably  the  weak  section  will  be  just  above  them ;  if 
stepped  at  intervals  up  the  back  of  the  wall,  the  surface  of  separa- 
tion between  the  wall  and  filling,  if  it  is  plane,  will  probably 
pass  through  the  first  step  and  incline  forward  as  much  as  pos- 
sible without  cutting  the  wall. 

450.  Straight  Earth-profile. — The  general  case  can  be  simpli- 
fied as  follows  (the  earth-profile  BD  being  straight,'  at  angle 
=  C  with  vertical,  =  DET) :  Since  the  triangles  ABC'  and 


7- 


u  \ 


N-\ 

y^ 


FIG.  499. 


C'AF'  are  equal,  from  §  447,  and  A  C'  is  common,  therefore 
B8—F'H  (both  being  drawn  1  to  AC').  Draw  AE  and 
BM  ||  to  F'G'  (i.e.,  at  angle  d  with  AD\  cutting  DB,  pro- 
longed, in  ^.  We  have 


DE 


EA 


C'E 


and 


But  C'F'  =  ^Jf  (since  J?xS  =  H'F') ; 


therefore 


EA 


=      =;  i.e.,  DE.  BE  = 


(/'JS'       ^^T 


which  justifies  the  following  construction  for  locating  the  de- 
sired point  C'  on  BD,  and  thus  finding  AC'  —  L'  and  the 
angle  0':  Describe  'a  circle  on  ED  as  a  diameter,  and  draw 


580  MECHANICS    OF   ENGINEERING. 


~|  to  BD)  thus  fixing  X  in  the  curve.  With  centre  E 
describe  a  circular  arc  through  JT,  cutting  BD  in  (7',  required 

Having  AC'  (i.e.,  Z'),  <p'  is  known  ;  hence  from  eq.  (3)  we 
obtain  the  earth  thrust  or  pressure  P'\  or,  with  F'  as  centre 
and  radius  —  C'F',  describe  arc  G'J'\  then  the  triangle  G'F'J1 
is  the  base  of  a  prism  of  unity  height  whose  weight  =  P'  (as 
in  §  447). 

Centre  of  Pressure.  —  Applying  the  method  of  §  448,  Fig. 
498,  to  this  case,  we  find  that  the  successive  Z"s  are  propor- 
tional to  the  depths  ab,  ac9  ad,  etc.,  anc^  that  the  successive  P'B 
are  proportional  [see  (3)]  to  the  squares  of  the  depths  ;  hence 
the  area  in  Fig.  498  must  be  triangular  in  this  case,  and  the 
point  of  application  of  the  resultant  pressure  on  AB  is  one 
third  of  AB  from  A  :  just  as  with  liquid  pressure. 

451.  Resistance  of  Retaining  Walls.—  (Fig.  500.)  Knowing 
the  height  of  the  wall  we  can  find  its  weight,  =  Gl  ,  for  an  as- 
sumed thickness,  and  unity  width  ~]  to  paper.  The  resultant 
of  Gl  ,  acting  through  the  centre  of  gravity  of  wall,  and  P',  the 
thrust  of  the  embankment,  in  its  proper 
line  of  action,  should  cut  the  base  A  V 
within  the  middle  third  and  make  an 
angle  with  the  normal  (to  the  base)  less 
than  the  angle  of  friction. 

For  the  straight   wall   and   straight 
earth-profile  of  Fig.  499  and  §  450,  the 
Fm.5oo  length  Z',  =  AC',  can  be  expressed  in 

terms  of  the  (vertical)  height,  A,  of  wall,  thus  : 


cos  a 
-  oi\  h         sin  (C  — 


and  Z'  --  AC'  -  AB    *™  - 

UUU  JL/     —  J3.  \j      —  -£LX>  •    -j 7— — ,r —  —  •   — TT; T/V  > 

sin  (C  —  0 )        cos  OL     sin  (C  —  0 ) 
,•.  eq.  (3)  becomes 

__  Aa         Bin'G*  -  0Qsina(C-^)       ,        Aa 

tr  ~^~^  '    "~¥iTyin2(C'-  00  W cos9  a      ' W 

[A  representing  the  large  fraction  for  brevity.] 


RETAINING    WALLS. 


581 


This  equation  will  require,  for  a  wall  of  rectangular  section, 
that  the  thickness,  d,  increase  as  h,  in  order  that  its  weight  may 
increase  as  A2  (i.e.,  as  P')  and  that  its  resisting  moment  may 
increase  with  the  overturning  moment. 

By  this  equality  of  moments  is  meant  that  /"  a  =  GJ> ; 
where  a  and  J  are  the  respective  lever-arms  of  the  two  forces 
about  the  front  edge  of  the  middle  third.  (AB  is  the  back  of 
the  wall.)  In  other  words,  their  resultant  will  pass  through 
this  point. 

The  following  table  is  computed  on  the  basis  just  mentioned, 
viz.,  that  the  resultant  of  P'  and  G  shall  pass  through  the 
front  edge  of  the  midde  third. 

The  symbols  of  eq.  (5)  and  the  table  are  all  shown  in  Fig. 
499,  except  y,  0,  and  $.  y  =  weight  of  a  cubic  foot  of  earth, 
here  assumed  =  f  that  of  masonry  (e.g.,  if  earth  weighs 
100  Ibs.,  masonry  is  assumed  to  weigh  150  Ibs.  per  cubic  foot) ; 
6  =  angle  which  the  thrust  P'  makes  with  the  back  of  the 
wall ;  and  $  =  a  -f-  8,  =  6  in  this  case  as  the  wall  is  vertical, 
or  a  =  0.  d  is  the  proper  safe  thickness  to  be  given  to  the 
wall,  of  rectangular  section,  to  prevent  overturning,  as  stated 
above ;  h  is  the  altitude,  and  A  is  the  fraction  shown  in  eq.  (5). 

Whether  the  wall  is  safe  against  sliding  on  its  base,  and 
whether  a  safe  compression  per  unit  area  is  exceeded  on  the 
front  edge  of  the  base,  are  matters  for  separate  consideration. 
The  latter  will  seldom  govern  with  ordinary  retaining  walls. 

a  =  0;  i.e.,  wall  is  vertical;  also  density  of  wall  =  f  that  of  the  earth. 


I. 

II. 

III. 

<  =  90° 

£  =  90° 

£  =  0 

0  =  90° 

0=0 

0  =  0 

tan/3 

ft 

v 

A 

d 

i' 

A 

d 

*' 

^l 

d 

1.0 

45° 

22|° 

.17 

.34* 

26° 

.18 

.22* 

45°. 

.71 

.33* 

1.5 

56i° 

28° 

.29 

.44* 

33° 

.26 

.30* 

56° 

.83 

.43* 

2.0 

63^° 

31f° 

.38 

.51* 

38° 

.33 

.36* 

63° 

.89 

.51* 

40 

76° 

38° 

.61 

.64* 

45° 

.54 

.50* 

76° 

.97 

.65* 

Infinity 

90° 

45° 

1.00 

.82* 

90° 

1.00 

.82* 

90° 

1.00 

.82* 

In  Case  I  of    table,  since   a  =  0,  B  =  90°  and  C  =  90° ; 
.-.  d  =  90°,  and  hence  C'F'  of  Fig.  499  is  1  to  AD,  so  that 


582 


MECHANICS   OF   ENGINEERING. 


(since   the    area    of    &ABC'  =  kAC'F')    0'  must  = 
These  values,  in  (5),  give 

P1  =  \YK  tan8  %ft ;  i.e.,  A  =  tan'  %/3.  .     .     . 


(6) 


In  Case  II,  since  C  =  90°,  a  =  0  and  0  =  fa    .-.    d  =  /3  ; 
and  (5)  reduces  to 

sina(/?-0Q.    .  sina(/?-00         m 

^h  sin/?  cos'  0"    '  ~sln>cos'0'- 


In  Case  III,  C  =  /?  and  ^Z>  wiU  be  ||  to  ^1Z>,  D  being  at 

infinity.   See  Fig.  501.   Through 
^  draw  BH  ~\  to  ^^>,  and  RF" 


now  to  be  located  on  BD,  so  as 
to  make  (area  of)  &ABC'  = 
(area  of)  l^AC'F1  (according 
to  §  447),  the  angle  C '  F 'A  being 
=  d  =  <x-\-  0 ;  =  0,  in  this  case, 
and  hence  also  =  ft.  Conceive 
FIG.  SOL  B  and  F'  to  be  joined. 

Now  &AC'F'  =  &ABF"  +  &BF'F". 

But  &ABC'   =  &BF'F"  (equal  bases  and  altitudes). 

Hence  A  ABC'  cannot  =  &AC'F'  unless  Cf  is  moved  out 
to  infinity  ;  and  then  0r  becomes  =  ft,  and  eq.  (5)  reduces  to 


[Increasii 


Pf=  %yh*  sin  fi ;    i.e.,  A  =  sin  fi. 


(8) 


[ncreasing  a  from  zero  will  decrease  the  thickness  d ;  i.e., 
inclining  the  wall  inwards  will  decrease  the  required  thickness, 
but  diminish  the  frictional  stability  at  the  base,  unless  the  lat- 
ter be  ~l  to  AB.  The  back  of  the  wall  is  frequently  inclined 
outwards,  making  the  section  a  trapezoid,  to  increase  the  fric- 
tional stability  at  the  base  when  necessary,  as  with  timber 
walls  supporting  water.] 


EET AIDING    WALLS.  583 

452,  Practical  Considerations.  —  An  examination  of  the 
values  of  A  and  d  in  the  table  of  §  451  will  show  that  in  sup- 
porting quicksand  and  many  kinds  of  clay  which  are  almost 
fluid  under  the  influence  of  water,  it  is  important  to  know 
what  kind  of  drainage  can  be  secured,  for  on  that  will  depend 
the  thickness  of  the  wall.  With  well  compacted  material  free 
from  water-bearing  strata,  an  assumed  natural  slope  of  1£  to  1 
(i.e.,  1J  hor.  to  1  vert.)  will  be  safe ;  the  actual  pressure  below 
the  effect  of  frost  and  surface  water  will  be  that  due  to  a  much 
steeper  slope  on  account  of  cohesion  (neglected  in  this  theory). 

The  thrust  from  freshly  placed  material  can  be  reduced  by 
depositing  it  in  layers  sloping  back  from  the  wall.  If  it  is  not 
so  placed,  however,  the  natural  slope  will  seldom  be  flatter 
than  1%  to  1  unless  reduced  by  water.  In  supporting  material 
which  contains  water-bearing  strata  sloping  toward  the  wall 
and  overlain  by  strata  which  are  liable  to  become  semi-fluid 
and  slippery,  the  thrust  may  exceed  that  due  to  semi-fluid  ma- 
terial on  account  of  the  surcharge.  If  these  strata  are  under 
the  wall  and  cannot  be  reached  by  the  foundation,  or  if  resist- 
ance-to  sliding  cannot  be  obtained  from  the  material  in  front 
by  sheet-piling,  no  amount  of  masonry  can  give  security. 

Water  at  the  back  of  the  wall  will,  by  freezing,  cause  the 
material  to  exert  an  indefinitely  great  pressure,  besides  disinte- 
grating the  wall  itself.  If  there  is  danger  of  its  accumulation, 
drainage  should  be  provided  by  a  layer  of  loose  stone  at  the 
back  leading  to  "weep-holes"  through  the  wall. 

A  friction-angle  at  the  back  of  the  wall  equal  to  that  of  the 
filling  should  always  be  realized  by  making  the  back  rough  by 
steps,  or  projecting  stones  or  bricks.  Its  effect  on  the  required 
thickness  is  too  great  to  be  economically  ignored. 

The  resistance  to  slipping  at  the  base  can  be  increased,  when 
necessary,  by  inclining  the  foundation  inwards ;  by  stepping 
or  sloping  the  back  of  the  wall  so  as  to  add  to  its  effective 
weight  or  incline  the  thrust  more  nearly  to  the  vertical;  by 
sheet-piling  in  front  of  the  foundation,  thus  gaining  the  resist- 
ance offered  by  the  piles  to  lateral  motion  ;  by  deeper  founda- 
tions, gaining  the  resistance  of  the  earth  in  front  of  the  wall. 


584  MECHANICS    OF   ENGINEERING. 

The  coefficient  of  friction  on  the  base  ranges,  according  to 
Trautwine,  from  0.20  to  0.30  on  wet  clay ; 
"       .50  to    .66   "  dry  earth ; 
"       .66  to    .75    "  sand  or  gravel ; 
"       .60  on  a  dry  wooden  platform  ;  to  .75  on  a 
wet  one. 

If  the  wall  is  partially  submerged,  the  buoyant  effort  should 
be  subtracted  from  Gl ,  the  weight  of  wall. 

453.  Results  of  Experience. — (Trautwine.)  In  railroad  prac- 
tice, a  vertical  wall  of  rectangular  section,  sustaining  sand, 
gravel,  or  earth,  level  with  the  top  [p.  682  of  Civ.  Eng.  Pocket 
Book]  and  loosely  deposited,  as  when  dumped  from  carts,  cars, 
etc.,  should  have  a  thickness  d,  as  follows : 

If  of  cut  stone,  or  of  first-class  large  ranged  rubble,  in  mortar.  .  .  .  d  =  .35ft 

"    good  common  scabbled  mortar-rubble,  or  brick d  —  .40A 

"    well  scabbled  dry  rubble d  =  .50A 

Where  h  includes  the  total  height,  or  about  3  ft.  of  foundations. 
(d)  For  the  best  masonry  of  its  class  h  may  be  taken  from 
the  top  of  the  foundation  in  front. 

(b)  A.  mixture  of  sand  or  earth,  with  a  large  proportion  of 
round  boulders  or  cobbles,  will  weigh  more  than  the  backing 
assumed  above  ;  requiring  d  to  be  increased  from  one  eighth  to 
one  sixth  part. 

(c)  The  wall  will  be  stronger  by  inclining  the  back  inwards, 
especially  if  of  dry  masonry,  or  if  the  backing  is  put  in  place 
before  the  mortar  has  set. 

(d)  The  back  of  the  wall  should  be  left  rough  to  increase 
friction. 

(e)  Where  deep  freezing  occurs,  the  back  should  slope  out- 
ward for  3  or  4  feet  below  the  top  and  be  left  smooth. 

(/)  When  a  wall  is  too  thin,  it  will  generally  fail  by  bulging 
outward  at  about  one  third  the  height.  The  failure  is  usually 
gradual  and  may  take  years. 

(g)  Counterforts,  or  buttresses  at  the  back  of  the  wall,  usually 
of  rectangular  section,  may  be  regarded  as  a  waste  of  ma- 
sonry, although  considerably  used  in  Europe;  the  bond  will 


RETAINING    WALLS.  585 

seldom  hold  them  to  the  wall.  Buttresses  in  front  add  to  the 
strength,  but  are  not  common,  on  account  of  expense. 

(Ji)  Land-ties  of  iron  or  wood,  tying  the  wall  to  anchors  im- 
bedded below  the  line  of  natural  slope,  are  sometimes  used  to 
increase  stability. 

(?')  Walls  with  curved  cross-sections  are  not  recommended. 

454.  Conclusions  of  Mr.  B.  Baker. — ("  Actual  Lateral  Pressure 
of  Earthwork.")     Experience  has  shown  that  d  =  0.25A,  with 
batter  of  1  to  2  inches  per  foot  on  face,  is  sufficient  when 
backing  and  foundation  are  both  favorable ;  also  that  under  no 
ordinary  conditions  of  surcharge  or  heavy  backing,  with  solid 
foundation,  is  it  necessary  for  d  to  be  greater  than  0.50A. 

Mr.  Baker's  own  rule  is  to  make  d  =  0.33A  at  the  top  of 
the  footings,  with  a  face  batter  of  1J  inches  per  foot,  in  ground 
of  average  character;  and,  if  any  material  is  taken  out  to  form 
a  face-panel,  three  fourths  of  it  is  put  back  in  the  form  of  a 
pilaster.  The  object  of  the  batter,  and  of  the  panel  if  used,  is 
to  distribute  the  pressure  better  on  the  foundation.  All  the 
walls  of  the  "  District  Railway"  (London)  were  designed  on 
this  basis,  and  there  has  not  been  a  single  instance  of  settle- 
ment, of  overturning,  or  of  sliding  forward. 

455.  Experiments  with  Models. — Accounts   of   experiments 
with  apparatus  on  a  small  scale,  with  sand,  etc.,  may  be  found 
in  vol.  LXXI  of  Proceedings  of  Institution  of  Civil  Engineers, 
London,  England  (p.  350) ;  also  in  vol.  n  of  the  "  Annales  des 
Ponts  et  Chaussees"  for  1885  (p.  788). 

The  results  of  these  experiments,  and  the  results  of  experi- 
ence given  in  §§  453  and  454,  when  compared  with  the  table 
of  p.  581,  indicate  a  fairly  close  agreement  between  practice 
and  theory.  This  agreement  is  believed  to  be  close  enough 
BO  that  the  general  method  of  §§  447  and  451,  with  the  table 
of  p.  581,  can  be  relied  upon  in  practice.  The  greatest  value  of 
this  method  will,  of  course,  be  for  cases  of  exceptional  loading, 
inclined  walls,  etc.,  where  the  results  of  experience  do  not 
furnish  so  valuable  a  guide. 


CHAPTEE  IV. 

HYDROSTATICS  (Contmued)-lMKERSWN  AND  FLOTATION. 

456.  Rigid  Body  Immersed  in  a  Liquid.  Buoyant  Effort. — If 
any  portion  of  a  body  of  homogeneous  liquid  at  rest  be  con- 
ceived to  become  rigid  without  alteration  of  shape  or  bulk,  it 
would  evidently  still  remain  at  rest ;  i.e.,  its  weight,  applied  at 
its  centre  of  gravity,  would  be  balanced  by  the  pressures,  on  its 
bounding  surfaces,  of  the  contiguous  portions  of  the  liquid ; 
hence, 

If  a  rigid  body  or  solid  is  immersed  in  a  liquid,  loth  being 
at  rest,  the  resultant  action  upon  it  of  the  surrounding  liquid 
(or  fluid]  is  a  vertical  upward  force  called  the  "buoyant 
effort"  equal  in  amount  to  the  weight  of  liquid  displaced, 
and  acting  through  the  centre  of  gravity  of  the  volume  (con- 
sidered as  homogeneous)  of  displacement  (now  occupied  by  the 
solid).  This  point  is  called  the  centre  of  buoyancy,  and  is 
sometimes  spoken  of  as  the  centre  of  gravity  of  the  displaced 
water.  If  V  =  the  volume  of  displacement,  and  y  =  heavi- 
ness of  the  liquid,  then  the 

buoyant  effort  =  Vy (1) 

(By  "  Tolume  of  displacement"  is  meant,  of  course,  the  volume 
of  liquid  actually  displaced  when  the  body  is  immersed.) 

If  the  weight  Gr  of  the  solid  is  not  equal  to  the  buoyant 
effort,  or  if  its  centre  of  gravity  does  not  lie  in  the  same  verti- 
cal as  the  centre  of  buoyancy,  the  two  forces  form  an  unbal- 
anced system  and  motion  begins.  But  as  a  consequence  of 
ihis  very  motion  the  action  of  the  liquid  is  modified  in  a  man- 
ner dependent  on  the  shape  and  kind  of  motion  of  the  body. 

586 


IMMERSION. 


587 


Problems  in  this  chapter  are  restricted  to  cases  of  rest,  i.e., 
balanced  forces. 

Suppose  G'  =  V'y ;  then, 

If  the  centre  of  gravity  lies  in  the  same  vertical  line  as  the 
centre  of  buoyancy  and  underneath  the  latter,  the  equilibrium 
is  stable ;  i.e.,  after  a  slight  angular  disturbance  the  body  re- 
turns to  its  original  position  (after  several  oscillations) ;  while 
if  above  the  latter,  the  equilibrium  is  unstable.  If  they  coin- 
cide, as  when  the  solid  is  homogeneous  (but  not  hollow),  and 
of  the  same  heaviness  (§  7)  as  the  liquid,  the  equilibrium  is 
indifferent,  i.e.,  possible  in  any  position  of  the  body. 

The  following  is  interesting  in  this  connection  : 

In  an  account  of  the  new  British  submarine  boat  "  Nautilus," 
a  writer  in  Chambers^  Journal  remarked  [1887]  :  a  At  each 
side  of  the  vessel  are  four  port-holes,  into  which  fit  cylinders 
two  feet  in  diameter.  When  these  cylinders  are  projected 
outwards,  as  they  can  be  by  suitable  gearing,  the  displacement 
of  the  boat  is  so  much  increased  that  the  vessel  rises  to  the 
surface;  but  when  the  cylinders  are  withdrawn  into  their 
sockets,  it  will  sink." 

As  another  case  in  point,  large  water-tight  canvas  "air-bags" 
have  recently  been  used  for  raising  sunken  ships.  They  are 
sunk  in  a  collapsed  state,  attacked  by  divers  to  the  submerged 
vessel,  and  then  inflated  with  air  from  pumps  above,  which  of 
course  largely  augments  their  displacement  while  adding  no 
appreciable  weight. 

457.  Examples  of  Immersion. — Fig.  502.  At  (a)  is  an  ex- 
ample of  stable  equi- 
librium, the  centre  of 
buoyancy  B  being  above 
the  centre  of  gravity  (7, 
and  the  buoyant  effort 

V'y  =  Gf  =  the  weight 
of  the  solid  ;  at  (#'),  con- 
versely, we  have  un- 
stable equilibrium,  with 

V'y  still  =  G'.     At  (bj  the  buoyant  effort  V'y  is  >  G',  and 


FIG.  502.1 


588  MECHANICS    OF   ENGINEERING. 

to  preserve  equilibrium  the  body  is  attached  by  a  cord  to  the 
bottom  of  the  vessel.  The  tension  in  this  cord  is 

*=  V'y-0'.   ......   (i) 

At  (c)  V'y  is  <  #',  and  the  cord  must  be  attached  to  a 
support  above,  and  its  tension  is 

8.  =  G'  -  V'y (2) 

If  in  eq.  (2)  [(<?)  in  figure]  we  call  Se  the  apparent  weight  of 
the  immersed  body,  and  measure  it  by  a  spring-  or  beam-bal- 
ance, we  may  say  that 

The  apparent  weight  of  a  solid  totally  immersed  in  a  liquid 
equals  its  real  weight  diminished  by  that  of  the  amount  of 
liquid  displaced  ;  in  other  words,  the  loss  of  weight  =  the 
weight  of  displaced  liquid. 

EXAMPLE  1. — How  great  a  mass  (not  hollow)  of  cast-iron  can 
be  supported  in  water  by  a  wrought-iron  cylinder  weighing 
140  Ibs.,  if  *he  latter  contains  a  vacuous  space,  and  displaces 

3  cub.  feet  of  water,  both  bodies  being  completely  immersed  ? 
[Ft.,  lb.,  sec.] 

The  buoyant  effort  on  the  cylinder  is 

V'y  =  3  X  62.5  =  187.5  Ibs., 

leaving  a  residue  of  47.5  Ibs.  upward  force  to  buoy  the  cast- 
iron,  whose  volume  V"  is  unknown,  while  its  heaviness  (§  7) 
is  y"  =  450  Ibs.  per  cub.  foot.  The  direct  buoyant  effort  of 
the  water  on  the  cast-iron  is  V'y  =  {V"  X  62.5]  Ibs., 
and  the  problem  requires  that  this  force  -f-  47.5  Ibs.  shall 
=  V"y"  =  the  weight  G"  of  the  cast-iron  ; 

.-.     V"  X  62.5  +  47.5  =  V"  X  450  ; 

.-.    V"  =  0.12  cub.  ft,  while  0.12  X  450=  54  Ibs.  of  cast-iron. 

Ans. 

EXAMPLE  2. — Required  the  volume  V ',  and  heaviness  y', 
of  a  homogeneous  solid  which  weighs  6  Ibs.  out  of  water  and 

4  Ibs.  when  immersed  (apparent  weight)  (ft.,  lb.,  sec). 


IMMERSION.  589 

From  eq.  (2),  4  =  6  -  V  X  62.5  ;     /.  V  —  0.032  cub.  feet  ; 
.-.  Y'  =  &  •*-  V  =  6  +-  °-032  =  m-5  lbs-  Per  cub-  ffc-> 


and  the  ratio  of  y'  to  y  is  187.5  :  62.5  =  3.0  (abstract  num- 
ber) ;  i.e.,  the  substance  of  this  solid  is  three  times  as  dense, 
or  three  times  as  heavy,  as  water.  [The  buoyant  effort  of  the 
air  has  been  neglected  in  giving  the  true  weight  as  6  lbs.] 

458,  Specific  Gravity.  —  By  specific  gravity  is  meant  the  ratio 
of  the  heaviness  of  a  given  homogeneous  substance  to  that  of 
a  standard  homogeneous  substance  ;  in  other  words,  the  ratio 
of  the  weight  of  a  certain  volume  of  the  substance  to  the 
weight  of  an  equal  volume  of  the  standard  substance.  Dis- 
tilled water  at  the  temperature  of  maximum  density  (4°  Centi- 
grade) under  a  pressure  of  14.7  lbs.  per  sq.  inch  is  sometimes 
taken  as  the  standard  substance,  more  frequently,  however,  at 
62°  Fahrenheit  (16°.6  Centigrade).  Water,  then,  being  the 
standard  substance,  the  numerical  example  last  given  illustrates 
a  common  method  of  determining  experimentally  the  specific 
gravity  of  a  homogeneous  solid  substance,  the  value  there  ob- 
tained being  3.  The  symbol  cr  will  be  used  to  denote  specific 
gravity,  which  is  evidently  an  abstract  number.  The  standard 
substance  should  always  be  mentioned,  and  its  heaviness;/; 
then  the  heaviness  of  a  substance  whose  specific  gravity  is  <r  is 


and  the  weight  G'  of  any  volume  V  of  the  substance  may  be 
written 

G'  =  Vy  =  V'o-y  ......     (2) 

Evidently  a  knowledge  of  the  value  of  y'  dispenses  with  the 
use  of  cr,  though  when  the  latter  can  be  introduced  into  prob- 
lems involving  the  buoyant  effort  of  a  liquid  the  criterion  as 
to  whether  a  homogeneous  solid  will  sink  or  rise,  when  im- 
mersed in  the  standard  liquid,  is  more  easily  applied,  thus  : 
Being  immersed,  the  volume  V  of  the  body  =  that,  V9  of 
displaced  liquid.  Hence, 


590  MECHANICS    OF   ENGINEERING. 

if  &  is  >  V'r,  i.e.,  if   V'y'  is  >  V'y,  or  cr  >  1,  it  sinks ; 
while  if  G'  is  <  V 'y, or  cr  <  1,  it  rises ; 

i.e.,  according  as  the  weight  G'  is  >  or  <  than  the  buoyant 
effort. 

Other  methods  of  determining  the  specific  gravity  of  solids, 
liquids,  and  gases  are  given  in  works  on  Physics. 

459.  Equilibrium  of  Flotation. — In  case  the  weight  G'  of  an 
immersed  solid  is  less  than  the  buoyant  effort  V'y  (where  V  is 
the  volume  of  displacement,  and  y  the  heaviness  of  liquid)  the 
body  rises  to  the  surface,  and  after  a  series  of  oscillations  comes 
to  rest  in  such  a  position,  Fig.  503,  that  its  centre  of  gravity  C 
and  the  centre  of  buoyancy  B  (the  new  B,  belonging  to  the 
new  volume  of  displacement,  which  is  limited  above  by  the 
horizontal  plane  of  the  free  surface  of  the  liquid)  are  in  the 
same  vertical  (called  the  axis  of  flotation,  or  line  of  support), 
and  that  the  volume  of  displacement  has  diminished  to  such  a 
new  value  V,  that 

Vy=G' (1) 

In  the  figure,   V  =  vol.  AND,  below  the  horizontal  plane 
AN,  and  the  slightest  motion  of  the  body  will  change  the  form 
of  this  volume,  in  general  (whereas  with 
complete  immersion  the  volume  of  dis- 
placement remains  constant).     For  stable 
equilibrium  it  is  not  essential  in  every 
case  that  C  (centre  of  gravity  of  body) 
should  be  below  B  (the  centre  of  buoy- 
ancy) as  with  complete  immersion,  since  if 
503.  "  the  solid  is  turned,  B  may  change  its  posi- 

tion in  the  body,  as  the  form  of  the  volume  AND  changes. 

There  is  now  no  definite  relation  between  the  volume  of 
displacement  Fand  that  of  the  body,  V,  unless  the  latter  is 
homogeneous,  and  then  for  G'  we  may  write  V'y'^  i.e. 

V'y'  =  Vy  (for  a  homogeneous  solid) ;    .     .     (2) 
or,  the  volumes  are  inversely  proportional  to  the  heavinesses. 


FLOTATION. 


591 


The  buoyant  effort  of  the  air  on  the  portion  ANE  may  be 
neglected  in  most  practical  cases,  as  being  insignificant. 

If  the  solid  is  hollow,  the  position  of  its  centre  of  gravity  0 
may  be  easily  varied  (by  shifting  ballast,  e.g.)  within  certain 
limits,  but  that  of  the  centre  of  buoyancy  B  depends  only  on 
the  geometrical  form  of  the  volume  of  displacement  AND, 
below  the  horizontal  plane  AN. 

EXAMPLE.— (Ft.,  lb.,  sec.)  Will  a  solid  weighing  G  =  400 
Ibs.,  and  having  a  volume  V  —  8  cub.  feet,  without  hollows 
or  recesses,  float  in  water?  To  obtain  a  buoyant  effort  of 
400  Ibs.,  we  need  a  volume  of  displacement,  see  eq.  (1),  of 

V      G'       400 

V  =  —  =  -  —  =  only  6.4  cub.  ft. 
y       62.5 

Hence  the  solid  will  float  with  8  —  6.4,  or  1.6,  cub.  ft.  pro- 
jecting above  the  water  level. 

Query :  A  vessel  contains  water,  reaching  to  its  brim,  and 
also  a  piece  of  ice  which  floats  without  touching  the  vessel. 
When  the  ice  melts  will  the  water  overflow  ? 

460.  The  Hydrometer  is  a  floating  instrument  for  determin- 
ing the  relative  heavinesses  of  liquids.  Fig.  504  shows  a  sim- 
ple form,  consisting  of  a  bulb  and  a  cylin- 
drical stem  of  glass,  so  designed  and 
weighted  as  to  float,  upright  in  all  liquids 
whose  heavinesses  it  is  to  compare.  Let  F 
denote  the  uniform  sectional  area  of  the 
stem  (a  circle),  and  suppose  that  when  float- 
ing in  water  (whose  heaviness  =  y)  the 
water  surface  marks  a  point  A  on  the  stem  ; 
and  that  when  floating  in  another  liquid, 
say  petroleum,  whose  heaviness,  =  yp,  we 
wish  to  determine,  it  floats  at  a  greater 
depth,  the  liquid  surface  now  marking  A 
on  the  stem,  a  height  =  x  above  A.  G'  is 
the  same  in  both  experiments;  but  while  the  volume  of  dis- 
placement in  water  is  F,  in  petroleum  it  is  F  +  Fx.  There- 
fore from  eq.  (1),  §  459, 


592  MECHANICS   OF  ENGINEERING. 

in  the  water         G'  =  Vy,        .....    (1) 

and  in  the  petroleum,  G'  =  (  V-\-Fx)yp\       .     .     (2) 

from  which,  knowing  G',  F>  ae,  and  y,  we  find  Fand  yp,  i.e., 

F=fland     X.-..     ...     (3) 


[N.B.  —  FIB  best  determined  by  noting  the  additional  dis- 
tance, =  I,  through  wliich  the  instrument  sinks  in  water  under 
an  additional  load  P,  not  immersed  /  for  then 

&  +  P  =(  F+  Fl)Y,  or    F  =  - 


EXAMPLE.  —  [Using  the  inch,  ounce,  and  second,  in  which 
system  y  =  1000  -r-  1728  =  0.578  (§  409).]  With  £'  =  a 
ounces,  and  ^^  0.10  sq.  inch,  x  being  observed,  on  the 
graduated  stem,  to  be  5  inches,  we  have  for  the  petroleum 


3X0.578 
T0^5^0578  = 

=  56.7  Ibs.  per  cub.  foot. 


Temperature  influences  the  heaviness  of  most  liquids  to 
some  extent. 

In  another  kind  of  instrument  a  scale-pan  is  fixed  to  the  top 
of  the  stem,  and  the  specific  gravity  computed  from  the  weight 
necessary  to  be  placed  on  this  pan  to  cause  the  hydrometer  to 
sink  to  the  same  point  in  all  liquids  for  which  it  is  used. 

461.  Depth  of  Flotation. — If  the  weight  and  external  shape 
of  the  floating  body  are  known,  and  the  centre  of  gravity  so 
situated  that  the  position  of  flotation  is  known,  the  depth  of 
the  lowest  point  below  the  surface  may  he  determined. 


FLOTATION. 


593 


CASE  I.  Right  prism  or  cylinder  with  its  axis  vertical. — 
Fig.  505.  (For  stability  in  this  position, 
see  §  464#.)  Let  G'  —  weight  of  cylin- 
der, /'"'the  area  of  its  cross-section  (full 
circle),  h'  its  altitude,  and  h  the  un- 
known  depth  of  flotation  (or  draught] ; 
then  from  eq.  (1),  §  426, 


6^ 

FY 


in  which  y  =  heaviness  of  the  liquid. 
If    the   prism    (or  cylinder)   is  homo- 
geneous (and  then  (7,  at  the  middle  of  h',  is  higher  than  J3) 
and  y'  its  heaviness,  we  then  have 


FIG.  505. 


_  Fh'y'  _  y'h'  _ 


(2) 


in  which  cr  =  specific  gravity  of  solid  referred  to  the  liquid  as 
standard.     (See  §  458.) 

CASE  II.  Pyramid  or  cone  with  axis  vertical  and  vertex 
down. — Fig.  506.     Let  V  =.  volume  of 
whole  pyramid  (or  cone),  and  V=  vol- 
ume   of    displacement.     From    similar 
pyramids, 


A" 


.h'. 


But  G'  =  Vy ;  or,  V=  — ;  whence 


FIG.  506. 


h  =  h' 


(3) 


594 


MECHANICS    OF    ENGINEERING. 


CASE  III.  Ditto,  but  vertex  up. — Fig.  507.     Let  the  nota« 
T~~         tion  be  as  before,  for  V  and  V.     The 
part  out  of  water  is  a  pyramid  of  volume 
=  V"  =  V  —  F,  and  is  similar  to  the 
whole  pyramid  ; 


'-  V:  V  ::h"*  :  h'\ 


Also, 


FIG.  507. 


.:,  finally,  h  =  A'[l  -  .J/l  -  [6"  -  F'y]].    ...    (4) 
CASE  IV.  Sphere.  —  Fig.  508.     The  volume  immersed  is 


V  = 


/  l"~      ™n 

(%rz  —  z*)dz  =  ?rAa   r  —  —     $ 
v  o    I 

and  hence,  since  Vy  =  ^  =  weight 
of  sphere, 


„       ?rA3       G' 

nrh  —  =  — , 

3         v 


(5) 


From  which  cubic  equation  A  may  be 
^^^  obtained  by  successive  trials  and  ap- 

"  :=^    proximations. 
Fl°-  508-  [An  exact  solution  of  (5)  for  the 

unknown  h  is  impossible,  as  it  falls  under  the  irreducible  case 
of  Cardan's  Kule.] 

CASE  V.  Right  cylinder  with  axis  horizontal.  —  Fig.  509. 

lers.  J 


°f 


X 


—  —        hence,  since 

Ffa.  509. 

lr*\ct  —  \  sin 


Y 
— 


(6) 


EXAMPLES   OF  FLOTATION.  595 

From  this  transcendental  equation  we  can  obtain  <*5  by  trial, 
*n  ™adians  (see  example  in  §  428),  and  finally  A,  since 

h  =  r(l  —  cos  at) (7) 

EXAMPLE  1. — A  sphere  of  40  inches  diameter  is  observed  to 
have  a  depth  of  flotation  h  =  9  in.  in  water.  Required  its 
weight  G '.  From  eq.  (5)  (inch,  lb.,  sec.)  we  have 

Gf  =  [62.5  -r-  1728]7r92[20  -  £  X  9]  =  156.5  Ibs. 

The  sphere  may  be  hollow,  e.g.,  of  sheet  metal  loaded  with 
shot ;  constructed  in  any  way,  so  long  as  Gf  and  the  volume 
V  of  displacement  remain  unchanged.     But  if  the  sphere  ifl 
homogeneous,  its  heaviness  (§  7)  yf  must  be 

=  Q'  -=-  V  =  Gf  -r-  f  nr*  =  (156.5)  -f-  f  7r20" 
=  .00466  Ibs.  per  cubic  inch, 

and  hence,  referred  to  water,  its  specific  gravity  is  cr  =  about 
u!3. 

EXAMPLE  2. — The  right  cylinder  in  Fig.  509  is  homogeneous 
and  10  inches  in  diameter,  and  has  a  specific  gravity  (referred 
to  water)  of  <r  =  0.30.  Required  the  depth  of  flotation  A. 

Its  heaviness  must  be y'  =  cry ;  hence  its  weight 

0'  =  V'o-r=7rr*l(rr; 
bonce,  from  eq.  (6), 

r*l[a  —  J  sin  2#]  =  TT^'Zcr,  /.  a  —  \  sin  2<x  =  TTCT 

(involving  abstract  numbers  only).  Trying  a  =  60°  (  =  \it  in 
radians),  we  have 

|TT  —  £  sin  120°  =  0.614 ;  whereas  no-  =  .9424 

For  a  =  TO0,  1.2217  -  i  sin  140°  =  0.9003 ; 

For  a  =  71°,  1.2391  -  -J-  sin  142°  =  0.9313  ; 

Vor  a  =  71°  22',  1.2455  -  J  sin  142°  44r  =  0.9428,  which 

fcp  considered  sufficiently  close.     Now  from  eq.  (7), 

A  =  (5  in.)  (1  -  cos  71°  22')  =  3.40  in.— An* 


•596  MECHANICS   OF   ENGINEER! NG. 

462,  Draught  of  Ships. — In  designing  a  ship,  especially  if  01 
a  new  model,  the  position  of  the  centre  of  gravity  is  found  by 
eq.  (3)  of  §  23  (with  weights  instead  of  volumes)  ;  i.e.,  the  sum 
of  the  products  obtained  by  multiplying  the  weight  of  each 
portion  of  the  hull  and  cargo  by  the  distance  of  its  centre  of 
gravity  from  a  convenient  reference-plane  (e.g.,  the  horizontal 
plane  of  the  keel  bottom)  is  divided  by  the  sum  of  the  weights, 
and  the  quotient  is  the  distance  of  the  centre  of  gravity  of  the 
whole  from  the  reference-plane. 

Similarly,  the  distance  from  another  reference-plane  is  de- 
termined. These  two  co-ordinates  and  the  fact  that  the  centre 
of  gravity  lies  in  the  median  vertical  plane  of  symmetry  of  the 
ship  (assuming  a  symmetrical  arrangement  of  the  framework 
ind  cargo)  fix  its  location.  The  total  weight,  6?/,  equals,  of 
course,  the  sum  of  the  individual  weights  just  mentioned.  The 
centre  of  buoyancy,  for  any  assumed  draught  and  correspond- 
ing position  of  ship,  is  found  by  the  same  method ;  but  more 
simply,  since  it  is  the  centre  of  gravity  of  the  imaginary  homo- 
geneous volume  between  the  water-line  plane  and  the  wetted 
surface  of  the  hull.  This  volume  (of  "displacement")  is 
divided  into  an  even  number  (say  4  to  8)  of  horizontal  laminae 
of  equal  thickness,  and  Simpson's  Rule  applied  to  find  the  vol- 
ume (i.e.,  the  V  of  preceding  formulae),  and  also  (eq.  3,  §  23) 
the  height  of  its  centre  of  gravity  above  the  keel.  Similarly, 
by  division  into  (from  8  to  20)  vertical  slices,  1  to  keel  (an 
even  number  and  of  equal  thickness\  we  find  the  distance  of 
the  centre  of  gravity  from  the  bow.  Thus  the  centre  of  buoy- 
ancy is  fixed,  and  the  corresponding  buoyant  effort  Vy  (tech- 
nically called  the  displacement  and  usually  expressed  in  tons') 
computed,  for  any  assumed  draught  of  ship  (upright).  That 
position  in  which  the  "  displacement"  =  G'  —  weight  of  ship 
is  the  position  of  equilibrium  of  the  ship  when  floating  up- 
right in  still  water,  and  the  corresponding  draught  is  noted. 
As  to  whether  this  equilibrium  is  stable  or  unstable,  the  fol- 
lowing will  show. 

In  most  ships  the  centre  of  gravity  O  is  several  feet  above 
the  centre  of  buoyancy,  B,  and  a  foot  or  more  below  the  water 
line. 


DRAUGHT   OF   SHIPS. 


597 


After  a  ship  is  afloat  and  its  draught  actually  noted  its  total 
weight  6r',  =  Vy,  can  be  computed,  the  values  of  Vfor  dif- 
ferent draughts  having  been  calculated  in  advance.  In  this 
way  the  weights  of  different  cargoes  can  also  be  measured. 

EXAMPLE. — A  ship  having  a  displacement  of  5000  tons  is 
itself  5000  tons  in  weight,  and  displaces  a  volume  of  salt  water 
•  Y=  G'-i-  y  =  10,000,000  Ibs.  -*-  64  Ibs.  per  cub.  ft.  =  156250 
cub.  ft. 

463.  Angular  Stability  of  Ships. — If  a  vessel  floating  upright 
were  of  the  peculiar  form  and  position  of 
Fig.  510  (the  water-line  section  having  an 
area  =  zero)   its   tendency    to   regain   that 
position,  or  depart  from  it,  when  slightly 
inclined  an  angle  0  from  the  vertical  is  due 
to  the  action  of  the  couple  now  formed  by 
the  equal  and  parallel  forces  Vy  and  G', 
which  are  no  longer  directly  opposed.     This 
couple  is  called  a  righting  couple  if  it  acts 
to  restore  the  first  position  (as  in  Fig.  511, 
where  C  is  lower  than  JB)9  and  an 
upsetting    couple   if   the   reverse,  C 
above  B.     In    either  case  the  mo- 
ment of  the  couple  is 

=  Vy  .  BC  sin  0  =  Vye  sin  0,       ^ 

and  the  centre  of  buoyancy  B  does  not 
change  its  position  in  the  vessel,  since 
the  water-displacing  shape  remains 
the  same ;  i.e.,  no  new  portions  of 
the  vessel  are  either  immersed  or 
raised  out  of  the  water. 

But  in  a  vessel  of  ordinary  form,  when  turned  an  angle  0  from 
the  vertical,  Fig.  512  (in  which  ED  is  a  line  which  is  vertical 
when  the  ship  is  upright),  there  is  a  new  centre  of  buoyancy, 
B^ ,  corresponding  to  the  new  shape  A^N^D  of  the  displacement- 
volume,  and  the  couple  to  right  the  vessel  (or  the  reverse) 


FIG.  510. 


FIG.  511. 


598 


MECHANICS    OF 


consists  of  the  two  forces  Gf  at  O  and  Vy  at  Bl ,  and  has  a 

moment  (which  we  may  call  M,  or 
/E  moment  of   stability)   of  a   value 

(§28) 


I    / 


jzM -  i^-l        _X_      Jf  =  Vy  .  mC  sin  0. 


(1) 


FIG.  512. 


C.rL=_~  Now  conceive  put  in  at  B  (centre 
^^=r- of  buoyancy  of  the  upright  posi- 
tion) two  vertical  and  opposite 
forces,  each  =  Vy  =  G' ,  calling 
them  P  and  P,  (see  §  20),  Fig.  512. 
"We  can  now  regard  the  couple  [6r'5  Vy]  as  replaced  by  the 
two  couples  [#',  P]  and  [P,,  Vy~\\  for  evidently 

Vy  .  mC  sin   0  =  Vy  .  BG  sin  0  +  F/  .  mJ?  sin  0 ; 
(§§33  and  34;) 

.-.  J/  =  Vy  Wsin  0  +  Vy  ~mB  sin  0.    .     .     (2) 

But  the  couple  [#',  P]  would  be  the  only  one  to  right  the 
vessel  if  no  new  portions  of  the  hull  entered  the  water  or 
emerged  from  it,  in  the  inclined  position ;  hence  the  other 


couple 


Flo.  513. 


Vy}  owes  its  existence  to  the  emersion  of  the 
wedge  AOA19  and  the  immersion 
of  the  wedge  NON^  i.e.,  to  the 
loss  of  a  buoyant  force  Q  =  (vol- 
ume A  #4i)  X  y  on  one  side,  and  the 
£-=.  gain  of  an  equal  buoyant  force  on 
the  other;  therefore  this  couple 
[P0  Vy]  is  the  equivalent  of  the 
couple  [Q,Q],  Fig.  51L,  formed  by 
putting  in  at  the  centre  of  buoyancy 
of  each  of  the  two  wedges  a  vertical 
force 


Q  =  (vol.  of  wedge)  X  r  —  Vwy.    (See  figure.) 


STABILITY   OF   SHIPS.  599 

If  a  denotes  the  arm  of  this  couple,  we  may  write 

Vy  TmB  sin  0,  [of  eq.  (2)],  =  Vwya  ;     .     .     (3) 

and  hence,  denoting  J?6Yby  e,  we  have 

M  =  ±  Vye  sin  0  +  Vwya ;    ....     (4) 

the  negative  sign  in  which  is  to  be  used  when  C  is  above  B 
(as  with  most  ships).  0,  the  intersection  of  ED  and  AN, 
does  not  necessarily  lie  on  the  new  water-line  plane  AlJVl. 

EXAMPLE. — If  a  ship  of  ( Vy  =)  3000  tons  displacement 
with  C  4  ft.  above  B  (i.e.,  e  —  —  4  ft.)  is  deviated  10°  from 
the  vertical,  in  salt  water,  for  which  angle  the  wedges  A  OAl  and 
NON^l  have  each  a  volume  of  4000  cubic  feet,  while  the  hori- 
zontal distance  a  between  their  centres  of  buoyancy  is  18  feet, 
the  moment  of  the  acting  couple  will  be,  from  eq.  (4)  (ft.-ton- 
sec.  system,  in  which  y  of  salt  water  =  0.032), 

M  =  —  3000  X  4  X  0.1736  +  4000  X  0.032  X 18  =  220.8  ft.  tons, 
which  being  -f-  indicates  a  righting  couple. 

464.  Remark. — If  with  a  given  ship  and  cargo  this  moment 
of  stability,  M,  be  computed,  by  eq.  (4),  for  a  number  of  values 
of  0,  and  the  results  plotted  as  ordinates  (to  scale)  of  a  curve, 
0  being  the  abscissa,  the  curve  ob- 
tained is  indicative  of  the  general 
stability  of  the  ship.  See  Fig.  514. 
For  some  value  of  0=  OK  (as  well 
as  for  0  —  0)  the  value  of  M  is 
zero,  and  for  0  >  OK,  M  is  nega- 
tive, indicating  an  upsetting  couple.  FlG- 514- 
That  is,  for  0=0  the  equilibrium  is  stable,  but  for  0  =  OK, 
unstable  ;  and  M  =  0  in  both  positions.  From  eq.  (4)  we  see 
why,  if  Cis  above  B,  instability  does  not  necessarily  follow. 

464a.  Metacentre  of  a  Ship. — Kef  erring  again  to  Fig.  512, 
we  note  that  the  entire  couple  \G' ,  Vy]  will  be  a  righting 
couple,  or  an  upsetting  couple,  according  as  the  point  m  (the 


—<p- 


600  MECHANICS    OF    ENGINEERING. 

intersection  of  the  vertical  through  Bl ,  the  new  centre  of 
buoyancy,  with  BC  prolonged)  is  above  or  below  the  centre 
of  gravity  C  of  the  ship.  The  location  of  this  point  m  changes 
with  0;  but  as  0  becomes  very  small  (and  ultimately  zero)  m 
approaches  a  definite  position  on  the  line  DE,  though  not  oc- 
cupying it  exactly  till  0  =  0.  This  limiting  position  of  m  is 
called  the  metacentre^  and  accordingly  the  following  may  be 
stated  :  A  ship  floating  upright  is  in  stable  equilibrium  if  its 
metacentre  is  above  its  centre  of  gravity  /  and  vice  versa. 
In  other  words,  for  a  slight  inclination  from  the  vertical  a 
righting,  and  not  an  upsetting,  couple  is  called  into  action,  if 
m  is  above  C.  To  find  the  metacentre,  by  means  of  the  dis- 
tance Bm,  we  have,  from  eq.  (3), 


Vy  sin  0 


and   wish   ultimately  to  make   0  =  0.      Now   the    moment 
( Vwy)a  =  the  sum  of  the  moments  about  the  horizontal  fore- 
and-aft  water-line  axis   OL,  Fig.  515,  of  the  buoyant  efforts 
<pzdfy  due  to  the  immersion   of   the 

N>sx      ______  „ A— ^-v---  L    separate     vertical     elementary 

<?J  VX?A^F^~T        .-''  JP  prisms  of  the  wedge  OLNJtf. 

plus  the  moments  of  those  lost, 
from  emersion,  in  the  wedge 
OLA.A.  Let  OA.LN,  be  the 
new  water-line  section  of  the 
ship  when  inclined  a  small 
FIO.  515.  angle  0  from  the  vertical 

(0  =  NO.N^,  and  OALN  the  old  water-line.  Let  z  =  the 
"1  distance  of  any  elementary  area  dF  of  the  water-line  section 
from  OL  (which  is  the  intersection  of  the  two  water-line 
planes).  Each  dFis  the  base  of  an  elementary  prism,  with 
altitude  =  02,  of  the  wedge  N^OLN  (or  of  wedge  A.OLA 
when  z  is  negative).  The  buoyant  effort  of  this  prism  =  (its 
vol.)  X  y  =  yz<pdF,  and  its  moment  about  OL  is 
Hence  the  total  moment,  =  Qa,  or  Vwya,  of  Fig.  513, 

=  </>yfs*dF=  y<t>  X  JOL 


THE   METACENTRE.  601 

of  water-line  section,  in  which  IOL  denotes  the  "  moment  of 
inertia5'  (§  85)  of  the  plane  figure  OALNO  about  the  axis  OL. 
Hence  from  (5),  putting  0  =  sin  0(true  when  0  —  0),  we  have 
mB  =  IOL  -r-  F;  and  therefore  the  distance  mC\  of  the  meta- 
centre  m  above  (7,  the  centre  of  gravity  of  the  ship,  Fig.  512,  is 


=  AM  ,  =  .   ±  e,  .    .    . 


in  which  e  =  BC  =  distance  from  the  centre  of  gravity  to  the 
centre  of  buoyancy,  the  negative  sign  being  used  when  C  is 
above  B  ;  while  F=  whole  volume  of  water  displaced  by  the 
ship* 

We  may  also  write,  from  eqs.  (6)  and  (1),  for  small  values 


Mom.  of  righting  couple  =  M  =  Vy  sin  0    ~  ±  e  ,  .    (7) 


or 


Ve].     ....     (7)' 


Eqs.  (7)  and  (7)'  will  give  close  approximations  for  0  <  10°  or 
15°  with  ships  of  ordinary  forms, 

EXAMPLE   1.  —  A    homogeneous    right   parallelepiped,    of 
heaviness    y',  floats    upright    as  in  ^ 

Fig.      516.        Find      the     distance  ^ 

mO  =  hm  for  its  metacentre  in  this     A  p-f 
position,  and  whether  the  equilibrium  ~^=  /| 
is  stable.     Here  the  centre  of  gravity,  :^\=;~f 
C,  being  the  centre  of  figure,  is  of  ~=^ 
course  above  B,  the  centre  of  buoy-  ir£ 
ancy  ;  hence  e  is  negative.     B  is  the   ^-1^1 


centre  of  gravity  of  the  displacement,     ^^  ^ 

and  is  therefore  a  distance  %h  below 

the  water-line.     We  here  assume  that  I  is  greater  than 

From  eq.  (2),  §  461, 


602  MECHANICS   OF   ENGINEERING. 

and  since  CD  =  }  A',  and  BD  r=  JA,  .-.  e  =  i(A'- 


while  (§  90)  /OL,  of  the  water-line  section  AN, 
Also, 


and  hence,  from  eq.  (6),  we  have 


Hence  if  &'a  is  >  6A'8^  (l  —  ¥  ^  the  position  in  Fig.  516  is 

one  of  stable  equilibrium,  and  vice  versa.     E.g.  ,  if  yf  =  f^, 
Jr  =  12  inches  and  hr  =.  6  inches,  we  have  (inch,  pound,  sec.) 


-  6  X 


=  2.5  in. 


The  equilibrium  will  be  unstable  if,  with  yf  =  J^,  5r  is  made 
less  than  1.225  h'  \  for,  putting  m(7  =  0,  we  obtain  £/  = 
1.225  h'. 

EXAMPLE  2.  —  (Ft.,  lb.,  sec.)    Let  Fig.  517  represent  the  half 
water-line  section  of  a  loaded  ship  of  G'  =  Vy  =  1010  tons 


olML. 


-I 4— 0..J L L L— °-_5j| 

23*5678 


FIG.  Sir. 


displacement ;  required  the  height  of  the  metacentre  above  the 
centre  of  buoyancy,  i.e.,  mB  =  ?  (See  equation  just  before  eq. 
(6).)  Now  the  quantity  7OL,  of  the  water-line  section,  may, 
from  symmetry,  (see  §  93?)  be  written 


(1) 


METACENTRE.  608 

in  which  y  =  the  ordinate  1  to  the  axis  OL  at  any  point;  and 
this,  again^  by  Simpson's  Rule  for  approximate  integration, 
OL  being  divided  into  an  even  nnmber,  n,  of  equal  parts,  and 
ordinates  erected  (see  figure),  may  be  written 


From  which,  by  numerical  substitution  (see  figure  for  dimen- 
sions ;  n  =  8), 

2      160 


-/or,  =  -^  • 


3    3X8 


[(0.5s  +  4(5"  + 128  + 133  +  Y3) 


125 

1728  729 

or>  2197  2744 

343  1331 

IOL  =  ^[0.125+4  X  4393  +  2  X  4804+  0.125] 

Inr.  120801 


, 
=  3.8  feet. 

That  is,  the  metacentre  is  3.8  feet  above  the  centre  of  buoyancy, 
and  hence,  if  JSC  =  2  feet,  is  1.90  ft.  above  the  centre  of 
gravity.  [See  Johnson's  Cyclopaedia,  article  Naval  Architec- 
ture.'] 

465.  Metacentre  for  Longitudinal  Stability, — If  we  consider 
the  stability  of  a  vessel  with  respect  to  pitching,  in  a  manner 
similar  to  that  just  pursued  for  rolling,  we  derive  the  position 
of  the  metacentre  for  pitching  or  for  longitudinal  stability — 
and  this  of  course  occupies  a  much  higher  position  than  that 
for  rolling,  involving  as  it  does  the  moment  of  inertia  of  the 
water-line  section  about  a  horizontal  gravity  axis  ~]  to  the  keel. 
With  this  one  change,  eq.  (6)  holds  for  this  case  also.  In 
large  ships  the  height  of  this  metacentre  above  the  centre  of 
gravity  of  the  ship  may  be  as  great  as  90  feet. 


CHAPTEK  Y. 

HYDROSTATICS  (Continued)— GASEOUS  FLUIDS. 

466.  Thermometers. — The  temperature,  or  "  hotness"  of 
liquids  has,  within  certain  limits,  but  little  influence  on  their 
statical  behavior,  but  with  gases  must  always  be  taken  into 
account,  since  the  three  quantities,  tension,  temperature,  and 
volume,  of  a  given  mass  of  gas  are  connected  by  a  nearly  in- 
variable law,  as  will  be  seen. 

An  air-thermometer,  Fig.  518,  consists  of  a  large  glass  bulb 
filled  with  air,  from  which  projects  a  tine  straight  tube  of 

even  bore  (so  that  equal  lengths 
represent  equal  volumes).  A 
small  drop  of  liquid,  A,  sepa- 
rates the  internal  from  the  ex- 
ternal  air,  both  of  which  are 
at  a  tension  of  (say)  one  at- 
mosphere (14.7  Ibs.  per  sq.  inch).  When  the  bulb  is  placed 
in  melting  ice  (freezing-point)  the  drop  stands  at  some  point  F 
in  the  tube ;  when  in  boiling  water  (boiling  under  a  pressure 
of  one  atmosphere),  the  drop  is  found  at  B,  on  account  of  the 
expansion  of  the  internal  air  under  the  influence  of  the  heat 
imparted  to  it.  (The  glass  also  expands,  but  only  about  y^ 
as  much ;  this  will  be  neglected.)  The  distance  FB  along  the 
tube  may  now  be  divided  into  a  convenient  number  of  equal 
parts  called  degrees.  If  into  one  hundred  degrees,  it  is  found 
that  each  degree  represents  a  volume  equal  to  the  y^nnnr 
(.0036Y)  part  of  the  total  volume  occupied  by  the  air  at  freez- 
ing-point ;  i.e.,  the  increase  of  volume  from  the  temperature  of 
freezing-point  to  that  of  the  boiling-point  of  water  —  0.36T  of  the 
volume  at  freezing,  the  pressure  being  the  same,  and  even  having 
any  value  whatever  (as  well  as  one  atmosphere),  within  ordi- 
nary limits,  so  long  as  it  is  the  same  both  at  freezing  and  boil- 

604 


THEKMOMETEKS.  605 

ing.  It  must  be  understood,  however,  that  by  temperature  of 
boiling  is  always  meant  that  of  water  boiling  under  one  at- 
mosphere pressure.  Another  way  of  stating  the  above,  if  one 
hundred  degrees  are  used  between  freezing  and  boiling,  is  as 
follows  :  That  for  each  degree  increase  of  temperature  the  in- 
crease of  volume  is  ^  of  t..e  total  volume  at  freezing;  2T3 
being  the  reciprocal  of  .00367. 

As  it  is  not  always  practicable  to  preserve  the  pressure  con- 
stant under  all  circumstances  with  an  air- thermometer,  we  use 
the  common  mercurial  thermometer  for  most  practical  pur- 
poses. In  this,  the  tube  is  sealed  at  the  outer  extremity,  with 
a  vacuum  above  the  column  of  mercury,  and  its  indications 
agree  very  closely  with  those  of  the  air-thermometer.  That 
equal  absolute  increments  of  volume  should  imply  equal  incre- 
ments of  heat  imparted  to  these  thermometric  fluids  (under 
constant  pressure)  could  not  reasonably  be  asserted  without 
satisfactory  experimental  evidence.  This,  however,  is  not  al- 
together wanting,  so  that  we  are  enabled  to  say  that  within  a 
moderate  range  of  temperature  equal  increments  of  heat  pro- 
duce equal  increments  of  volume  in  a  given  mass  not  only  of 
atmospheric  air,  but  of  the  so-called  "  perfect"  or  "permanent" 
gases,  oxygen,  nitrogen,  hydrogen,  etc.  (so  named  before  it  was 
found  that  they  could  be  liquefied).  This  is  nearly  true  for 
mercury  also,  and  for  alcohol,  but  not  for  water.  Alcohol 
freezes  at  —  200°  Fahr.,  and  hence  is  used  instead  of  mercury 
as  a  thermometric  substance  to  measure  temperatures  below 
the  freezing-point  of  the  latter. 

The  scale  of  a  mercurial  thermometer  is  fixed  ;  but  with  an 
air-thermometer  we  should  have  to  use  a  new  scale,  and  in  a 
new  position  on  the  tube,  for  each  value  of  the  pressure. 

467.  Thermometric  Scales.— In  the  Fahrenheit  scale  the  tube 
between  freezing  and  boiling  is  marked  off  into  180  equal 
parts,  and  the  zero  placed  at  32  of  these  parts  below  the  freez- 
ing point,  which  is  hence  +  32°,  and  the  boiling-point  +  212°. 

The  Centigrade,  or  Celsius,  scale,  which  is  the  one  chiefly 
used  in  scientific  practice,  places  its  zero  at  freezing,  and  100° 
at  boiling-point.  Hence  to  reduce 


606 


MECHANICS    OF   ENGINEERING. 


Fahr.  readings  to  Centigrade,  subtract  32°  and  multiply  by  $; 
Cent.  "  "  Fahrenheit,  multiply  by  £  and  add  32°. 

468.  Absolute  Temperature. — Experiment  also  shows  that  if 
a  mass  of  air  or  other  perfect  gas  is  confined  in  a  vessel  whose 
volume  is  but  slightly  affected  by  changes  of  temperature, 
equal  increments  of  temperature  (and  therefore  equal  incre- 
ments of  heat  imparted  to  the  gas,  according  to  the  preceding 
paragraph)  produce  equal  increments  of  tension  (i.e.,  pressure 
per  unit  area) ;  or,  as  to  the  amount  of  the  increase,  that  when 
the  temperature  is  raised  by  an  amount  1°  Centigrade,  the  ten- 
sion is  increased  ^^  of  its  value  at  freezing-point.  Hence, 
theoretically,  an  ideal  barometer  (containing  a  liquid  unaffected 
by  changes  of  temperature)  communicating  with  the  confined 
gas  (whose  volume  practically  remains  constant)  would  by 
its  indications  serve  as  a  thermometer, 
Fig.  519,  and  the  attached  scale  could  be 
graduated  accordingly.  Thus,  if  the  col- 
imn  stood  at  A  when  the  temperature 
was  freezing,  A  would  be  marked  0°  on 
the  Centigrade  system,  and  the  degree 
spaces  above  and  below  A  would  each 
FIG.  519.  _  ^i^  Q£  ^jie  hejg]^  AB^  an(j  therefore 

the  point  B  (cistern  level)  to  which  the  column  would  sink  if 
the  gas-tension  were  zero  would  be  marked  —  273°  Centi- 
grade. 

But  a  zero-pressure,  in  the  Kinetic  Theory  of  Gases  (§  408), 
signifies  that  the  gaseous  molecules,  no  longer  impinging 
against  the  vessel  walls  (so  that  the  press.  =  0),  have  become 
motionless;  and  this,  in  the  Mechanical  Theory  of  Heat,  or 
Thermodynamics,  implies  that  the  gas  is  totally  destitute  of  heat. 
Hence  this  ideal  temperature  of  —  273°  Centigrade,  or  —  460° 
Fahrenheit,  is  called  the  Absolute  Zero  of  Temperature,  and  by 
reckoning  temperatures  from  it  as  a  starting-point,  our  formulas 
will  be  rendered  much  more  simple  and  compact.  Tempera- 
ture so  reckoned  is  called  absolute  temperature,  and  will  be 
denoted  by  the  letter  T.  Hence  the  following  rules  for  re- 
duction : 


GASES   AND   VAPOKS.  607 

Absol.  temp.  T  in  Cent,  degrees  ==  Ordinary  Cent.  +  273°  ; 
Absol.  temp.  T  in  Fahr.  degrees  =  Ordinary  Fahr.  -f  460°. 
For  example,  for  20°  Cent.,  T  =  293°  Abs.  Cent. 

469.  Distinction  Between  Gases  and  Vapors.  —  All  known 
gases  can  be  converted  into  liquids  by  a  sufficient  reduction  of 
temperature  or  increase  of  pressure,  or  both.  ;  some,  however, 
with  great  difficulty,  such  as  atmospheric  air,  oxygen,  hydro- 
gen, nitrogen,  etc.,  these  having  been  but  recently  (1878)  re- 
duced to  the  liquid  form.  A  vapor  is  a  gas  near  the  point  of 
liquefaction,  and  does  not  show  that  regularity  of  behavior 
under  changes  of  temperature  and  pressure  characteristic  of  a 
gas  when  at  a  temperature  much  above  the  point  of  liquefac- 
tion. All  gases  treated  in  this  chapter  (except  steam)  are  sup- 
posed in  a  condition  far  removed  from  this  stage.  The  fol- 
lowing will  illustrate  the  properties  of  vapors.  See  Fig.  520. 
Let  a  quantity  of  liquid,  say  water,  be  intro-  THrRM. 
duced  into  a  closed  space,  previously  vacuous, 
of  considerably  larger  volume  than  the  water, 
and  furnished  with  a  manometer  and  ther- 
mometer. Vapor  of  water  immediately  be- 
gins to  form  in  the  space  above  the  liquid,  and 
continues  to  do  so  until  its  pressure  attains  a 
definite  value  dependent  on  the  temperature, 
and  not  on  the  ratio  of  the  volume  of  the  vessel  and  the  origi- 
nal volume  of  water  ;  e.g.,  if  the  temperature  is  70°  Fahren- 
heit, the  vapor  ceases  to  form  when  the  tension  reaches  a  value 
of  0.36  Ibs.  per  sq.  inch.  If  heat  be  gradually  applied  to  raise 
the  temperature,  more  vapor  will  form  (with  ebullition  ;  i.e., 
from  the  body  of  the  liquid,  unless  the  heat  is  applied  very 
slowly),  but  the  tension  will  not  rise  above  a  fixed  value  for 
each  temperature  (independent  of  size  of  vessel)  so  long  as 
there  is  any  liquid  left.  Some  of  these  corresponding  values, 
for  water,  are  as  follows  :  For  a 

Fahr.  temp.      =   70°     100°     150°     212°     220°     287°    300° 
=  °'36    0'93     3'69     14<T     17'2     55'°     67'2 


=  one  atm. 
At  any  such  stage  the  vapor  is  said  to  be  saturated. 


608  MECHANICS   OF   ENGINEERING. 

Finally,  at  some  temperature,  dependent  on  the  ratio  of  the 
original  volume  of  water  to  that  of  the  vessel,  all  of  the  water 
will  have  been  converted  into  vapor  (i.e.,  steam);  and  if  the 
temperature  be  still  further  increased,  the  tension  also  increases 
and  no  longer  depends  on  the  temperature  alone,  but  also  on 
the  heaviness  of  the  vapor  when  the  water  disappeared.  The 
vapor  is  now  said  to  be  superheated,  and  conforms  more  in  its 
properties  to  perfect  gases. 

470.  Critical  Temperature. — From  certain  experiments  there 
seems  to  be  reason  to  believe  that  at  a  certain  temperature^ 
called  the  critical  temperature,  different  for  different  liquids, 
all  of  the  liquid  in  the  vessel  (if  any  remains,  and  supposing 
the  vessel  strong  enough  to  resist  the  pressure)  is  converted 
into  vapor,  whatever  be  the  size  of  the  vessel.  That  is,  above 
the  critical  temperature  the  substance  is  necessarily  gaseous, 
in  the  most  exclusive  sense,  incapable  of  liquefaction  by  pres- 
sure alone ;  while  below  this  temperature  it  is  a  vapor,  and  lique- 
faction will  begin  if,  by  compression  in  a  cylinder  and  conse- 
quent increase  of  pressure,  the  tension  can  be  raised  to  a  value 
corresponding,  for  a  state  of  saturation,  to  the  temperature 
(in  such  a  table  as  that  just  given  for  water).  For  example,  if 
vapor  of  water  at  220°  Fahrenheit  and  tension  of  10  Ibs.  per 
sq.  inch  (this  is  superheated  steam,  since  220°  is  higher  than 
the  temperature  which  for  saturation  corresponds  to  ^>  —  10 
Ibs.  per  sq.  inch)  is  compressed  slowly  (slowly,  to  avoid  change 
of  temperature)  till  the  tension  rises  to  17.2  Ibs.  per  sq.  in., 
which  (see  above  table)  is  the  pressure  of  saturation  for  a  tem- 
perature of  220°  Fahrenheit  for  water-vapor,  the  vapor  is  satu- 
rated, i.e..  liquefaction  is  ready  to  begin,  and  during  any  fur- 
ther slow  reduction  of  volume  the  pressure  remains  constant 
and  some  of  the  vapor  is  liquefied. 

By  "  perfect  gases,"  or  gases  proper,  we  may  understand, 
therefore,  those  which  cannot  be  liquefied  by  pressure  unac- 
companied by  great  reduction  of  temperature;  i.e.,  whose 
"  critical  temperatures"  are  very  low.  The  critical  temperature 
of  N2O,  or  nitrous  oxide  gas,  is  between  —  11°  and  -(-  8°  Cen- 
tigrade, while  that  of  oxygen  is  said  to  be  at  —  118°  Centi- 


LAW   OF    CHAKLES.  609 

grade.  [See  p.  471,  vol.  122  of  the  Journal  of  the  franklin 
Institute.  For  an  account  of  the  liquefaction  of  oxygen,  etc., 
see  the  same  periodical,  January  to  June,  1878.] 

471,  Law  of  Charles  (and  of  Gay  Lussac).  —  The  mode  of  gradu- 
ation of  the  air-thermometer  may  be  expressed  in  the  follow- 
ing formula,  which  holds  good  (for  practical  purposes)  within 
the  ordinary  limits  of  experiment  for  a  given  mass  of  any 
perfect  gas,  the  tension  remaining  constant  : 

y  =  7o  _j_  0.0036T  V.t  =  F.(l  +  .0036ft)  ;..(!) 

in  which  V0  denotes  the  volume  occupied  by  the  given  mass 
at  freezing-point  under  the  given  pressure,  V  its  volume  at 
any  other  temperature  t  Centigrade  under  the  same  pressure. 
Now,  273  being  the  reciprocal  of  .00367,  we  may  write 

v_  F(2W-M).  ie     y     .  T  (press.)  .  ,« 

y'  ~~273~"         e''  T.  ~  Tt    '         1  const.  \  '   (2> 

(see  §  468  ;)  in  which  T0  =  the  absolute  temperature  of  freezing- 
point.  =  273°  absolute  Centigrade,  and  T  the  absolute  tem- 
perature corresponding  to  t  Centigrade.  Eq.  (2)  is  also  true 
when  T  and  T0  are  both  expressed  in  Fahrenheit  degrees  (from 
absolute  zero,  of  course).  Accordingly,  we  may  say  that,  the 
pressure  remaining  the  same,  the  volume  of  a  given  mass  of 
gas  varies  directly  as  the  absolute  temperature. 

Since  the  weight  of  the  given  mass  of  gas  is  invariable  at  a 
given  place  on  the  earth's  surface,  we  may 

always  use  the  equation    Vy  =  V0y0  ,   .......     (3) 

pressure  constant  or  not,  and  hence  (2)  may  be  rewritten 

Y         T 


.  const.);    .    (4) 
Y         ^o 

i.e.,  if  the  pressure  is  constant,  the  heaviness  (and  therefore 
the  specific  gravity)  varies  inversely  as  the  absolute  tempera- 
ture. 


610 


MECHANICS   OF   ENGINEERING. 


Experiment  also  shows  (§468)  that  if  the  volume  [and  there- 
fore the  heaviness,  eq.  (3)]  remains  constant,  while  the  tem- 
perature varies,  the  tension  p  will  change  according  to  the 
following  relation,  in  which  p0  =  the  tension  when  the  tem- 
perature is  freezing : 


(5) 


t  denoting  the  Centigrade  temperature.     Hence  transforming, 
as  before,  we  have 

£  =  —       \  voL>  and  •'•  I  • 
p9    "  T0 '       (  heav.,  const.  (  ' 


(6) 


or,  the  volume  and  heaviness  remaining  constant,  the  tension 
of  a  given  mass  of  gas  varies  directly  as  the  absolute  tempera- 
ture. This  is  called  the  Law  of  Charles  (or  of  Gay  Lussac). 

472.  General  Formulae  for  any  Change  of  State  of  a  Perfect  Gas. 
— If  any  two  of  the  three  quantities,  viz.,  volume  (or  heavi- 
ness), tension,  and  temperature,  are  changed,  the  new  value  of 
the  third  is  determinate  from  those  of  the  other  two,  according 
to  a  relation  proved  as  follows  (remember- 
ing that  henceforth  the  absolute  temperature 
only  will  be  used,  T,  §  468) :  Fig.  521. 
At  A  a  certain  mass  of  gas  at  a  tension  of 
^?0,  one  atmosphere,  and  absolute  tempera- 
ture T0  (freezing),  occupies  a  volume  V0 . 
Let  it  now  be  heated  to  an  absolute  temp. 
=  T',  without  change  of  tension  (expanding 
behind  a  piston,  for  instance).  Its  volume  will  increase  to  a 
value  V  which  from  (2)  of  §  471  will  satisfy  the  relation 


Fio.  521. 


.- 

V  ~~  T  ' 

'  9  -*<> 


(See  B  in  figure.) 

Let  it  now  be  heated  without  change  of  volume  to  an  abso- 
lute temperature  T  (C  in  figure).     Its  volume  is  still  V,  but 


LAWS   OF  PERFECT  GASES.  611 

the  tension  has  risen  to  a  value  p,  such  that,  on  comparing  B 
and  0  by  eq.  (6),  we  have 

*-£ •  • « 

Combining  (T)  and  (8),  we  obtain  for  any  state  in  which  the 
tension  is  z>,  volume  V,  and  absolute  temperature  T,  in 


(General)  .    .    .    ^~  =&£* ;    or^-  =  a  constant ;  .    (9) 

or 

(General).    .    .    .    ^™    m  =  ^n    n , (10) 

which,  since 

(General).    .     Vy  =  V,y,  =  VmYm  =  Vnyn ,    .    .    .    (11) 

is  true  for  any  change  of  state,  we  may  also  write 

(General)  ....      -^  =  -^-, (12) 


or 

Pm.  Pn 

~ 


/-(  o\ 

(    } 


m-m  n-n 


These  equations  (9)  to  (13),  inclusive,  hold  good  for  any  state 
of  a  mass  of  any  perfect  gas  (most  accurately  for  air).  The 
subscript  0  refers  to  the  state  of  one-atmosphere  tension  and 
freezing-point  temperature,  m  and  n  to  any  two  states  what- 
ever (within  practical  limits)  ;  y  is  the  heaviness,  §§  Y  and  409, 
and  T  the  absolute  temperature,  §  468. 

If  p,  V,  and  7"of  equation  (9)  be  treated  as  variables,  and 
laid  off  to  scale  as  co-ordinates  parallel  to  three  axes  in  space, 
respectively,  the  surface  so  formed  of  which  (9)  is  the  equation 
is  a  hyperbolic  paraboloid. 

473.  Examples.  —  EXAMPLE  1.  —  What  cubic  space  will  be 
occupied  by  2  Ibs.  of  hydrogen  gas  at  a  tension  of  two  atmos- 
pheres and  a  temperature  of  27°  Centigrade? 


612  MECHANICS    OF   ENGINEERING. 

With  the  inch-lb.-sec.  system  we  have  p0  =  14.7  Ibs.  per  sq. 
inch,  y0  =  [.0056  -=- 1728]  Ibs.*  per  cubic  inch,  and  T0  =  273° 
absolute  Centigrade,  when  the  gas  is  at  freezing-point  at  one 
atmosphere  (i.e.,  in  state  sub-zero).  In  the  state  mentioned  in 
the  problem,  we  have^?  =  2  X  14.7  Ibs.  per  sq.  in., 

T  =  273  +  27  =  300°  absolute  Centigrade, 
while  y  is  required.     Hence,  from  eq.  (12), 

2  X  14.7      14.7 

Y  300     ~  (.0056  -*- 1728)273  ' 

/.  Y  =  '    OQ  Ibs.  per  cub.  in.  =  .0102  Ibs.  per  cub.  foot ;  and  if 

1728 

the  total  weight,  =  G,  =  Vy,  is  to  be  2  Ibs.,  we  have  (ft.,  lb., 
sec.)  V  =  2  -f-  0102  =  196  cubic  ieet.—Ans. 

EXAMPLE  2. — A  mass  of  air  originally  at  24°  Centigrade 
and  a  tension  indicated  by  a  barometric  column  of  40  inches 
of  mercury  has  been  simultaneously  reduced  to  half  its 
former  volume  and  heated  to  100°  Centigrade;  required  its 
tension  in  this  new  state,  which  we  call  the  state  n,  m  being  the 
original  state.  Use  the  inch,  lb.,  sec.  We  have  given,  there- 
fore, pm  =  f  £  X  14.7  Ibs.  per  sq.  inch,  Tm  =  273  +  24  =  297° 
absolute  Centigrade,  the  ratio 

Vm  :  Vn  =  2  :  1,  and  Tn  =  273  +  100  =373°  Abs.  Cent; 
while  pn  is  the  unknown  quantity.     From  eq.  (10),  hence, 

pn  =  ^.^..pm=  2  X  m .  «  X  U.r  =  4»^21l 

'  n      -*-  m 

which  an  ordinary  steam-gauge  would  indicate  as 
(49.22  -  14.7)  =  34.52  Ibs.  per  sq.  inch. 

(That  is,  if  the  weather  barometer  indicated  exactly  14.7  Ibs. 
per  sq.  inch.) 

*  See  table  on  p.  517. 


EXAMPLES.      PERFECT   GASES. 


618 


EXAMPLE  3. — A  mass  of  air,  Fig.  522,  occupies  a  rigid  closed 
vessel  at  a  temperature  of  15°  Centigrade  (equal  to  that  of  sur- 
rounding objects)  and  a  tension 
of  four  atmospheres  [state  m~\. 
J3y  opening  a  stop-cock  a  few 
seconds,  thus  allowing  a  portion 
of  the  gas  to  escape  quickly,  and  0 
then  shutting  it,  the  remainder  FIG.  522. 

of  the  air  [now  in  state  ri\  is  found  to  have  a  tension  of  only 
2.5  atmospheres  (measured  immediately) ;  its  temperature  can- 
not be  measured  immediately  (so  much  time  being  necessary 
to  affect  a  thermometer),  and  is  less  than  before.  To  compute 
this  temperature,  Tn,  we  allow  the  air  now  in  the  vessel  to 
come  again  to  the  same  temperature  as  surrounding  objects 
(15°  Centigrade) ;  find  then  the  tension  to  be  2.92  atmospheres. 
Call  the  last  state,  state  r  (inch,  lb.,  sec.).  The  problem  then 
stands  thus : 


pm  =4x14.7 
ym  =  1 

Tm  =  288°  Abs.  Cent. 


pn  =  2.5  X  14.7 

rp  _  o  j  principal 
n        1  unknown 


j9r  =  2.92  X  14.7 

yr  =  7n  (since  Fr=  Vn) 

Tr  =  Tm  =  288°  Abs.  Cent, 


In  states  n  and  r  the  heaviness  is  the  same  ;  hence  an  equa- 
tion like  (6)  of  §  4:71  is  applicable,  whence 


288=  246°  Ab,  Cent. 


or  —  27°  Centigrade  ;  considerably  lelow  freezing,  as  a  result  of 
allowing  the  sudden  escape  of  a  portion  of  the  air,  and  the  con- 
sequent sudden  expansion,  and  reduction  of  tension,  of  the  re- 
mainder. In  this  sudden  passage  from  state  m  to  state  n,  the 
remainder  altered  its  heaviness  (and  its  volume  in  inverse  ratio) 
in  the  ratio  (see  eqs.  (11)  and  (10)  of  §  472) 


Yn    _ 


=  Pn      '^n  = 

'  v*. '  Tn  ~ 


2.5  X  14.7   288 
4  X  14.7  '  246 


Now  the  heaviness  in  state  m  (see  eq.  (12),  §  472)  was 


614 


MECHANICS   OF  ENGINEERING. 


_Prn_       Y.T.  _  4  X  14.7 

on  288 


Yin  rri 

•*•  m 


.0807     273  _    .306 
1728  '  14.7  ~~  1728 


Ibs.  per  cub.  in.  =  .306  Ibs.  per  cub.  ft. 

.-.  yn  =  0.73  X  ym  =  0.223  Ibs.  per  cub.  ft., 

and  also,  since  Vm  =  0.73  Vn ,  about  -gfa  of  the  original  quan- 
tity of  air  in  vessel  has  escaped. 

[NOTE. — By  numerous  experiments  like  this,  the  law  of 
cooling,  when  a  mass  of  gas  is  allowed  to  expand  suddenly  (as, 
e.g.,  behind  a  piston,  doing  work)  has  been  determined ;  and 
vice  versa,  the  law  of  heating  under  sudden  compression  ;  see 
§487.] 

474.  The  Closed  Air-manometer. — If  a  manometer  be  formed 
of  a  straight  tube  of  glass,  of  uniform  cylindrical  bore,  which 
is  partially  filled  with  mercury  and  then  inverted  in  a  cistern 
of  mercury,  a  quantity  of  air  having  been  left  between  the 

mercury  and  the  upper  end  of  the 
tube,  which  is  closed,  the  tension  of 
this  confined  air  (to  be  computed 
from  its  observed  volume  and  tem- 
perature) must  be  added  to  that  due 
to  the  mercury  column,  in  order  to 
obtain  the  tension^/  to  be  measured. 
See  Fig.  523.  The  advantage  of  this 
kind  of  instrument  is,  that  to  meas- 
ure great  tensions  the  tube  need  not 
be  very  long.  Let  the  temperature 
T7,  of  whole  instrument,  and  the  tension  p,  of  the  air  or  gas 
in  the  cistern,  be  known  when  the  mercury  in  the  tube  stands 
at  the  same  level  as  that  in  the  cistern.  The  tension  of  the 
air  in  the  tube  must  now  be  j?x  also,  its  temperature  Tt ,  and  its 
volume  is  Vl  =  Fhl ,  T^being  the  sectional  area  of  the  bore  of 
the  tube ;  see  on  left  of  figure.  When  the  instrument  is  used, 
gas  of  unknown  tension  p'  is  admitted  to  the  cistern,  the  tem- 
perature of  the  whole  instrument  being  noted  (=  T\  and  the 
heights  h  and  h"  are  observed  (h  +  h"  cannot  be  put  =  hl 


v\ 

P" 

T'' 


CLOSED  AIR-MANOMETEK.  615 

unless  the  cistern  is  very  large),    p'  is  then  computed  as  fol- 
lows (eq.  (2),  §  413)  : 

.......    (1) 


in  which  p  =  the  tension  of  the  air  in  the  tube,  and  ym  the 

heaviness  of   mercury.     But  from   eq.  (10),  §  472,  putting 

7,  =  ^  and  V= 


*^T'T  =  JT*  ......   <2> 

Hence  finally,  from  (1)  and  (2), 

p>  =  h"Ym+hf.^pt  ......    (3) 

Since  Tl9p^^  and  A,  are  fixed  constants  for  each  instrument, 
we  may,  from  (3),  compute  p1  for  any  observed  values  of  h  and 
T  (N.B.  T  and  Tt  are  absolute  temperatures),  and  construct 
a  series  of  tables  each  of  which  shall  give  values  of  p'  for  a 
range  of  values  of  A,  and  one  special  value  of  T* 

EXAMPLE.  —  Supposing  the  fixed  constants  of  a  closed  air- 
manometer  to  be  (in  inch-lb.-sec.  system)  p^  =  14.7  (or  one 
atmosphere),  T,  =  285°  Abs.  Cent,  (i.e.,  12°  Centigrade),  and 
Aj  =  3'  4"  =  40  inches  ;  required  the  tension  in  the  cistern 
indicated  by  h"  —  25  inches  and  h  —  15  inches,  when  the 
temperature  is  —  3°  Centigrade,  or  T  =  270°  Abs.  Cent. 

For  mercury,  ym  =  [848.7  -T-  1728]  (§  409)  (though  strictly 
it  should  be  specially  computed  for  the  temperature,  since  it 
varies  about  .00002  of  itself  for  each  Centigrade  degree). 
Hence,  eq.  (3), 


1)8.  per  sq.  inch,  or  nearly  3£  atmospheres  [steam-gauge  would 
read  34.7  Ibs.  per  sq.  in.]. 

475.  Mariotte's  Law,  (or  Boyle's,)  Temperature  Constant  ;  Le., 
Isothermal  Change.  —  If   a  mass  of  gas  be  compressed,  or  al- 


616  MECHANICS   OF   ENGINEERING. 

lowed  to  expand,  isothermally,  i.e.,  without  change  of  tern* 
perature  (practically  this  cannot  be  done  unless  the  walls  of  the 
vessel  are  conductors  of  heat,  and  then  the  motion  must  be 
slow),  eq.  (10)  of  §  472  now  becomes  (since  Tm  =  Tn) 

( MariottJs    Law,  \        y        __  y          or  ^  =  E?i  •     m 
(      Temp,  constant )  pn    '  Vm  ' 

i.e.,  the  temperature  remaining  unchanged,  the  tensions  are 
inversely  proportional  to  the  volumes,  of  a  given  mass  of  a 
perfect  gas  ;  or,  the  product  of  volume  by  tension  is  a  constant 
quantity.  Again,  since  Vmym  =  Vnyn  for  any  change  of 
state, 

j  MariottJs     LaW,  \  Pm  __.  Ym        Qr     Pm  _  Pn  .  /m 

\      Temp,  constant  \  pn  ""  yn9          ym      yn  9  ^  ' 

i.e.,  the  pressures  (or  tensions  are  directly  proportional  to  the 
(first  power  of  the)  heavinesses,  if  the  temperature  is  the  same. 
This  law,  which  is  very  closely  followed  by  all  the  perfect 
gases,  was  discovered  by  Boyle  in  England  and  Mariotte  in 
France  more  than  two  hundred  years  ago,  but  of  course  is  only 
a  particular  case  of  the  general  formula,  for  any  change  of 
state,  in  §  472.     It  may  be  verified  experimen- 
tally in  several  ways.     E.g.,  in  Fig.  524,  the 
tube  0 M  being  closed  at  the  top,  while  PN  is 
open,  let  mercury  be  poured  in  at  P  until  it 
reaches  the  level  A ' Ef.     The  air  in  OA  is  now 
at  a  tension  of  one  atmosphere.     Let  more  mer- 
cury be  slowly  poured  in  at  P,  until  the  ail 
confined  in  0  has  been  compressed  to  a  volume 
,,p»     ,,-jA*'    OAn  =  i  of  OA,  and  the  height  B"E"  then 
A  ~         measured  ;  it  will  be  found  to  be  30  inches ;  i.e., 


A     the  tension  of  the  air  in  0  is  now  two  atmos- 
.M    pheres  (corresponding  to  60  inches  of  mercury). 
FIG.  524.         Again,  compress  the  air  in  0  to  %  its  original 
volume  (when  at   one  atmosphere),  i.e.,  to  volume   OA'"  =z 
\OA'>  and  the  mercury  height  B'"E'"  will  be  60  inches,  show- 
ing a  tension  of  three  atmospheres  in  the  confined  air  at  0  (90 


MAEIOTTE'S  LAW. 


617 


inches  of  mercury  in  a  barometer).  It  is  understood  fhat  the 
temperature  is  the  same,  i.e.,  that  time  is  given  the  compressed 
air  to  acquire  the  temperature  of  surrounding  objects  after 
being  heated  by  the  compression,  if  sudden. 

[NOTE.  —  The  law  of  decrease  of  steam  -pressure  in  a^steam- 
englne  cylinder,  after  the  piston  has  passed  the  point  of  "  cut- 
off "  and  the  confined  steam  is  expanding,  does  not  materially 
differ  from  Mariotte's  law,  which  is  often  applied  to  the  case 
of  expanding  steam  ;  see  §  479.] 

While  Mariotte's  law  may  be  considered  exact  for  practical 
purposes,  it  is  only  approximately  true,  the  amount  of  the 
deviations  being  different  at  different  temperatures.  Thus, 
for  decreasing  temperatures  the  product  Vp  of  volume  by 
tension  becomes  smaller,  with  most  gases. 

EXAMPLE  1.  —  If  a  mass  of  compressed  air  expands  in  a 
cylinder  behind  a  piston,  having  a  tension  of  60  Ibs.  per  sq. 
inch  (45.3  by  steam-gauge)  at  the  beginning  of  the  expansion, 
which  is  supposed  slow  (that  the  temperature  may  not  fall)  ; 
then  when  it  has  doubled  in  volume  its  tension  will  be  only 
30  Ibs.  per  sq.  inch  ;  when  it  has  tripled  in  volume  its  tension 
will  be  only  20  Ibs.  per  sq.  inch,  and  so  on. 

EXAMPLE  2.  Diving-bell.  —  Fig.  525.  If  the  cylindrical 
diving-bell  AB  is  10  ft.  in  height,  in  what 
depth,  h  =  ?,  of  salt  water,  can  it  be  let  down 
to  the  bottom,  without  allowing  the  water  to 
rise  in  the  bell  more  than  a  distance  a  =  4  ft.  ? 
Call  the  horizontal  sectional  area,  F.  The 
mass  of  air  in  the  bell  is  constant,  at  a  constant 
temperature.  First,  algebraically  ;  at  the 
surface  this  mass  of  air  occupied  a  volume 
Vm  =  Fh"  at  a  tension  pm  =  14.7  X  144  Ibs. 
per  sq.  ft.,  while  at  the  depth  mentioned  it  is 
compressed  to  a  volume  Vn  =  F(ji"  —  a\  and 
is  at  a  tension  pn  =  pm  -\-  (h  —  a)yW9  in  which 
yw  =  heaviness  of  salt  water.  Hence,  from 


'h= 


618  MECHANICS   OF  ENGINEERING. 


hence,  numerically,  (ft.,  lb.,  sec.,) 


476.  Mixture  of  Gases.  —  It  is  sometimes  stated  that  if  a  vessel 
is  occupied  by  a  mixture  of  gases  (between  which  there  is  no 
chemical  action),  the  tension  of  the  mixture  is  equal  to  the  sum 
of  the  pressures  of  each  of  the  component  gases  present  ;  or, 
more  definitely,  is  equal  to  the  sum  of  the  pressures  which  the 
separate  masses  of  gas  would  exert  on  the  vessel  if  each  in  turn 
occupied  it  alone  at  the  same  temperature. 

This  is  a  direct  consequence  of  Mariotte's  law,  and  may  be 
demonstrated  as  follows  : 

Let  the  actual  tension  be  j?,  and  the  capacity  of  the  vessel  V. 
Also  let  Vl  ,  "P,  ,  etc.,  be  the  volumes  actually  occupied  by  the 
separate  masses  of  gas,  so  that 

F,  +  F,+  ...=  F;      .....    (1) 


and  pl9  j9a,  etc.,  the  pressures  they  would  individually  exert 
when  occupying  the  volume  V  alone  at  the  same  tempera- 
ture. Then,  by  Mariotte's  law, 


etc.;     ...    (2) 
whence,  by  addition,  we  have 


i.e.,    p=pl+p9+  ........    (3) 

Of  course,  the  same  statement  applies  to  any  number  of 
separate  parts  into  which  we  may  imagine  a  mass  of  homo- 
geneous gas  to  be  divided. 

For  numerical  examples  and  practical  questions  in  the  solu- 
tion of  which  this  principle  is  useful,  see  p.  239,  etc.,  Ean- 
kine's  Steam-engine.  (Rankine  uses  0.365,  where  0.36T  has 
been  used  here.) 


BAKOMETBIC   LEVELLING.  619 

477,  Barometric  Levelling. — By  measuring  with  a  barometer 
the  tension  of  the  atmosphere  at  two  different  levels,  simul- 
taneously, and  on  a  still  day,  the  two  localities  not  being  widely 
separated  horizontally,  we  may  compute  their  vertical  distance 
apart  if  the  temperature  of  the  stratum  of  air  between  them 
is  known,  being  the  same,  or  nearly  so,  at  both  m 

stations.     Since  the  heaviness  of   the  air  is 
different  in  different   layers   of  the  vertical    .; 
column  between  the  two  elevations  ^Vand  J/,    : 
Fig.  526,  we  cannot  immediately  regard  the     ' 
whole  of  such  a  column  as  a  free  body  (as  was 


done  with  a  liquid,  §  412),  but  must  consider 


f 


h: 


a  horizontal  thin  lamina,  Z,  of  thickness 
=  dz  and  at  a  distance  =  z  (variable)  below 
M,  the  level  of  the  upper  station,  N  being 
the  lower  level  at  a  distance,  A,  from  J/. 

The  tension,  p,  must  increase  from  M 
downwards,  since  the  lower  laminae  have  to  support  a  greater 
weight  than  the  upper ;  and  the  heaviness  y  must  also  increase, 
proportionally  toj9,  since  we  assume  that  all  parts  of  the  col- 
umn are  at  the  same  temperature,  thus  being  able  to  apply 
Mariotte's  law.  Let  the  tension  and  heaviness  of  the  air  at 
the  upper  base  of  the  elementary  lamina,  Z,  be  p  and  y  re- 
spectively. At  the  lower  base,  a  distance  dz  below  the  upper, 
the  tension  is  p  +  dp.  Let  the  area  of  the  base  of  lamina  be 
F\  then  the  vertical  forces  acting  on  the  lamina  are  Fp^  down- 
ward ;  its  weight  yFdz  downward  ;  and  F(p  +  dp)  upward. 
For  its  equilibrium  ^(vert.  compons.)  must  =  0 ; 

.-.  F(p  +  dp)  —  Fp  —  Fydz  —  0 ; 

i.e.,  dp  =  ydz, (1) 

which  contains  three  variables.     But  from  Mariotte's  law, 
§  4T5,  eq.  (2),  if  pn  and  yn  refer  to  the   air  at  N,  we  may 

substitute  y  =  ?-^p  and  obtain,  after  dividing  by  p,  to  separate 
the  variables^  and  0, 


620  MECHANICS    OF   ENGINEERING. 


Summing  equations  like  (2),  one  for  each  lamina  between 
M  (where  p  =pm  and  2  =  0)  and  N  (where  p  =pn&nd.2  =  h\ 
we  have 


Yn 


which  gives  A,  the  difference  of  level,  or  altitude,  between  M. 
and  N)  in  terms  of  the  observed  tensions  pn  and  pm,  and  of  yn  , 
the  heaviness  of  the  air  at  JV,  which  may  be  computed  from 
eq.  (12),  §  472,  substituting  from  which  we  have  finally 


in  which  the  subscript  0  refers  to  freezing-point  and  one  at- 
mosphere tension  ;  Tn  and  T0  are  absolute  temperatures.  For 
the  ratio  pn  :  pm  we  may  put  the  equal  ratio  hn  :  hm  of  the 
actual  barometric  heights  which  measure  the  tensions.  The 
log.  e(or  Naperian,  or  natural,  or  hyperbolic,  log.)  =  (common 
log.  to  base  10)  X  2.30258.  From  §  409,  y0  of  air  =  0.08076 
Ibs.  per  cub.  ft.,  andj?0  =  14.701  Ibs.  per  sq.  inch  ;  TQ  =  273° 
Abs.  Cent. 

If  the  temperatures  of  the  two  stations  (both  in  the  shade) 
are  not  equal,  a  mean  temp.  =  ^(Tm-\-  Tn)  may  be  used  for 
Tn  in  eq.  (4),  for  approximate  results.  Eq.  (4)  may  then  be 
written 

h  (in  feet)  =  26213^.  log.  .  f^l  •     ...     (5) 

*i  LPm_\ 

The  quantity  ^  =  26213  ft.,  just   substituted,  is   called   the 

Ko 

height  of  the  homogeneous  atmosphere,  i.e.,  the  ideal  height 
which  the  atmosphere  would  have,  if  incompressible  and  non- 


BAROMETRIC   LEVELLING  —  ADIAEATIC   LAW.  621 

expansive  like  a  liquid,  in  order  to  exert  a  pressure  of  14.701 
Ibs.  per  sq.  inch  upon  its  base,  being  throughout  of  a  constant 
heaviness  =  .08076  Ibs.  per  cub.  foot. 
By  inversion  of  eq.  (4)  we  may  also  write 


_ 
'* 


(6) 


where  e  =  2.71828  =  the  Naperian  Base,  which  is  to  be  raised 

/77 

to  the  power  whose  index  is  the  abstract  number  —  .  -—•  .  h, 

Po       J-n 

and  the  result  multiplied  by  j?OT  to  obtain  pn. 

EXAMPLE.  —  Having  observed  as  follows  (simultaneously)  : 

At  lower  station  N,  hn  =  30.05  in.  mercury  ;  temp.  =  77.6°  F.  ; 
"  upper      "     M,  hm  =  23.66  "          "  «      =  70.4°  F.  ; 

required  the  altitude  h.  From  these  figures  we  have  a  mean 
absolute  temperature  of  460°  +  £(77.6  +  70.4)  =  534°  Abs. 
Fahr.  ;  hence,  from  (5), 


h  =  26213  X  «f  X  2.30258  X  log.  10  =  6787.9  ft. 

(Mt.  Guanaxuato,  in  Mexico,  by  Baron  von  Hum  bold  t.) 
Strictly,  we  should  take  into  account  the  latitude  of  the  place, 
since  y0  varies  with  g  (see  §  76),  and  also  the  decrease  in  the 
intensity  of  gravitation  as  we  proceed  farther  from  the  earth's 
centre,  for  the  mercury  in  the  barometer  weighs  less  per  cubic 
inch  at  the  upper  station  than  at  the  lower. 

Tables  for  use  in  barometric  levelling  can  be  found  in  Traut- 
wine's  Pocket-book,  and  in  Searles's  Field-book  for  Railroad 
Engineers,  as  also  tables  of  boiling-points  of  water  under  dif- 
ferent atmospheric  pressures,  forming  the  basis  of  another 
method  of  determining  heights. 

478.  Adiabatic  Change  —  Poisson's  Law.  —  By  an  adidbatio 
change  of  state,  on  the  part  of  a  gas,  is  meant  a  compression 
or  expansion  in  which  work  is  done  upon  the  gas  (in  compress- 

V-- 

OF  THE 

UNIVERSITY 

Of 

C.JI     IV.^Mfeltlk 


622  MECHANICS   OF   ENGINEERING. 

ing  it)  or  ~by  the  gas  (in  expanding  against  a  resistance)  when 
there  is  no  transmission  of  heat  between  the  gas  and  enclosing 
vessel,  or  surrounding  objects,  by  conduction  or  radiation. 
This  occurs  when  the  volume  changes  in  a  vessel  of  non-con- 
ducting material,  or  when  the  compression  or  expansion  takes 
place  so  quickly  that  there  is  no  time  for  transmission  of  heat 
to  or  from  the  gas. 

The  experimental  facts  are,  that  if  a  mass  of  gas  in  a  cylinder 
be  suddenly  compressed  to  a  smaller  volume  its  temperature  is 
raised,  and  its  tension  increased  more  than  the  change  of  vol- 
ume would  call  for  by  Mariotte's  law  ;  and  vice  versa,  if  a  gas 
at  high  tension  is  allowed  to  expand  in  a  cylinder  and  drive  a 
piston  against  a  resistance,  its  temperature  falls,  and  its  tension 
diminishes  more  rapidly  than  by  Mariotte's  law. 

Again  (see  Example  3,  §  473),  if  -gfa  of  the  gas  in  a  rigid 
vessel,  originally  at  4  atmos.  tension  and  temperature  of 
15°  Cent.,  is  allowed  to  escape  suddenly  through  a  stop-cock 
into  the  outer  air,  the  remainder,  while  increasing  its  volume 
in  the  ratio  100  :  73,  is  found  to  have  cooled  to  —  27°  Cent., 
and  its  tension  to  have  fallen  to  2.5  atmospheres ;  whereas,  by 
Mariotte's  law,  if  the  temperature  had  been  kept  at  288°  Abs. 
Cent.,  the  tension  would  have  been  lowered  to  y7^  of  4,  i.e., 
to  2.92  atmospheres  only. 

The  reason  for  this  cooling  during  sudden  expansion  is,  ac- 
cording to  the  Kinetic  Theory  of  Gases,  that  since  the  "  sensi- 
ble heat"  (i.e.,  that  perceived  by  the  thermometer),  or  "  fiat- 
ness"  of  a  gas  depends  on  the  velocity  of  its  incessantly  moving 
molecules,  and  that  each  molecule  after  impact  with  a  receding 
piston  has  a  less  velocity  than  before,  the  temperature  neces- 
sarily falls ;  and  vice  versa,  when  an  advancing  piston  com- 
presses the  gas  into  a  smaller  volume. 

If,  however,  a  mass  of  gas  expands  without  doing  work,  as 
when,  in  a  vessel  of  two  chambers,  one  a  vacuum,  the  other 
full  of  gas,  communication  is  opened  between  them,  and  the 
gas  allowed  to  fill  both  chambers,  no  cooling  is  noted  in  the 
mass  as  a  whole  (though  parts  may  have  been  cooled  tem- 
porarily). 

By  experiments  similar  to  that  in  Example  3,  §  473,  it  has 


ADIABATIC   CHANGE  —  EXAMPLES. 

been  found  that  for  air  and  the  "  perfect  gases,"  in  an  adiabatic 
change  of  volume  [and  therefore  of  heaviness],  the  tension 
varies  inversely  with  the  1.41th  power  of  the  volume.  This 
is  called  Poissorfs  Law.  For  ordinary  ^purposes  (as  Weisbach 
suggests)  we  may  use  f  instead  of  1.41,  and  hence  write 

Adiabat.  \  Pm  _  (Y*tf  orpm 

Change]  K      W'    '  S 

and  combining  this  relation  with  the  general  eqs.  (10)  and  (13), 
§  472,  we  have  also 


Adiabat. 


Pm  _  I  ^mY  /o\ 

pn  "  I  Tj  ' 


i.e.,  the  tension  varies  directly  as  the  cube  of  the  absolute  tem- 
perature; also, 

Adiabat.  I        lZA-(Tn 
Change}        \Vj-\T 

i.e.,  the  volume  is  inversely,  and  the  heaviness  directly,  as  the 
square  of  the  absolute  temperature. 

Here  in  and  n  refer  to  any  two  adiabatically  related  states. 
Tis  the  absolute  temperature. 

EXAMPLE  1.  —  Air  in  a  cylinder  at  20°  Cent,  is  suddenly 
compressed  to  ^  its  original  volume  (and  therefore  is  six  times 
as  dense,  i.e.,  has  six  times  the  heaviness,  as  before).  To  what 
temperature  is  it  heated  ?  Let  m  be  the  initial  state,  and  n  the 
final.  From  eq.  (3)  we  have 


=  A        ;    .-.  Tn  =  718°  Abs.  Cent, 


293       V    1 


or  nearly  double  the  absolute  temperature  of  boiling  water. 

EXAMPLE  2. — After  the  air  in  Example  1  has  been  given 
time  to  cool  again  to  20°  Cent,  (temperature  of  surrounding 
obiects)  it  is  allowed  to  resume,  suddenly,  its  first  volume,  i.e., 

*  That  is,  for  cases  where  the  larger  tension  is  not  greater  than  four  to 
six  times  the  smaller. 


624  MECHANICS    OF    ENGINEEETXG. 

to  increase  its  volume  sixfold  by  expanding  behind  a  piston, 
To  what  temperature  has  it  cooled  ?  Here  Tm  —  293°  Abs. 
Cent.,  the  ratio  Vm  :  Vn  =  £,  and  Tn  is  required.  Hence, 
from  (3), 


=  ;   .-.  Tn  =  293  -  V^  119.5°  Abs.  Cent., 


or  =  —  154°  Cent.,  indicating  extreme  cold. 

From  these  two  examples  the  principle  of  one  kind  of  ioe- 
making  apparatus  is  very  evident.  As  to  the  work  necessary 
to  compress  the  air  in  Example  1,  see  §  483.  It  is  also  evident 
why  motors  using  compressed  air  expansively  have  to  encoun- 
ter the  difficulty  of  frozen  watery  vapor  (present  in  the  air  to 
some  extent). 

EXAMPLE  3.  —  What  is  the  tension  of  the  air  in  Example  1 
(suddenly  compressed  to  %  ^ts  original  volume)  immediately 
after  the  compression,  if  the  original  tension  was  one  atmos- 
phere? That  is,  with  Vn  :  Vm  =  1  :  6,  and^?m  =  14.7  Ibs.  per 
sq.  inch,  pn  =  ?  From  eq.  (1),  (in.,  lb.,  sec.,) 


pn  =  14.T  X  6  =  14.7    26  =  216 

Ibs.  per  sq.  inch  ;  whereas,  if,  after  compression  and  without 
change  of  volume,  it  cools  again  to  20°  Cent.,  the  tension  is 
only  14.7  X  6  =  88.2  Ibs.  per  sq.  inch  (now  using  Mariotte's 
law). 

479.  Work  of  Expanding  Steam  following  Mariotte's  Law,—* 
Although  gases  do  not  in  general  follow  Mariotte's  law  in  ex- 
panding behind  a  piston  (without  special  provision  for  sup- 
plying heat),  it  is  found  that  the  tension  of  saturated  steam 
(i.e.,  saturated  at  the  beginning  of  the  expansion)  in  a  steau* 
engine  cylinder,  when  left  to  expand  after  the  piston  has 
passed  the  point  of  "  cut-off"  diminishes  very  nearly  in 
accordance  with  Mariotte's  law,  which  may  therefore  be  ap- 
plied in  this  case  to  find  the  work  done  per  stroke,  and  thence 
the  power.  In  Fig.  527  a  horizontal  steam-cylinder  iff 


EXPANDING   STEAM. 


shown  in  which  the  piston  is  making  its  left-to-right  stroke. 

The   "  back-  pressure"   is  con- 

stant  and  —  Fq,  F  being  the 

area  of  the  piston  and  y  the 

intensity  (i.e.,  per  unit  area) 

of  the  back  or  exhaust  pres- 

sure on  the  right  side  of  the 

piston  ;     while    the    forward 

pressure  on  the  left  face  of  the 

piston  =  Fp,  in  which  ^?  is  the 

steam-pressure  per  unit  area, 

and  is   different   at   different 

points  of  the  stroke.  While  the 

piston  is  passing  from  O"  to 

D",p  is  constant,  being  =pb  =  the  boiler-pressure,  since  the 

steam-port  is  still  open.     Between  D"  and  C",  however,  the 

steam  being  cut  off  (i.e.,  the  steam-port  is  closed)  at  D",  a  dis- 

tance a  from  0"  ,  p  decreases  with  Mariotte's  law  (nearly),  and 

its  value  is  (Fa  -r-  Fx)pb  at  any  point  on  C"  D"  ,  x  being  the 

distance  of  the  point  from  On  '. 

Above  the  cylinder,  conceive  to  be  drawn  a  diagram  in 
which  an  axis  OX\&  \\  to  the  cylinder-axis,  OY  an  axis  1  to 
fche  same,  while  0  is  vertically  above  the  left-hand  end  of  the 
cylinder.  As  the  piston  moves,  let  the  value  of  p  correspond- 
ing to  each  of  its  positions  be  laid  off,  to  scale,  in  the  vertical 
immediately  above  the  piston,  as  an  ordinate  from  the  axis  X. 
Make  OD'  =  q  by  the  same  scale,  and  draw  the  horizontal 
D  '  C  '.  Then  the  effective  work  done  on  the  piston-rod  while 
it  moves  through  any  small  distance  dx  is 


d  W  =  force  X  distance  =  F(p  — 


and  is  proportional  to  the  area  of  the  strip  RS,  whose  width  is 
rfa?  and  length  =j>  —  #;  so  that  the  effective  work  of  one 
stroke  is 


626  MECHANICS    OF   ENGHSTEEK1NG. 

and  is  represented  graphically  by  the  area  A'ARBC'D'A'* 
From  0"  to  D"  p  is  constant  and  —  pb  (while  q  is  constant  at 
all  points),  and  x  varies  from  0  to  a  ; 

which  may  be  called  the  work  of  entrance ,  and  is  represented 
by  the  area  of  the  rectangle  A'ADD' . 

From  D"  to  G" p  is  variable  and,  by  Mariotte's  law,  =  — pb  ; 


ie., 


:s  the  work  of  expansion,  adding  which  to  that  of  entrance, 
we  have  for  the  total  effective  work  of  one  stroke 


By  effective  work  we  mean  that  done  upon  the  piston-rod 
and  thus  transmitted  to  outside  machinery.  Suppose  the 
engine  to  be  "  double-acting" ;  then  at  the  end  of  the  stroke  a 
communication  is  made,  by  motion  of  the  proper  valves,  be- 
tween the  space  on  the  left  of  the  piston  and  the  condenser  of 
the  engine ;  and  also  between  the  right  of  the  piston  and  the 
boiler  (that  to  the  condenser  now  being  closed).  On  the  return 
stroke,  therefore,  the  conditions  are  the  same  as  in  the  forward 
stroke,  except  that  the  two  sides  of  the  piston  have  changed 
places  as  regards  the  pressures  acting  on  them,  and  thus  the 
same  amount  of  effective  work  is  done  as  before. 

If  n  revolutions  of  the  fly-wheel  are  made  per  unit  of  time 
(two  strokes  to  each  revolution),  the  effective  work  done  per 
unit  of  time,  i.e.,  the  power  of  the  engine,  is 


L  =  2/i  W=  27i^^l  +  log..  - 


WOKK   OF   STEAM-ENGINES.  627 

For  simplicity  the  above  theory  has  omitted  the  considera- 
tion of  "  clearance"  that  is,  the  fact  that  at  the  point  of  "  cut- 
off "  the  mass  of  steam  which  is  to  expand  occupies  not  only 
the  cylindrical  volume  Fa,  but  also  the  "  clearance"  or  small 
space  in  the  steam-passages  between  the  valve  and  the  entrance 
of  the  cylinder,  the  space  between  piston  and  valve  which  is 
never  encroached  upon  by  the  piston.  "  Wire-drawing"  has 
also  been  disregarded,  i.e.,  the  fact  that  during  communication 
with  the  boiler  the  steam-pressure  on  the  piston  is  a  little  less 
than  boiler-pressure.  For  these  the  student  should  consult 
special  works,  and  also  for  the  consideration  of  water  mixed 
with  the  steam,  etc.  Again,  a  strict  analysis  should  take  into 
account  the  difference  in  the  areas  which  receive  fluid-pressure 
on  the  two  sides  of  the  piston. 

EXAMPLE  1. — A  reciprocating  steam-engine  makes  120  revo- 
lutions per  minute,  the  boiler«pressure  is  40  Ibs.  by  the  gauge 
(i.e.,  j?&  =  40  +  14.7  =  54.7  Ibs.  per  sq.  Inch),  the  piston  area 
is  F=  120  sq.  in.,  the  length  of  stroke  I  =  16  in.,  the  steam 
being  "cut  off"  at  J  stroke  (.*.  a  =  4  in.,  and  I  :  a  =  4.00), 
and  the  exhaust  pressure  corresponds  to  a  "vacuum  of  25 
inches"  (by  which  is  meant  that  the  pressure  of  the  exhaust 
steam  will  balance  5  inches  of  mercury),  whence  q  =  -f$  of 
14.7  =  2.45  Ibs.  per  sq.  inch.  Eepired  the  work  per  stroke, 
W9  and  the  corresponding  power  Z. 

Since  I :  a  =  4,  we  have  log.,  4  =  2.302  X  .60206  =  1.386, 
and  from  eq.  (4),  (foot,  lb.,  see ,) 

W=m  (54.7  X  144) .  £ .  [2.386]  -  tf% (2.45  X  144) .  f 

=  5165.86  —  392.0  =  4773.868  ft.  Ibs.  of    work  per  stroke, 
and  therefore  the  power  at  2  rev.  per  sec.  (eq.  5)  is 

L  =  2  X  2  X  4773.87  =  19095.5  ft.  Ibs.  per  second. 
Hence  in  horse-powers,  which,  in  ft.,-lb.-sec.  system,  =Z-r-55fy 

Power  =  19095.5  -f-  550  =  34.7  H.  P. 
EXAMPLE  2. — Required  the  weight  of  steam  consumed  per 


628  MECHANICS   OF   ENGINEERING. 

second  by  the  above  engine  with  given  data;  assuming  with 
Weisbach  that  the  heaviness  of  saturated  steam  at  a  definite 
pressure  (and  a  corresponding  temperature,  §  469)  is  about  |-  of 
that  of  air  at  the  same  pressure  and  temperature. 

The  heaviness  of  air  at  54.7  Ibs.  per  sq.  in.  tension  and 
temperature  287°  Fahr.  (see  table,  §  469)  would  be,  from  eq- 
(12)  of  §  472  (see  also  §  409), 

-y.y.     P=.  0807X492    54.7 
T    >0        460  +  287    '14,7" 

Ibs.  per  cub.  foot,  f  of  which  is  0.1237  Ibs.  per  cub.  ft.  Now 
the  volume*  of  steam,  of  this  heaviness,  admitted  from  the 
boiler  at  each  stroke  is  V—  Fa  =  |ff  .  %  —  0.2777  cub.  ft., 
and  therefore  the  weight  of  steam  used  per  second  is 

4  X  .2777  X  0.1237  =  0.1374  Ibs. 

Hence,  per  hour,  0.1374  X  3600  =  494.6  Ibs.  of  feed-water 
tire  needed  for  the  boiler. 

If,  with  this  same  engine,  the  steam  is  used  at  full  boiler 
pressure  throughout  the  whole  stroke,  the  power  will  be 
greater,  viz.  =  %nFl(pb  —  q)  —  33440  ft.  Ibs.  per  sec.,  but 
the  consumption*  of  steam  will  be  four  times  as  great;  and 
hence  in  economy  of  operation  it  will  be  only  0.44  as  efficient 
(nearly). 

480.  Graphic  Representation  of  any  Change  of  State  of  a  Con- 
fined Mass  of  Gas. — The  curve  of  expansion  AB  in  Fig.  527  is 
an  equilateral  hyperbola,  the  axes  JTand  Y  being  its  asymp- 
totes. If  compressed  air  were  used  instead  of  steam  its  ex- 
pansion curve  would  also  be  an  equilateral  hyperbola  if  its 
temperature  could  be  kept  from  falling  during  the  expansion 
(by  injecting  hot-water  spray,  e.g.),  and  then,  following 
Mariotte's  law,  we  would  have,  as  for  steam,  (§  475,)^?  V=  con- 
stant, i&.,pFx  =  constant,  and  therefore  px  =  constant,  which 
is  the  equation  of  a  hyperbola,  p  being  the  ordinate  and  x  the 
abscissa.  This  curve  (dealing  with  a  perfect  gas)  is  also  called 
an  isothermal,  the  x  and  y  co-ordinates  of  its  points  being  pro- 

*  We  here  neglect  the  practical  fact  that  a  portion  of  the  fresh  steam 
entering  the  cylinder  is  condensed  prematurely,  so  that  the  actual  con- 
sumption is  somewhat  greater  than  as  here  derived. 


GKAPHICS   OF   CHANGE   OF   STATE   OF   GAS. 


629 


portional  to  the  volume  and  tension,  respectively,  of  a  mass  of 
air  (or  perfect  gas)  whose  temperature  is  maintained  constant. 
Hence,  in  general,  if  a  mass  of  gas  be  confined  in  a  rigid 
cylinder  of  cross-sec- 
tion F  (area),  provided 
with  an  air-tight  pis- 
ton, Fig.  528,  its  vol- 
ume, Fx,  is  propor- 
tional to  the  distance 
OD  —-  x  (of  the  piston 
from  the  closed  end  of 
the  cylinder)  taken  as 
an  abscissa,  while  its  o 
tension  p  at  the  same 
instant  may  be  laid  off 
as  an  ordinate  from  D. 
Thus  a  point  A  is  fixed.  Describe  an  equilateral  hyperbola 
through  A,  asymptotic  to  X  and  Y,  and  mark  it  with  the  ob- 
served temperature  (absolute)  of  the  air  at  this  instant.  In  a 
similar  way  the  diagram  can  be  filled  up  with  a  great  number 
of  equilateral  hyperbolas,  or  isothermal  curves,  each  for  its 
own  temperature.  Any  point  whatever  (i.e.,  above  the  critical 
temperature)  in  the  plane  angular  space  YO  X  will  indicate  by 
its  co-ordinates  a  volume  and  a  tension,  while  the  correspond 
ing  absolute  temperature  T  will  be  shown  by  the  hyperbola 
passing  through  the  point,  since  these  three  variables  always 
satisfy  the  relation  (§  472) 


.    .    .    (1)' 


FIG  528. 


Any  change  of  state  of  the  gas  in  the  cylinder  may  now  be 
represented  by  a  line  in  the  diagram  connecting  the  two  points 
corresponding  to  its  initial  and  final  states.  Thus,  a  point 
moving  along  the  line  AB,  a  portion  of  the  isothermal  marked 
293°  Abs.  Cent.,  represents  a  motion  of  the  piston  from  D  to 
E,  and  a  consequent  increase  of  volume,  accompanied  by  just 
sufficient  absorption  of  heat  by  the  gas  (from  other  bodies)  to 
maintain  its  temperature  at  that  figure  (viz.,  its  temperature  at 


630  MECHANICS    OF    ENGINEERING. 

A).  If-  the  piston  move  from  D  to  E,  without  transmission 
of  heat,  i.e.,  adiabatically  (§  478),  the  tension  falls  more 
rapidly,  and  a  point  moving  along  the  line  AB1  represents  the 
corresponding  continuous  change  of  state.  AB'  is  a  portion 
of  an  adiabatic  curve,  whose  equation,  from  §  478,  is 


=  const  5 


in  which  pK  and  XK  refer  to  the  point  K  where  this  particular 
adiabatic  curve  cuts  the  isothermal  of  freezing-point.  Evi- 
dently an  adiabatic  may  be  passed  through  any  point  of  the 
diagram.  The  mass  of  gas  in  the  cylinder  may  change  its 
state  from  A  to  B'  by  an  infinite  number  of  routes,  or  lines  on 
the  diagram,  the  adiabatic  route,  however,  being  that  most  likely 
to  occur  for  a  rapid  motion  of  the  piston.  For  example,  we 
may  cool  it  without  allowing  the  piston  to  move  (and  hence 
without  altering  its  volume  nor  the  abscissa  x)  until  the  pres- 
sure falls  to  a  value  pB>  =  DL  =  EBf,  and  this  change  is  rep- 
resented by  the  vertical  path  from  A  to  L  ;  and  then  allow  it 
to  expand,  and  push  the  piston  from  D  to  E  (i.e.,  do  external 
work),  during  which  expansion  heat  is  to  be  supplied  at  just 
such  a  rate  as  to  keep  the  tension  constant,  =%>&  =pL,  this 
latter  change  corresponding  to  the  horizontal  path  LB'  from 
L  to  B'. 

It  is  further  noticeable  that  the  work  done  by  the  expanding 
gas  upon  the  near  face  of  the  piston  (or  done  upon  the  gas  when 
compressed)  when  the  space  dx  is  described  by  the  piston,  is 
=  Fpdx,  and  therefore  is  proportional  to  the  area  pdx  of  the 
small  vertical  strip  lying  between  the  axis  X  and  the  line  or 
route  showing  the  change  of  state  ;  whence  the  total  work  done 
on  the  near  piston-face,  being  —  Ffpdx,  is  represented  by  the 
area  fpdx  of  the  plane  figure  between  the  initial  and  final 
ordinates,  the  axis  X  and  the  particular  route  followed  be- 
tween the  initial  and  final  states  (N.B.  We  take  no  account 
here  of  the  pressure  on  the  other  side  of  the  piston,  the  latter 
depending  on  the  style  of  engine).  For  example,  the  work 
done  on  the  near  face  of  the  piston  during  adiabatic  expansion 


WOEK   OF   ADIABATIC   EXPANSION". 


631 


from  D  to  E  is  represented  by  the  plane  figure  AB'EDA^ 
and  is  measured  by  its  area. 

The  mathematical  relations  between  the  quantities  of  heat 
imparted  or  rejected  by  conduction  and  radiation,  and  trans- 
formed into  work,  in  the  various  changes  of  which  the  con- 
fined gas  is  capable,  belong  to  the  subject  of  Thermodynamics^ 
which  cannot  be  entered  upon  here. 

It  is  now  evident  how  the  cycle  of  changes  which  a  definite 
mass  of  air  or  gas  experiences  when  used  in  a  hot-air  engine, 
compressed-air  engine,  or  air-compressor,  is  rendered  more  in- 
telligible by  the  aid  of  such  a  diagram  as  Fig.  528 ;  but  it 
must  be  remembered  that  during  the  entrance  into,  or  exit 
from,  the  cylinder,  of  the  mass  of  gas  used  in  one  stroke,  the 
distance  x  does  not  represent  its  volume,  and  hence  the  locus 
of  the  points  in  the  diagram  determined  by  the  co-ordinates^? 
and  x  during  entrance  and  exit  does  not  indicate  changes  of 
state  in  the  way  just  explained  for  the  mass  when  confined  in 
the  cylinder.  However,  the  work  done  by  or  upon  the  gas 
during  entrance  and  exit  will  still  be  represented  by  the  plane 
figure  included  by  that  locus  (usually  a  straight  horizontal 
line,  pressure  constant)  and  the  axis  of  JT  and  the  terminal 
ordinates. 

481.  Adiabatic  Expansion  in  an  Engine  using  Compressed  Air. 
— Fig.  529.  Let  the  compressed  air  at  a  tension  pm  and  an 
absolute  temperature  Tm  be  supplied 
from  a  reservoir  (in  which  the  loss  is 
continually  made  good  by  an  air-com- 
pressor). Neglecting  the  resistance  of 
the  port,  its  tension  and  temperature 
when  behind  the  piston  are  still  pm  and 
Tm.  Let  xn  =  length  of  stroke,  and  Q  1°  1  !  j 

Jet  the  cut-off  (or  closing  of  communi- 
cation with  the  reservoir)  be  made  at 
eome  point  D  where  x  =  xm ,  the  posi- 
tion of  D  being  so  chosen  (i.e.,  the  ratio 
:  xn  so  computed)  that  after  adia- 


x 


FIG.  529. 


batic  expansion  from  D  to  E  the  pres- 
sure shall  have  fallen  from  pm  at  M  (state  in)  to  a  value  pn  =  pa 


632  MECHANICS   OF   ENGINEERING. 

=  one  atmosphere  at  N  (state  ri),  at  the  end  of  stroke  ;  so  that 
when  the  piston  returns  the  air  will  be  expelled  (ki  exhausted  ") 
at  a  tension  equal  to  that  of  the  external  atmosphere  (though 
at  a  low  temperature).  Hence  the  back-pressure  at  all  points 
either  way  will  be  —  pn  per  unit  area  of  piston,  and  hence  the 
total  back-pressure  =  Fpn,  F  being  the  piston  area. 

From  0  to  D  the  forward  pressure  is  constant  and  =  Fpmy 
and  the  effective  work,  therefore,  or  work  on  piston-rod  from 
0  to  D,  is 

V=  F\_pm  —  P^Xm,          .       .       (1) 

represented  by  the  rectangle  M'MLN'.  The  cut-off  being 
made  at  Z>,  the  volume  of  gas  now  in  the  cylinder,  viz., 
Vm  =  Fxm,  is  left  to  expand.  Assuming  no  device  adopted 
(such  as  injecting  hot-water  spray)  for  preventing  the  cooling 
and  rapid  decrease  of  tension  during  expansion,  the  latter  is 
adidbatic,  and  hence  the  tension  at  any  point  P  between  M 
and  N  will  be 


.    .    [see  §478;  V=Fx\;  .    .  (a) 

oc  i 

.   Work  of  expansion 

®  =  FJ^pdx  -  Fpn(xn  -  #w),  (2) 
and  is  represented  by  the  area  MPNL. 
But    £pdx  =p^£x-*fa=  -  S^ipl)*-  (1)*]  ; 

-Q*].    .   .  (3) 

Now  substitute  (3)  in  (2)  and  then  add  (2)  to  (1),  noting  that 


COMPKESSED-AIR  ENGINE.  633 

-  Fpn(xn  -  aw)  = 


which  furthermore,  since  n  and  m  are  adiabatically  related 
[see  (a)],  can  be  reduced  to 


and  we  have  finally  : 

Total  work  on  piston-)       m-      nw          |~1       ^™V~I 
rod  per  stroke         f=T    =  3^^»-  [     "  W  J' 

But  ,?fem  =  Fm  ,  and  the  adiabatic  relation  holds  good, 


therefore  we  may  also  write 


in  which  Vm  =  the  volume  which  the  mass  of  air  used  per 
stroke  occupies  in  the  state  m,  i.e.,  in  the  reservoir,  where  the 
tension  isj?m  and  the  absolute  temperature  =  Tm. 

To  find  the  work  done  per  pound  of  air  used  (or  other  unit 
of  weight),  we  must  divide  W  by  the  weight  G  =  Vmym  of 
the  air  used  per  stroke,  remembering  (eq.  (13),  §  472)  that 


Work  per  unit  of  weight  of\_^T    p0  r«      IP*  \*"| 
lir  used  in  adiabatic  working  j  ~        m  y0T0[_        \pm)  J 


air 


The  back-pressurej?n  —pa  =  one  atmosphere. 
In  (6)  y0  =  .0807  Ibs.  per  cub.  foot,  p,  =  14.7  Ibs.  per  sq, 
inch,  and  T  -=-  ^3°  Abs,  Cent,  or  492°  Abs.  Fahr. 


634  MECHANICS   OF   ENGINEERING. 

It  is  noticeable  in  (6)  that  for  given  tensions  pm  and  pni  the 
work  per  unit  of  weight  of  air  used  is  proportional  to  the  ab- 
solute temperature  Tm  of  the  reservoir.  The  temperature  Tn 
to  which  the  air  has  cooled  at  the  end  of  the  stroke  is  obtained 
as  in  Example  2,  §478,  and  may  be  far  below  freezing-point 
unless  Tm  is  very  high  or  the  ratio  of  expansion,  xm  :  xn  ,  large. 

EXAMPLE.  —  Let  the  cylinder  of  a  compressed-air  engine  have 
a  section  of  F  •=-  108  sq.  in.  and  a  stroke  xn  =  15  inches.  The 
compressed  air  entering  the  cylinder  is  at  a  tension  of  2  atmos. 
(i.e.,  pm  =  29.4  Ibs.  per  sq.  in.,  and  pn  -7-pm  =  £),  and  at  a 
temperature  of  27°  Cent,  (i.e.,  Tm  '—  300°  Abs.  Cent.).  Ee- 
quired  the  proper  point  of  cut-off,  or  xm  =  ?  ,  in  order  that  the 
tension  may  fall  to  one  atmosphere  at  the  end  of  the  stroke  ; 
also  the  work  per  stroke,  and  the  work  per  pound  of  air.  Use 
the  foot,  pound,  and  second. 

From  eq.  (a),  above,  we  have 


P 


and  hence  the  volume  of  air  in  state  m,  used  per  stroke  [eq. 
(5)]  is 


Vm  =  Fxm  =  -Hi  X  0.7875  =  0.5906  cubic  feet; 
while  the  work  per  stroke  is 

W=  3  X  0.5906  X  29.4  X  144  X  [1  -  (i)»  ]  =  1545  ft.  Ibs., 
and  the  work  obtained  from  each  pound  of  air,  eq.  (6), 


ft.  Ibs.  per  pound  of  air  used. 

The  temperature  to  which  the  air  has  cooled  at  the  end  of 
stroke  [eq.  (2),  §  478]  is 


COMPRESSED-AIR  ENGINE.  635 

Tn  =  Tm  A/—-  =  300  X  VT=  300  X  .794  =  238°  Abe.  C. ; 

i.e.,  —  35°  Centigrade. 

482.  Remarks  on  the  Preceding, — This  low  temperature  is 
objectionable,  causing,  as  it  does,  the  formation  and  gradual 
accumulation  of  snow,  from  the  watery  vapor  usually  found 
in  small  quantities  in  the  air,  and  the  ultimate  blocking  of  the 
ports.  By  giving  a  high  value  to  Tm ,  however,  i.e.,  by  heat- 
ing the  reservoir,  Tn  will  be  correspondingly  higher,  and  also 
the  work  per  pound  of  air,  eq.  (6).  If  the  cylinder  be  encased 
in  a  u  jacket"  of  hot  water,  or  if  spray  of  hot  water  be  injected 
behind  the  piston  during  expansion,  the  temperature  may  be 
maintained  nearly  constant,  in  which  event  Mariotte's  law  will 
hold  for  the  expansion,  and  more  work  will  be  obtained  per 
pound  of  air ;  but  the  point  of  cut-off  must  be  differently 
placed.  Thus  if,  in  eq.  (4),  §  479,  we  make  the  back-pressure, 
which  =  (Fa  -f-  Fl)pb ,  equal  to  the  value  to  which  the  air- 
pressure  has  fallen  at  the  end  of  the  stroke  by  Mariotte's  law, 
we  have 

Work  per  stroke  with  )     _  17      -i        / 1\_  y      i        ( ^\   n\ 
isotherm,  expans.    j  ~~     a^>b  °f*'*  \a/~    ^M^^-S?  \ar 

and  hence 

Work  per  unit  of  weight  of  air,  )  _1  -77     PO     i         (l\       /9\ 
with  isothermal  expansion         j  ~~     m  y  f     £'e  \^/*  *  \.) 

Applying  these  equations  to  the  data  of  the  example,  we 
obtain 

Work  per  unit  of  weight  of  air  with  iso-  1  _  A  69  T7    P* 
thermal  expansion  \  "     '        my0f ' 

whereas,  with  adiabatic  expansion,  work  |  _  ~  „„  ~j    p. 
per  unit  of  weight  of  air  is  only  )  =  y~T' 


636  MECHANICS   OF   E1STGINEEKING. 

483.  Double-acting  Air-compressor,  with  Adiabatic  Compres- 
sion.— This  is  the  converse  of  §  481.  In  Fig.  530  we  have  the 
piston  moving  from  right  to  left,  compressing  a  mass  of  air 
which  at  the  beginning  of  the  stroke  fills  the  cylinder.  This  is 
brought  about  by  means  of  an  external 
motor  (steam-engine  or  turbine,  e.g.) 
which  exerts  a  thrust  or  pull  along  the 
^  piston-rod,  enabling  it  with  the  help 

<<  of  the    atmospheric  pressure   of  the 

"T j  1      fresh  supply  of  air  flowing  in  behind 
*  '   ^    x   it,  to  first  compress  a  cylinder-full  of 

.__ j         air  to  the  tension  of  the  compressed 

air  in   the   reservoir,   and   then,   the 


X—- ,1 


I  •'"  ''•'  I  I     port   or  valve   opening  at  this  stage, 
to  force  or  deliver  it  into  the  reservoir. 
— 1     Let  the  temperature  and  tension  of  the 


cylinder-full  of  fresh  air  be  THl  and 

pni ,  and  the  tension  in  the  reservoir  be  pmi .  Suppose  the 
compression  adiabatic.  As  the  piston  passes  from  E  toward 
the  left,  the  air  on  the  left  has  no  escape  and  is  compressed,  its 
tension  and  temperature  increasing  adiabatically  until  it  reaches 
a  value  pmi  =  that  in  reservoir,  at  which  instant,  the  piston 
being  at  some  point  D,  a  valve  opens  and  the  further  progress 
of  the  piston  simply  transfers  the  compressed  air  into  the  re- 
servoir without  further  increasing  its  tension.  Throughout 
the  whole  stroke  the  piston-rod  has  the  help  of  one  atmosphere 
pressure  on  the  right  face,  since  a  new  supply  of  air  is  entering 
on  the  right  to  be  compressed  in  its  turn  on  the  return  stroke. 
The  work  done  from  EtvD  may  be  called  the  work  of  com- 
pression  •  that  from  D  to  0,  the  work  of  delivery. 

[Since,  here,  dx  and  dW(or  increment  of  work)  have  con- 
trary signs,  we  introduce  the  negative  sign  as  shown.] 

/»z> 
The  work  of  compression  = —JEF(p —pni)dx.  .    .    .    (Ic) 

r° 

The  worl  tf  delivery       =  —JDF(pmi  —pn^dxa     .     .     (Id) 


AIR-COMPRESSOR.  637 

In  these  equations  only^  and  x  are  variables.  In  the  sum- 
mation indicated  in  (10)  p  change?  »4fek»  tically ;  in  (ld)j?  ii 
constant  =pmj  as  now  written. 

In  the  adiabatic  compression  the  air  passes  from  the  state  nt 
to  the  state  ml  (see  N^  and  Ml  in  figure). 

The  summations  in  these  equations  being  of  the  same  form 
as  those  in  equations  (1)  and  (2)  of  §  481,  but  with  limits  in- 
verted, we  may  write  immediately, 


Work  per  stroke  =  W=  3  Vmipmi  |~1  -(^Yl  .  .    (2) 

\Pml  1  -I 

ind 

Work  per  unit  of  weight 1  =  37*      P*    f~i  _(_^iV~~|         /3\ 
of  air  compressed          j  ~         mi  ^T7'  L        \^>m  /  J 


The  value  of  Tnil ,  at  the  immediate  end  of  the  sudden  com- 
pression, by  eq.  (2)  of  §  (478),  is 


The  temperature  of  the  reservoir  being  Tm,  as  in  §481 
(usually  much  less  than  Tmi),  the  compressed  air  entering  it 
cools  down  gradually  to  that  temperature,  Tm ,  contracting  in 
volume  correspondingly  since  it  remains  at  the  same  tension 
jpmi .  The  mechanical  equivalent  of  this  heat  is  lost. 

Let  us  now  inquire  what  is  the  efficiency  of  the  combination 
of  air-compressor  and  compressed-air  engine,  the  former  sup- 
plying air  for  the  latter,  both  working  adiabatically,  assuming 
that  no  tension  is  lost  by  the  compressed  air  in  passing  along 
fche  reservoir  between,  i.e.,  that^>OTl=^?m.  Also  assume  («s 
already  implied,  in  fact)  that^?^  =pn  =  one  atmos.,  and  that 
the  temperature,  Tni ,  of  the  air  entering  the  compressor  cyl- 
inder is  equal  to  that,  Tm ,  of  the  reservoir  and  transmission- 
pipe. 

To  do  this  we  need  only  find  the  ratio  of  the  amount  of 
work  obtained  from  one  pound  (or  other  unit  of  weight)  in  the 
eompressed-air  engine  to  the  amount  spent  in  compressing  onr 
of  air  in  the  compressor.  Calling  this  ratio  77,  the 


638  MECHANICS   OF   ENGINEERING. 

efficiency,  and  dividing  eq.  (6)  of  §  481  by  eq.  (3)  of  this  para- 
graph, we  have,  with  substitutions  just  mentioned, 

_  Tm    _  Abs.  temp,  of  outer  free  air  t 


_     m    _         .          ,  t  ft» 

p,  of  air  at  end  ' 
den  compression, 


Tmi        (  Abs.  temp,  of  air  at  end  ' 
\      of  sudd 


or,  substituting  from  eq.  (4),  and  remembering  that  TUl  =  T 
we  have  also 


(6) 

JS  *jsmi 

also,  since 


we  may  write 

_  7^  __  Ab.  tern,  air  leaving  eng.  cyl.  ,„. 

Tm  Ab.  tein.  outer  free  air. 

For  practical  details  of  the  construction  and  working  of 
engines  and  compressors,  and  the  actual  efficiency  realized,  the 
student  may  consult  special  works,  as  they  lie  somewhat  be- 
yond the  scope  of  the  present  work. 

EXAMPLE  1.  —  In  the  example  of  §  445,  the  ratio  of  pm  to  pn 
was  =  J-.  Hence,  if  compressed  air  is  supplied  to  the  reser- 
voir under  above  conditions,  the  efficiency  of  the  system  is, 
from  eq.  (6),  ij  =  V~%  =  0.794,  about  80  per  cent. 

z>       1 
EXAMPLE  2.  —  If  the  ratio  of  the  tensions  is  as  small  as  —  —  ^, 

Pm         6 

the  efficiency  would  be  only  (-J)*  =  0.55  ;  i.e.,  45  per  cent  of 
the  energy  spent  in  the  compressor  is  lost  in  heat. 

EXAMPLE  3.  —  What  horse-power  is  required  in  a  blowing 
engine  to  furnish  10  Ibs.  of  air  per  minute  at  a  pressure  of 
4  atmos.,  with  adiabatic  compression,  the  air  being  received 
by  the  compressor  at  one  atmosphere  tension  and  27°  Cent. 
(ft.-lb.-sec.  system).  Since  27°  C.  =  300°  Abs.  C.  =  Tn,  ,  we 
have,  from  eq.  (4), 


Tmi  =  300  (±)i  =  477°  Abs.  Cent; 
and  hence,  eq.  (3), 


HOT-AIK   ENGINES.     Z.P*C  639 


14  7  X  144 
The  work  per  pound  of  air  =  3X  477  ' 


=  50870  ft.  Ibs.  per  pound  of  air.  Hence  10  Ibs.  of  air  will 
require  508700  ft.  Ibs.  of  work  ;  and  if  this  is  done  every  min- 
ute we  have  the  req.  H.  P.  =  WW  =  15-4  H-  P- 

NOTE.  —  If  the  compression  could  be  made  isothermal,  an 
approximation  to  which  is  obtained  by  injecting  a  spray  of 
cold  water,  we  would  have,  from  eqs.  (1)  and  (2)  of  §  482  : 

Worker  )_T     p0  ,       /  pmi  \  _300  X  14.7  X  144 
II.  air     }  -Tn^T^'\l^J  -       .0807  X  273 

=  39950  ft.  Ibs.  per  lb.,  and  the  corresponding  H.  P.  =  12.1  ; 
a  saving  of  about  25  per  cent,  compared  with  the  former. 
The  difference  was  employed  in  heating  the  air  in  the  air-com- 
pressor with  adiabatic  compression,  and  was  lost  when  that 
extra  heat  was  dissipated  in  the  reservoir  as  the  air  cooled 
again.  This  difference  is  easily  shown  graphically  by  compar- 
ing in  the  same  diagram  the  areas  representing  the  work  done 
in  the  two  cases.* 

484.  Hot--air  Engines.  —  Since  we  have  seen  that  the  tension 
of  air  and  other  gases  can  be  increased  by  heating,  if  the  vol- 
ume be  kept  the  same,  a  mass  of  air  thus  treated  can  after- 
wards be  allowed  to  expand  in  a  working  cylinder,  and  thus 
become  a  means  of  converting  heat  into  work.  In  Stirling's 
hot-air  engine  a  definite  confined  mass  of  air  is  used  indefinitely 
without  loss  (except  that  occasional  small  supplies  are  needed 
to  make  up  for  leakage),  and  is  alternately  heated  and  cooled. 
A  displacement-plunger,  or  piston,  fitting  loosely  in  a  bell-like 
chamber,  is  so  connected  with  the  piston  of  the  working- 
cylinder  and  the  fly-wheel,  that  its  forward  stroke  is  made 
while  the  other  piston  waits  at  the  beginning  of  its  stroke. 
In  this  motion  the  plunger  causes  the  confined  air  to  pass  in  a 
thin  sheet  over  the  top  and  sides  of  the  furnace  dome,  thus 
greatly  increasing  its  tension.  The  air  then  expands  behind 
the  working  piston  with  falling  tension  and  temperature,  and, 

*  See  Eng.  Neics,  pp.  234  and  297,  Oct.  and  Nov.  1897,  for  an  account 
of  a  "  four-stage  "  compressor  and  test  of  same. 


640  MECHANICS    OF   ENGINEERING. 

while  that  piston  pauses  at  the  end  of  its  forward  stroke,  is 
again  shifted  in  position,  though  without  change  of  volume, 
by  the  return  stroke  of  the  plunger,  in  such  a  way  as  to  pass 
through  a  coil  of  pipes  in  which  cold  water  is  flowing.  This 
reduces  both  its  temperature  and  tension,  and  hence  its  resist- 
ance to  the  piston  on  the  return  stroke  is  at  first  less  than  at- 
mospheric, but  is  gradually  increased  by  the  compression. 
This  cycle  of  changes  is  repeated  indefinitely,  and  is  easily 
traced  on  a  diagram  like  that  in  Fig.  528,  and  computations 
made  accordingly. 

A  special  invention  of  Stirling's  is  the  "  regenerator"  or  box 
filled  with  numerous  sheets  of  wire  gauze,  in  its  passage 
through  which  the  working  air,  after  expansion,  deposits  some 
of  its  heat,  which  it  re-absorbs  to  some  extent  when,  after 
further  cooling  in  the  "  refrigerator"  or  pipe  coil  and  com- 
pression by  the  return  stroke  of  the  piston,  it  is  made  to  pass 
backward  through  the  regenerator  to  be  further  heated  by  the 
furnace  in  readiness  for  a  forward  stroke.  This  feature,  how- 
ever, has  not  realized  all  the  expectations  of  its  inventor  and 
improvers,  as  to  economy  of  heat  and  fuel. 

In  Ericsson's  hot-air  engine,  of  more  recent  date,  the  dis- 
placement-plunger fits  its  cylinder  air-tight,  but  valves  can  be 
opened  through  its  edges  when  moving  in  one  direction,  thus 
causing  it  to  act  temporarily  as  a  loose  plunger,  or  shifter. 
The  two  pistons  move  simultaneously  in  the  same  direction  in 
the  same  cylinder,  but  through  different  lengths  of  stroke,  so 
that  the  space  between  them  is  alternately  enlarged  and  con- 
tracted. The  working  piston  also  has  valves  opening  through 
it  for  receiving  a  fresh  supply  of  air  into  the  space  between 
the  two  pistons.  During  the  forward  stroke  a  fresh  instal- 
ment from  the  outer  air  enters  through  the  working  piston  into 
the  space  between  it  and  the  other,  whose  valves  are  now 
closed  and  which  is  now  expelling  from  its  further  face, 
through  proper  valves,  the  air  used  in  the  preceding  stroke ; 
no  work  is  done  in  this  stroke.  On  the  return  stroke  this 
fresh  supply  of  air  is  free  to  expand  behind  the  now  retreating 
working  piston,  while  its  tension  is  greatly  increased  by  its 
being  shifted  (at  least  a  large  portion  of  it)  over  the  furnace 


GAS-ENGI1STES.  641 

dome  through  the  valves  (now  open)  of  the  plunger  piston,  by 
the  motion  of  the  latter,  which  now  acts  as  a  loose  plunger. 
The  engine  is  therefore  only  single-acting,  no  work  being  done 
in  each  forward  stroke.  For  further  details,  see  Goodeve's 
and  Rankine's  works  on  the  steam-engine ;  also  the  article 
"  Hot-air  Engine"  in  Johnson's  Cyclopaedia  by  Fres.  Barnard, 
and  Rontgen's  Thermodynamics. 

485.  Brayton's  Petroleum-engine. — Although  a  more  recent 
invention  than  the  gas-engines  to  be  mentioned  presently,  this 
motor  is  more  closely  related  to  hot-air  engines  than  the  latter. 
By  a  slow  combustion  of  petroleum  vapor  the  gaseous  products 
of  combustion,  while  under  considerable  tension,  are  enabled 
to  follow  up  a  piston  with  a  sustained  pressure,  being  left  to 
expand  through  the  latter  part  of  the  stroke.     Thus  we  have 
the   furnace   and  working   cylinder   combined  in  one.     The 
gradual  combustion  is  accomplished  by  making  use  of  the 
principle  of  the  Davy  safety-lamp  that  flame  will  not  spread 
through  layers  of  wire  gauze  of  proper  fineness. 

486.  Gas-engines. — We  again  have  the  furnace  and  working" 
cylinder  in  one  in  a  "gas-engine"  where  illuminating  gas  and 
atmospheric  air  are  introduced  into  the  working  cylinder  in 
proper  proportions  (about  ten  parts  of  air  to  one  of  gas,  by 
weight)  to  form  an  explosive  mixture  of  more  or  less  violence  and 
exploded  at  a  certain  point  of  the  stroke,  causing  a  very  sudden 
rise  of  temperature  and  tension,  after  which  the  mass  expands 
behind  the  piston  with  falling  pressure.     On  the  return  stroke 
the  products  of  combustion  are  expelled,  and  no  work  done, 
these  engines  being  single-acting.     In  some  forms  the  mixture 
is  compressed  before  explosion,  since  it  has  been  found  that 
under  this  treatment  a  mixture  containing  a  larger  proportion 
of  air  to  gas  can  be  made  to  ignite,  and  that  then  the  resulting 
pressure  is  more  gradual  and  sustained,  like  that  of  steam  or  of 
the  mixture  in  the  Brayton  engine.     That  is,  the   effect  ia 
analogous  to  that  of  "  slow-burning  powder"  in  a  gun. 

In  the  "  Otto  Silent  Gas-engine"  the  explosion  occurs  only 
every  fourth  stroke,  and  one  side  of  the  piston  is  always  open 


642  MECHANICS    OF   ENGINEERING. 

to  the  air.  The  action  on  the  other  side  of  the  piston  is  as 
follows :  (1)  In  the  forward  stroke  a  fresh  supply  of  explosive 
mixture  is  drawn  into  the  cylinder  at  one  atmosphere  tension. 
(2)  The  next  (backward)  stroke  compresses  the  mixture  into 
about  one  fourth  of  its  original  bulk,  this  operation  occurring 
at  the  expense  of  the  kinetic  energy  of  the  fly-wheel.  (3)  The 
mixture  is  ignited,  the  pressure  rises  to  8  or  10  atmospheres, 
and  work  is  done  on  the  piston  through  the  next  (forward) 
stroke,  the  tension  of  the  products  of  combustion  having 
fallen  to  about  two  atmospheres  at  the  end  of  the  stroke. 
(4)  In  the  next  (backward)  stroke  the  products  of  combus- 
tion are  expelled  and  no  work  is  done. 

The  Atkinson  "Cycle  Gas-engine,"  an  English  invention  of 
recent  date  (see  the  London  Engineer  for  May  1887 ;  pp.  361 
and  380)  also  makes  an  explosion  every  fourth  stroke,  but  the 
link  work  connecting  the  piston  and  fly-wheel  is  of  such  de- 
sign that  the  latter  makes  but  one  revolution  during  the  four 
strokes.  Also  the  length  of  the  expansion  or  working  stroke 
is  greater  than  that  of  the  compression  stroke  and  the  products 
of  combustion  are  completely  expelled.  Consequently  the  effi- 
ciency of  this  motor  is  at  present  greater  than  that  of  any  other 
gas-engine.*  See  §  487. 

One  of  the  most  simple  gas-engines  is  made  by  the  Economic 
Motor  Company  of  New  York.  The  piston  has  no  packing, 
being  a  long  plunger  ground  to  fit  the  cylinder  accurately  and 
kept  well  lubricated.  As  with  most  gas-engines  the  cylinder  is 
encased  in  a  water-jacket  to  prevent  excessive  heating  of  the 
working  parts  and  consequent  decomposition  of  the  lubricant. 

For  further  details  on  these  motors,  see  Rankine's  "  Steam- 
engine,"  Clark's  "Gas-engines"  in  Yan  BTostrand's  Science 
Series,  and  article  "  Gas-engine"  in  Johnson's  Cyclopaedia ;  also 
Prof.  Thurston's  report  on  Mechanical  Engineering  at  the 
Vienna  Exhibition  of  1873,  and  proceedings  of  the  "Society 
of  Engineers"  (England)  for  1881. 

487.  Efficiency  of  Heat-engines. — According  to  the  mechan- 
ical theory  of  heat,  the  combustion  of  one  pound  of  coal,  pro- 

*  Later :  The  Diesel  petroleum-engine  (practically  a  gas-engine)  greatly 
exceeds  the  Atkinson  engine  in  performance  (see  Engineering  News, 
Mar.  1898,  p.  172,  and  Dec.  1898,  p.  422),  a  heat  efficiency  of  35  per  cent 
having  been  attained  (in  the  cylinder  ;  25  on  the  brake). 


GAS-ENGINES.  643 

ducing,  as  it  does,  about  14,000  heat-units  (British  Thermal 
units ;  see  §  149,  Mechanics)  should  furnish 

14,000  X  772  =  10,808,000  ft.  Ibs.  of  work, 

if  entirely  converted  into  work.  Let  us  see  how  nearly  this  is 
accomplished  in  the  performance  of  the  most  recent  and 
economical  marine  engines  of  the  present  day,  viz.,  the  triple 
expansion  engines  of  some  Atlantic  steamers,  which  are  claimed 
to  have  consumed  per  hour  only  1.25  Ibs.  of  coal  for  each 
measured  (u  indicated  ")  horse-power  of  effective  work  done  in 
their  cylinders.  The  work-equivalent  of  1.25  Ibs.  of  coal  per 
hour  is 

1.25  X  14,000  X  772  =  13,510,000  ft.  Ibs.  per  hour; 

while  the  actual  work  per  hour  implied  in  "  one  H.  P.  per 
hour"  is 

33000  X  60  =  1,980,000  ft.  Ibs.  per  hour. 

That  is,  the  engines  utilize  only  one  seventh  of  the  heat  of  com- 
bustion of  the  fuel. 

According  to  Prof.  Thurston,  this  is  a  rather  extravagant 
claim  (1.25),  the  actual  consumption  having  probably  been  1.4 
Ibs.  of  coal  per  H.  P.  per  hour. 

The  ordinary  compound  marine  engine  is  stated  to  use  as 
little  as  2.00  Ibs.  per  hour  for  each  H.  P. 

Most  of  the  heat  not  utilized  is  dissipated  in  the  condenser. 

Similarly,  the  water  jacket,  a  necessary  evil  in  the  operation 
of  the  gas-engine,  is  a  source  of  great  loss  of  heat  and  work. 
Still,  Mr.  Wm.  Anderson  in  his  recent  work,  "  Conversion  of 
Heat  into  Work"  (London,  1887),  mentions  a  motor  of  this 
class  as  having  converted  into  work  -J-  of  the  heat  of  combus- 
tion [an  Otto  "  Silent  Gas-engine,"  tested  at  the  Stevens 
Institute,  Hoboken,  N".  J.,  in  1883]  ;  while  Prof.  Unwin  found 
the  Atkinson  engine  (see  last  paragraph)  capable  of  returning 
(in  the  cylinder)  fully  \  of  the  heat-equivalent  of  the  gas  con- 
sumed. [This  latter  result  was  confirmed  in  Philadelphia  in 
Jan.  1889  by  Prof.  Barr,  under  direction  of  Prof.  Thurston.] 


644  MECHANICS    OF   ENGINEEKING. 

488.  Duty  of  Pumping-engines. — Another  way  (often   used 
in  speaking  of  the  performance  of  pumping-enginesj  of  ex- 
pressing the  degree  of  economy  attained  in  the  use  of  fuel  by 
the  combined  furnace,  boiler,  and  engine  is  to  give  the  num- 
ber of  foot-pounds  of  work  obtained  from  each  100  Ibs.  of  coal 
consumed  in  the  furnace,  calling  it  the  "  duty"  of  the  engine. 

For  example,  by  a  duty  of  99,000,000  ft.  Ibs.  it  is  meant  that 
from  each  pound  of  coal  990,000  ft.  Ibs.  of  work  are  obtained. 
From  this  we  gather  that,  since  one  horse-power  consists  of 
33,000  X  60  =  1,980,000  ft.  Ibs.  per  hour,  the  engine  men- 
tioned  must  use  each  hour 

1,980,000  -r-  990,000  =  2  Ibs.  of  coal  for  each  H.  P.  developed ; 

which  is  as  low  a  figure  as  that  attained  by  the  marine  engines 
last  quoted.* 

489.  Buoyant  Effort  of  the  Atmosphere. — In  the  case  of  a 
body  of  large  bulk  but  of  small  specific  gravity  the  buoyant 
effort  of  the  air  (due  to  the  same  cause  as  that  of  water,  see 
§  456)   becomes   quite   appreciable,   and  may   sometimes   be 
greater  than  the  weight  of  the  body.     This  buoyant  effort  is 
equal  to  the  weight  of  air  displaced,  i.e.,  =  Vy,  where  V  is 
the  volume  of  air  displaced,  and  y  its  heaviness. 

If  Gl  —  total  weight  of  the  body  producing  the  displace- 
ment, the  resultant  vertical  force  is 

p=Gl-  VY,  .......   (i) 

and  for  equilibrium,  or  suspension  in  the  air,  we  must  have 
P  =  0,  i.e., 

G,=  Vy (2) 

We  may  therefore  find  approximately  the  elevation  where 
a  given  balloon  will  cease  to  ascend,  by  determining  the  heavi- 
ness y  of  the  air  at  that  elevation  from  eq.  (2) ;  then,  know- 
ing approximately  the  temperature  of  the  air  at  that  elevation, 
we  may  compute  its  tension  p  [eq.  (13),  §  472],  and  finally, 
from  eqs.  (3),  (4),  or  (5)  of  §  477,  obtain  the  altitude  required. 

EXAMPLE. — The  car  and  other  solid  parts  of  a  balloon  weigh 

*  The  duty  of  the  engine  (as  distinct  from  that  of  the  furnace  and  boiler) 
lias  recently  been  defined  to  be  the  number  of  foot-pounds  of  work  rendered 
by  the  engine  for  each  million  heat-units  in  the  steam  furnished  to  it.  (Kent.) 


BALLOONS.  645 

400  Ibs.,  and  the  bag  contains  12,000  cub.  feet  of  illuminating 
gas  weighing  0.030  Ib.  per  cub.  foot  at  a  tension  of  one  at- 
mosphere and  temperature  of  15°  Cent.,  so  that  its  total 
weight  =12,000  X  0.030  =  360  Ibs. 

Hence  G-1  =  T60  Ibs.     We  may  also  write  with  sufficient 
accuracy  :  Whole  volume  of  displacement  =  F  =  12,000  cub.  ft. 

As  the  balloon  ascends  the  exterior  pressure  diminishes,  and 
the  confined  gas  tends  to  expand  and  so  in-  -M--Or  ________  -*—  m 

crease  the  volume  of  displacement  "P";  but      .'•'•.'•.';••'•;.'•.  jv§* 

this  we  shall    suppose    prevented   by   the 

strength  of  the  envelope.     At  the  surface 

of  the  ground  (station  n  of  Fig.  531  ;  see 

also  Fig.  526)  let  the  barometer  read  29.6 

inches  and  the  temperature  be  15°  Cent. 

Then  Tn  =  288°  Abs.  Cent.,  and  the  heavi-  FIG.  531. 

ness  of  the  air  at  n  is 

.0807  X  273    *fjf  x  14.7 
14.7  288~~ 

" 


At  the  unknown  height  A,  where  the  balloon  is  to  come  to 
vest,  i.e.,  at  J/",  Gl  must  =  Yy    [eq.  (2)]  ;  therefore 


T  =  nt:  =  -0633  lbs-  per  cnb-  ft  ; 


and  if  the  temperature  at  M  be  estimated  to  be  5°  Cent,  (or 
Tm  =  278°  Abs.  Cent.)  (on  a  calm  day  the  temperature  de- 
creases about  1°  Cent,  for  each  500  ft.  of  ascent),  we  shall 


-    YnT«_WU   288 
~ 


pm  ~7mTm~  •  0633 
and  hence,  from  eq.  (5),  §  477,  with  $(Tm+  Tn)  put  for  Tn, 
h  =  26213  X  ff  f  X  2.30258  X  log.,.  1.206  =  5088  ft. 


CHAPTEE  YL 


HYDRODYNAMICS  BEGUN— STEADY  FLOW  OF  LIQUIDS 
THROUGH  PIPES  AND  ORIFICES. 


489a.  The  subject  of  Water  in  Motion  presents  one  of  the 
most  unsatisfactory  branches  of  Applied  Mechanics,  from  a 
mathematical  stand-point.  The  internal  eddies,  cross-currents, 
and  general  intricacy  of  motion  of  the  particles  among  each 
other,  occurring  in  a  pipe  transmitting  a  fluid,  are  almost  en- 
tirely defiant  of  mathematical  expression,  though  the  flow  of 
water  through  a  circular  orifice  in  a  thin  plate  into  the  air  pre- 
sents a  simpler  case,  where  the  conception  of  "  stream  lines"  is 
probably  quite  close  to  the  truth.  In  most  practical  cases  we 
are  forced  to  adopt  as  a  basis  for  mathematical  investigation 
the  simple  assumption  that  the  particles  move  side  by  side  in. 
such  a  way  that  those  which  at  any  instant  form  a  lamina 
or  thin  sheet,  1  to  the  axis  of  the  pipe  or  orifice,  remain 
together  as  a  lamina  during  the  further  stages  of  the  flow. 
This  is  the  Hypothesis  of  Flow  in  Plane  Layers,  or  Laminated 
Flow.  Experiment  is  then  relied  on  to  make  good  the  discre- 
pancies between  the  indications  of  the  formulae  resulting  from 
this  theory  and  the  actual  results  of  practice ;  so  that  the  science 
of  Hydrodynamics  is  largely  one  of  coefficients  determined  by 
experiment. 

490.  Experimental  Phenomena  of  a  "  Steady  Flow." — As  pre- 
liminary to  the  analysis  on  which  the  formulae  of  this  chapter 
are  based,  and  to  acquire  familiarity  with  the  quantities  involved, 
it  will  be  advantageous  to  study  the  phenomena  of  the  appara 
tus  represented  in  Fig.  532.  A  large  tank  or  reservoir  _BCi& 
connected  with  another,  DE,  at  a  lower  level,  by  means  of  a 
rigid  pipe  opening  under  the  water-level  in  each  tank.  This 

646 


STEADY   FLOW    OF   A    LIQUID. 


647 


pipe  has  no  sharp  curves  or  bends,  is  of  various  sectional  areas 
at  different  parts,  the  changes  of  section  being  very  gradual, 
and  the  highest  point  N^  not  being  more  than  30  ft.  higher 
than  JSOj  the  surface-level  of  the  upper  tank.  Let  both  tanks 


FIG.  532. 

be  filled  with  water  (or  other  liquid),  which  will  also  rise  to  If 
and  to  ./Tin  the  pipe.  Stop  the  ends  L  and  N^  of  the  pipe, 
and  through  M,  a  stop-cock  in  the  highest  curve,  pour  in  water 
to  fill  the  remainder  of  the  pipe ;  then,  closing  M,  unstop  L 
and  NI  . 

If  the  dimensions  are  not  extreme  (and  subsequent  formulas 
will  furnish  the  means  of  testing  such  points)  the  water  will 
now  begin  to  flow  from  the  upper  tank  into  the  lower,  and 
all  parts  of  the  pipe  will  continue  full  of  water  as  the  flow 
goes  on. 

Further,  suppose  the  upper  tank  so  large  that  its  surface- 
level  sinks  very  slowly'  or  that  an  influx  at  A  continually 
makes  good  the  efflux  at  E\  then  the  flow  is  said  to  be  a  Steady 
Flow  ;  or,  a  state  of  permanency  is  said  to  exist ;  i.e.,  the  cir- 
cumstances of  the  flow  at  each  section  of  the  pipe  are  per- 
manent, or  steady. 


648  MECHANICS   OF   ENGINEERING. 

By  measuring  the  volume,  V,  of  water  discharged  at  E  in  a 
time  t,  we  obtain  the  volume  of  flow  per  unit  of  time,  viz., 

«  =  T»  ........    (1) 

while  the  weight  of  flow  per  unit  of  time  is 


(2) 


where  y  =  heaviness  (§  7)  of  the  liquid  concerned. 

Water  being  incompressible  and  the  pipe  rigid,  it  follows 
that  the  same  volume  of  water  per  unit  of  time  must  be  pass- 
ing at  each  cross-section  of  the  pipe.  But  this  is  equal  to  the 
volume  of  a  prism  of  water  having  F,  the  area  of  the  section, 
as  a  base,  and,  as  an  altitude,  the  mean  velocity  =  v  with  which 
the  liquid  particles  pass  through  the  section.  Hence  for  any 
section  we  have 

,    .    (3) 


in  which  the  subscripts  refer  to  different  sections.  If  the  flow 
were  unsteady,  e.g.,  if  the  level  BC  were  sinking,  this  would 
be  true  for  a  definite  instant  of  time  ;  but  when  steady,  we 
see  that  it  is  permanently  true;  e.g.,  Flvl  at  any  instant  =  JF[vt 
at  the  same  or  any  other  instant,  subsequent  or  previous.  In 
other  words,  in  a  steady  flow  tJie  velocity  at  a  given  section 
remains  unchanged  with  lapse  of  time.* 

[K.B.  We  here  assume  for  simplicity  that  the  different 
particles  of  water  passing  simultaneously  through  a  given  sec- 
tion (i.e.,  abreast  of  each  other)  have  equal  velocities,  viz.,  the 
velocity  which  all  other  particles  will  assume  on  reaching  this 
section.  Strictly,  however,  the  particles  at  the  sides  are  some- 
what retarded  by  friction  on  the  surface  of  the  pipe.  This  as- 
sumption is  called  the  Assumption  of  Parallel  Flow,  or  Flow 
in  Plane  Layers,  or  Laminated  Flow.~\ 

Let  us  suppose  Q  to  have  been  found  as  already  prescribed. 
We  may  then,  knowing  the  internal  sectional  areas  at  different 
parts  of  the  pipe,  N^  N^,  etc.,  compute  the  velocities 

*  The  flow  of  water  in  the  drive-pipe  of  a  hydraulic  ram  is  a  familiar  instance 
of  an  unsteady  flow.  The  water  in  this  pipe  is  permitted  to  flow  with  an 
accelerated  motion  for  a  short  time  and  then  suddenly  brought  to  rest}  this 
operation  being  repeated  indefinitely. 


STEADY   FLOW   OF   LIQUIDS.  649 

v*=Q  +  Fi,    v*=Q  +  F;    etc., 

which  the  water  must  have  in  passing  those  sections,  respec- 
tively. It  is  thus  seen  that  the  velocity  at  any  section  has  no 
direct  connection  with  the  height  or  depth  of  the  section  from 
the  plane,  BC,  of  the  upper  reservoir  surface.  The  fraction 

-  will  be  called  the  height  due  to  the  velocity  >  v,  or  simply 

t/ 
the  velocity  -head,  for  convenience. 

Next,  as  to  the  value  of  the  internal  fluid  pressure,  p,  per 
unit-area  (in  the  water  itself  and  against  the  side  or  wall  of 
pipe)  at  different  sections  of  the  pipe.  If  the  end  N^  of  the 
pipe  were  stopped,  the  problem  would  be  one  in  Hydrostatics, 
and  the  pressure  against  the  side  of  the  pipe  at  Nl  (also  at  Nt 
on  same  level)  would  be  simply 


measured  by  a  water  column  of  height 


in  which  pa  =  one  atmosphere,  and  b  =  34  ft.  =  height  of  an 
ideal  water  barometer,  and  y  =  62.5  Ibs.  per  cubic  foot  ;  and 
this  would  be  shown  experimentally  by  screwing  into  the  side 
of  the  pipe  at  Nl  a  small  tube  open  at  both  ends  ;  the  water 
would  rise  in  it  to  the  level  BC.  That  is,  a  column  of  water 
of  height  =  Aj  would  be  sustained  in  it,  which  indicates  that 
the  internal  pressure  at  N^  corresponds  to  an  ideal  water  col' 
umn  of  a  height 


But  when  a  steady  flow  is  proceeding,  the  case  being  now  one 
of  Hydrodynamics,  we  find  the  column  of  water  sustained  at 
rest  in  the  small  tube  (called  an  open  piezometer)  N^S  has  a 
height  yl  ,  less  than  A,  ,  and  hence  the  internal  fluid  pressure  is 


650  MECHANICS    OF   ENGINEERING. 

less  than  it  was  when  there  was  no  flow.     This  pressure  being 
called  j9j,  the  ideal  water  column  measuring  it  has  a  height 


~  = 


at  Nl  ,  and  will  be  called  the  pressure-head  at  the  section  re- 
ferred to.  "We  also  find  experimentally  that  while  the  flow  is 
steady  the  piezometer-height  y  (and  therefore  the  pressure- 
head  t>  +  y)  at  any  section  remains  unchanged  with  lapse  of 
time,  as  a  characteristic  of  a  steady  flow. 

[For  correct  indications,  the  extremity  of  the  piezometer 
should  have  its  edges  flush  with  the  inner  face  of  the  pipe 
wall,  where  it  is  inserted.] 

At  N^  although  at  the  same  level  as  JV,  ,  we  find,  on  in- 
serting an  open  piezometer,  TF,  that  with  F3  =  F^  (and  there- 
fore with  v3  =  vt)  2/3  is  a  little  less  than  yl  ;  while  if  Ft  <  F^ 
(so  that  v9  >  0,),  2/3  is  not  only  less  than  yl  ,  but  the  dif- 
ference is  greater  than  before.  We  have  therefore  found 
experimentally  that,  in  a  general  way,  when  water  is  flowing 
in  a  pipe  it  presses  less  against  the  side  of  the  pipe  than  it  did 
before  the  flow  was  permitted,  or  (what  amounts  to  the  same 
thing)  the  pressure  between  the  transverse  laminae  is  less  than 
the  hydrostatic  pressure  would  be. 

In  the  portion  HN^O  of  the  pipe  we  find  the  pressure  less 
than  one  atmosphere,  and  consequently  a  manometer  register- 
ing pressures  from  zero  upward  (and  not  simply  the  excess 
over  one  atmosphere,  like  the  Bourdon  steam-gauge  and  the 
open  piezometer  just  mentioned)  must  be  employed.  At  N~%  , 
e.g.,  we  find  the  pressure 

=  i  atmos.,  i.e.,  &  =  17  ft. 

Even  below  the  level  BC,  by  making  the  sections  quite  nar- 
row (and  consequently  the  velocities  great)  the  pressure  may  be 
made  less  than  one  atmosphere.  At  the  surface  .Z?6Ythe  pres- 
sure is  of  course  just  one  atmosphere,  while  that  in  the  jet  at 
N^  entering  the  right-hand  tank  under  water,  is  necessarily 

pt  =  1  atmos.  -f-  press,  due  to  col.  h'  of  water  practically  at  rest; 


STEADY   FLOW   OF   LIQUIDS.  651 

i.e.,  —  =  pressure-head  at  N^  =  b  -f-  A'; 

(whereas  if  N<  were  stopped  by  a  diaphragm,  the  pressure- 
head  just  on  the  right  of  the  diaphragm  would  be  ~b  -f-  A',  and 
that  on  the  left  ft  +  A4.) 

Similarly,  when  a  jet  enters  the  atmosphere  in  parallel  fila- 
ments its  particles  are  under  a  pressure  of  one  atmosphere,  i.e., 
their  pressure-head  =  b  =  34  ft.  (for  water)  ;  for  the  air  im- 
mediately around  the  jet  may  be  considered  as  a  pipe  between 
which  and  the  water  is  exerted  a  pressure  of  one  atmosphere. 

491.  Recapitulation  and  Examples.  —  We  have  found  experi- 
mentally, then,  that  in  a  steady  flow  of  liquid  through  a  rigid 
pipe  there  is  at  each  section  of  the  pipe  a  definite  velocity  and 
pressure  which  all  the  liquid  particles  assume  on  reaching  that 
section  ;  in  other  words,  at  each  section  of  the  pipe  the  liquid 
velocity  and  pressure  remain  constant  with  progress  of  time. 

EXAMPLE  1.  —  If  in  Fig.  532,  the  flow  having  become  steady, 
the  volume  of  water  flowing  in  3  minutes  is  found  on  meas- 
urement to  be  134  cub.  feet,  the  volume  per  second  is,  from 
eq.  (1),  §  490, 

Q  =  |||.  —  0.744  cub.  ft.  per  second. 


EXAMPLE  2.  —  If  the  flow  in  2  min.  20  sec.  is  386.4  Ibs.,  the 
volume  of  flow  per  second  is  [ft.,  lb.,  sec.  ;  eqs.  (1)  and  (2)] 

V       G  386.4      1 

Q  =  "  V  *  *'-''  "62T  '  140  = 

EXAMPLE  3.  —  In  Fig.  532  the  height  of  the  open  piezometer 
at  Nl  is  yl  =  9  feet  ;  what  is  the  internal  fluid  pressure  ? 
[Use  the  inch,  lb.,  and  sec.]  The  internal  pressure  is 

pl  =pa  +  yy  =  14.7  +  108  X  •£#&  =  18.6  Ibs.  per  sq.  inch. 

The  pressure  on  the  outside  of  the  pipe  is,  of  course,  one  at- 
mosphere, so  that  the  resultant  bursting  pressure  at  that  point 

is  3.9  Ibs.  per  sq.  in. 
EXAMPLE  4.  —  The  volume  of  flow  per  second  being  .0441 


662  MECHANICS   OF   ENGINEERING. 

cub.  ft.  per  sec.,  as  in  Example  1,  required  the  velocity  at  a 
section  of  the  (circular)  pipe  where  the  diameter  is  2  inches. 
[Use  ft.,  lb.,  and  sec.] 

O       0.0441       0  A0  , 


while  at  another  section  of  the  pipe  where  the  diameter  is  four 
inches  (double  the  former)  and  the  sectional  area,  F,  is  there- 
fore four  times  as  great,  the  velocity  is  £  of  2.02  =  0.505  ft. 
per  sec. 

492.  Bernoulli's  Theorem  for  Steady  Flow;  without  Friction.  — 
If  the  pipe  is  comparatively  short,  without  sudden  bends, 
elbows,  or  abrupt  changes  of  cross-section,  the  effect  of  friction 
of  the  liquid  particles  against  the  sides  of  the  pipe  and  against 
each  other  (as  when  eddies  are  produced,  disturbing  the  paral- 
lelism of  flow)  is  small,  and  will  be  neglected  in  the  present 
analysis,  whose  chief  object  is  to  establish  a  formula  for  steady 
flow  through  a  short  pipe  and  through  orifices. 

An  assumption,  now  to  be  made,  (A  flow  in  plane  layers,  or 
laminated  flow,  i.e.,  flow  in  laminae  "1  to  the  axis  of  the  pipe 
at  every  point,  may  be  thus  stated  :  (see  Fig.  533,  which  shows 
a  steady  flow  proceeding,  through  a 
pipe  CD  of  indefinite  extent.)  All  the 
liquid  particles  which  at  any  instant 
form  a  small  lamina,  or  sheet,  as  AB, 
~|  to  axis  of  pipe,  ~keep  company  as  a 
lamina  throughout  the  whole  flow. 
The  thickness,  ds',  of  this  lamina  re- 
mains constant  so  long  as  the  pipe  is  of  constant  cross-section, 
but  shortens  up  (as  at  C)  on  passing  through  a  larger  section, 
and  lengthens  out  (as  at  D)  in  a  part  of  the  pipe  where  the 
section  is  smaller  (i.e.,  the  sectional  area,  F,  is  smaller).  The 
mass  of  such  a  lamina  is  Fds'y  -f-  g  [§  55],  its  velocity  at  any 
section  will  be  called  v  (pertaining  to  that  point  of  the  pipe's 
axis),  the  pressure  of  the  lamina  just  behind  it  is  Fp,  upon  the 
rear  face,  while  the  resistance  (at  the  same  instant)  offered  by 
its  neighbor  just  ahead  is  F(  p  -f-  dp)  on  the  front  face  ;  also 


BERNOULLI'S  THEOREM. 


653 


FIG.  534. 


its  weight  is  the  vertical  force  Fds'y.     Fig.  534  shows,  as  a 
free  body,  the  lamina  which  at 
any  instant  is  passing  a  point 
A  of  the  pipe's  axis,  where  the 
velocity  is  v  and  pressure  p. 

Note  well  the  forces  acting ; 
the  pressures  of  the  pipe  wall 
on  the  edges  of  the  lamina  have 
no  components  in  the  direction 
of  v,  for  the  wall  is  considered 
smooth,  i.e.,  those  pressures  are 
1  to  wall;  in  other  words,  no 
friction  is  considered.  To  this  free  body  apply  eq.  (7)  of  §74, 
for  any  instant  of  any  curvilinear  motion  of  a  material  point 

vdv  =  (tang,  acceleration)  X  ds,  .    .     .     .     (1) 

in  which  ds  =  a  small  portion  of  the  path,  and  is  described  in 
the  time  dt.  Now  the  tang,  accel.  =  ^(tang.  compons.  of  the 
acting  forces)  ~-  mass  of  lamina,  i.e., 

Fp  —  F(p  4-  dp)  +  Fyds'  cos  0 

tana.  ace.  =  — T      ,     — -•     •     (2) 

Fyds'  -±  g 

Now,  Fig.  535,  at  a  definite  instant  of  time,  conceive  the 
volume  of  water  in  the  pipe  to  be  subdivided  into  a  great 
number  of  laminae  of  equal  mass  (which  implies  equal  volumes 


in  the  case  of  a  liquid,  but  not  with  gaseous  fluids),  and  let  the 
ds  just  mentioned  for  any  one  lamina  be  the  distance  from  its 
centre  to  that  of  the  one  next  ahead  ;  this  mode  of  subdivision 


654  MECHANICS   OF   ENGINEERING. 

makes  the  ds  of  any  one  lamina  identical  in  value  with  its 
thickness  ds',  i.e., 

ds  =  ds'  ........     (3) 

We  have  also 

ds  cos  0  =  —  dz,     or  ds'  cos  0  =  —  dz  ;  .     .     (4) 

z  being  the  height  of  the  centre  of  a  lamina  above  any  con- 
venient horizontal  datum  plane.  Substituting  from  (2),  (3), 
and  (4)  in  (1),  we  derive  finally 

-  vdv+  ^-dp  +  dz  =  Q.  .  (5) 

g          r  ' 

The  flow  being  steady,  and  the  subdivision  into  laminae 
being  of  the  nature  just  stated,  each  lamina  in  some  small  time 
dt  moves  into  the  position  which  at  the  beginning  of  dt  was 
filled  by  the  lamina  next  ahead,  and  acquires  the  same  velocity, 
the  same  pressures  on  its  faces,  and  the  same  value  of  z,  that 
the  front  lamina  had  at  the  beginning  of  dt. 

Hence,  considering  the  simultaneous  advance  made  by  all 
the  laminae  in  this  same  dt,  we  may  write  out  an  equation  like 
(5)  for  each  of  the  laminae  between  any  two  cross-sections  n  and 
m  of  the  pipe,  thus  obtaining  an  infinite  number  of  equations, 
from  which  by  adding  corresponding  terms,  i.e.,  ty  integra- 
tion, we  obtain 


whence,  performing  the  integrations  and  transposing, 

Vm     \Pm\         _  v*    ,   Pn  ,  ,  (  BernoulW  s  \  ,~ 

~fy~~~f~~*m-fy~~~^~~*n'     '(    Theorem    }'     '"W 

Denoting  by  Potential  Head  the  vertical  height  of  any  section 
of  the  pipe  above  a  convenient  datum  level,  we  may  state 
Bernoulli's  Theorem  as  follows  : 

In  steady  flow  without  friction,  the  sum  of  the  velocity- 
head,  pressure-head,  and  potential  head  at  any  section  of  the 
pipe  is  a  constant  quantity,  being  equal  to  the  sum  of  the  cor- 
responding heads  at  any  other  section. 


APPLICATIONS   OF   BERNOULLI'S   THEOREM. 


It  is  noticeable  that  in  eq.  (7)  each  of  the  terms  is  a  lineal 
quantity,  viz.,  a  height,  or  head,  either  actual,  such  as  zn  and 
zm ,  or  ideal  (all  the  others),  and  does  not  bring  into  account  the 
absolute  size  of  the  pipe,  nor  even  its  relative  dimensions  (vm 
and  vn,  however,  are  connected  by  the  equation  of  continuity 
Fmvm  —  Fnvn\  and  contains  no  reference  to  the  volume  of 
water  flowing  per  unit  of  time  [()]  or  the  shape  of.  the  pipe's 
axis.  When  the  pipe  is  of  considerable  length  compared  with 
its  diameter  the  friction  of  the  water  on  the  sides  of  the  pipe 
cannot  be  neglected  (§  512). 

It  must  be  remembered  that  Bernoulli's  Theorem  does  not 
hold  unless  the  flow  is  steady,  i.e.,  unless  each  lamina,  in  com- 
ing into  the  position  just  vacated  by  the  one  next  ahead  (of 
equal  mass),  comes  also  into  the  exact  conditions  of  velocity 
and  pressure  in  which  the  other  was  when  in  that  position. 

[N.B.  This  theorem  can  also  be  proved  by  applying  to  all 
the  water  particles  between  n  and  m,  as  a  collection  of  small 
rigid  bodies  (water  being  incompressible)  the  theorem  of  Work 
and  Energy  for  a  collection  of  Rigid  Bodies  in  §  142,  eq.  (xvi), 
taking  the  respective  paths  which  they  describe  simultaneously 
in  a  single  dt.~\ 

493.  First  Application  of  Bernoulli's  Theorem  without  Friction. 
— Fig.  536  shows  a  large  tank  from  which  a  vertical  pipe  of 
uniform  section  leads  to  another  tank  and  dips  below  the  sur- 
face of  the  water  in  the  latter.  Both  surfaces  are  open  to  the 
air.  The  vessels  and  pipe  being  filled  with 
water,  and  the  lower  end  m  of  the  pipe  un-  •.••'>«:"•:•. 

stopped,  a  steady  flow  is  established  almost  *~»" 
immediately,  the  surface  BO  being   very 
large  compared  with  F,  the  area  of  the  (uni- 
form) section  of  the  pipe. 

Given  F,  and  the  heights  A0  and  A,  re- 
quired the  velocity  vm  of  the  jet  at  m  and 
also  the  pressure,  pnl  at  n  (in  pipe  near  en- 
trance of  same),  m  is  in  the  jet,  just  clear 
of  the  pipe,  and  practically  in  the  water- 
level,  AD.  The  velocity  vm  is  unknown,  FIG.  535. 
but  the  pressure  pm  is  practically  =pa=  one  atmosphere,  since 


656 


MECHANICS   OF  ENGINEERING. 


the  pressure  on  the  sides  of  the  jet  is  necessarily  the  hydro- 
static pressure  due  to  a  slight  depth  below  the  surface  AD. 


.-.  Press.-head  at  m  is 


=     *  =  I  =  34  feet.     .     .     (§  423) 


Now  apply  Bernoulli's  Theorem  to  sections  m  and  n,  taking 
a  horizontal  plane  through  m  as  a  datum  plane  for  potential 
heads,  so  that  zn  =  h  and  zm  =  0,  and  we  have 


But,  assuming  that  the  section  of  the  pipe  is  filled  at  every 
point,  we  must  have 

tW=*«? 

for,  in  the  equation  of  continuity 


if  we  put  Fm  =  Fn ,  the  pipe  being  of  uniform  section,  we  ob- 
tain vm  =  vn .    Hence  eq.  (1)  reduces  to 


=  I  -  h  =  34  f t.  -  h. 


(2) 


Hence  the  pressure  at  n  is  less  than  one  atmosphere,  and  if  a 
small  tube  communicating  with  an  air-tight  receiver  full  of  air 
were  screwed  into  a  small  hole  at  n,  the  air  in 
the  receiver  would  gradually  be  drawn  off  until 
its  tension  had  fallen  to  a  value  pn.  [This  is  the 
principle  of  SprengePs  air-pump,  mercury,  how- 
ever, being  used  instead  of  water,  as  for  this 
heavy  liquid  b  =  only  30  inches.] 

If  h  is  made  >  l>  for  water,  i.e.  >  34  feet  (or 
>  30  inches  for  mercury),  pn  would  be  negative 
from  eq.  (2),  which  is  impossible,  showing  that, 
the  assumption  of  full  pipe-sections  is  not  borne 
out.  In  this  case,  h  >  5,  only  a  portion,  mn\ 
(in  length  somewhat  less  than  &,)  of  the  tube  will  be  kept  full 


%j& 

Fia.  537. 


APPLICATIONS   OF   BERNOULLI'S   THEOREM.  657 

during  the  flow  (Fig.  537);  while  in  the  part  Kn'  vapor  of 
water,  of  low  tension  corresponding  to  the  temperature 
(§  469),  will  surround  an  internal  jet  which,  does  not  fill  tne 
pipe.  As  for  the  value  of  vm,  Bernoulli's  Theorem,  applied 
to  BC  and  m,  in  Fig.  536,  gives  finally  vm  =  Vtyh,  . 

EXAMPLE.  —  If  h  =  20  feet,  Fig.  536,  and  the  liquid  is  water, 
the  pressure-head  at  n  is  (ft.,  lb.,  sec.) 

£»  =  b  -  h  =  34'  -  20'  =  14  ft., 

r 

and  therefore 
pn  —  14  X  62.5  =  875  Ibs.  per  sq.  ft.  =  6.07  Ibs.  per  sq.  in. 

494.  Second  Application  of  Bernoulli's  Theorem  without  Fric- 
tion. —  Knowing  by  actual  measurement  the  open  piezometer 
height  yn  at  the  section  n  in  ..-AIR  .  .  . 
Fig.  538  (so  that  the  pressure- 


—  =  o  -f-  yn  i  ^  n  is 

Y 
known) ;    knowing  also   the 

vertical  distance  hn  from  n 
to  m,  and  the  respective 
cross- sections  Fn  and  Fm  (Fm  being  the  sectional  area  of  the 

jet,  flowing  into  the  air,  so  that  *22  =  5),  required  the  volume 
of  flow  per  sec.;  i.e.,  required  Q,  which 


(1) 


The  pipe  is  short,  with  smooth  curves,  if  any,  and  friction 
will  therefore  be  neglected.  From  Bernoulli's  Theorem  [eq. 
(7),  §  492],  taking  m  as  a  datum  plane  for  potential  heads,  we 
have 


But  from  (1)  we  have 


658  MECHANICS    OF   ENGINEERING. 


substituting  which  in  (2)  we  obtain,  solving  for 

(8) 


and  hence  the  volume  per  unit  of  time  becomes  known,  viz., 

Q  =  f.vm (4) 

NOTE. — If  the  cross-section  Fm  of  the  nozzle,  or  jet,  is  >  Fn , 
<vm  becomes  imaginary  (unless  yn  is  negative  (i.e.,^?n  <  one  at- 
mos.),  and  numerically  >  hn] ;  in  other  words,  the  assigned 
cross-sections  are  not  filled  by  the  flow. 

EXAMPLE. — If  yn  =  17  ft.  (thus  showing  the  internal  fluid 
pressure  at  n  to  be  pn  =  y(yn  -\-  b)  =  1-J  atmos.),  hn  =  10  ft., 
and  the  (round)  pipe  is  4  inches  in  diameter  at  n  and  3  inches 
at  the  nozzle  m,  we  have  from  (3)  (using  ft.-lb.-sec.  system  of 
units  in  which  g  =  32.2) 


=     4/2x32-2(17-  -^  =  50<4  f  t 


pT.B.  Since  Fm  -f-  ^  is  a  ratio  and  therefore  an  abstract 
number,  the  use  of  the  inch  in  the  ratio  will  give  the  same 
result  as  that  of  the  foot.] 

Hence,  from  (4), 

Q  =  F^m  ^  i^(A)3  X  50.4  =  2.474  cub.  ft.  per  sec. 

495.  Orifict*  in  Thin  Plate,— Fig.  539.  When  efflux  takes 
place  through  an  orifice  in  a  thin  plate,  i.e.,  a  sharp-edged 
orifice  in  the  plane  wall  of  a  tank,  a  contracted  vein  (or  "  vena 


ORIFICE   IN   THIN   PLATE. 


659 


contracta")  is  formed,  the  filaments  of  water  not  becoming 

parallel    until    reaching    a    plane,  m, 

parallel  to  the   plane  of  vessel  wall, 

which  for  circular  orifices  is  at  a  dis- 

tance from  the  interior  plane  of  vessel 

wall  equal  to  the  radius  of  the  circular 

aperture  ;  and  not  until  reaching  this 

plane    does   the   internal   fluid   pres- 

sure become  equal  to  that  of  the  sur- 

rounding medium  (atmosphere,  here),  FlG-  539- 

i.e.,  surrounding  the  jet.     We   assume   that  the  width  of  the 

orifice   is   small   compared   with  h,  unless   the   vessel  wall  is 

horizontal. 

The  area  of  the  cross-section  of  the  jet  at  m,  called  the  con- 
tracted section,  is  found  on  measurement  to  be  from  .60  to  .64 
of  the  area  of  the  aperture  with  most  orifices  of  ordinary 
shapes,  even  with  widely  different  values  of  the  area  of  aper- 
ture and  of  the  height,  or  head,  A,  producing  the  flow.  Call- 
ing this  abstract  number  [.60  to  .64]  the  Coefficient  of  Con- 
traction, and  denoting  it  by  (7,  we  may  write 


in  which  F  =  area  of  the  orifice,  and  Fm  =  that  of  the  con- 
tracted section.  C  ranges  from  .60  to  .64  with  circular  orifices, 
but  may  have  lower  values  with  some  rectangular  forms.  (See 
table  in  §  503.) 

A  lamina  of  particles  of  water  is  under  atmospheric 
pressure  at  n  (the  free  surface  of  the  water  in  tank  or  reser- 
voir), while  its  velocity  at  n  is  practically  zero,  i.e.  vn  =  0 
(the  surface  at  B  being  very  large  compared  with  the  area  of 
orifice).  It  experiences  increasing  pressure  as  it  slowly  de- 
scends until  in  the  immediate  neighborhood  of  the  orifice, 
when  its  velocity  is  rapidly  accelerated  and  pressure  decreased, 
in  accordance  with  Bernoulli's  Theorem,  and  its  shape  length- 
ened out,  until  finally  at  m  it  forms  a  portion  of  a  filament  of 
a  jet,  its  pressure  is  one  atmosphere,  and  its  velocity,  =  vm, 
we  wish  to  determine.  The  course  of  this  lamina  we  call  a 


660 


MECHANICS    OF    ET^GINEEKING. 


"stream-line?  and  Bernoulli's  Theorem  is  applicable  to  it, 
just  as  ii  it  were  enclosed  in  a  frictionless  pipe  of  the  same 
form.  Taking  then  a  datum  plane  through  the  centre  of  m, 
we  have 


while 


=  0,     and    vm  =  \\ 


£*  also  =  b,    zn  —  h,    and    vn  =  0. 


Y 


Hence  Bernoulli's  Theorem  gives 


(1) 


and 


That  is,  the  velocity  of  the  jet  at  m  is  theoretically  the  same  as 
that  acquired  by  a  body  falling  freely  in  vacuo  through  a 
height  =h=  the  "  head  of  water."  We  should  therefore  ex- 
pect that  if  the  jet  were  directly  ver- 
tically upward,  as  at  m,  Fig.  540, 

a  height  -^- 

would  be  actually  attained.  [See 
§§  52  and  53.]  Experiment  shows 
that  the  height  of  the  jet  (at  m) 
does  not  materially  differ  from  h  if 
h  is  not  >  6  or  8  feet.  For  h  >  8  ft.,  however,  the  actual  height 
reached  is  <  A,  the  difference  being  not  only  absolutely  but 
relatively  greater  as  h  is  taken  greater,  since  the  resistance  of 
the  air  is  then  more  and  more  effective  in  depressing  and 
breaking  up  the  stream.  (See  §  578.) 

At  m',  Fig.  540,  we  have  a  jet,  under  a  head  =  h ',  directed 


FIG.  540. 


ORIFICE   IN   THIN   PLATE.  661 

at  an  angle  a0  with  the  horizontal.  Its  form  is  a  parabola 
(§  81),  and  the  theoretical  height  reached  is  Ji"  =  h!  sina  a9 
(§  80). 

The  jet  from  an  orifice  in  thin  plate  is  very  limpid  and  clear. 

From  eq.  (1),  we  have  theoretically 

vm=  V'Zgli 

(an  equation  we  shall  always  use  for  efflux  into  the  air  through 
orifices  and  short  pipes  in  the  plane  wall  of  a  large  tank  whose 
water-surface  is  very  large  compared  with  the  orifice,  and  is. 
open  to  the  air),  but  experiment  shows  that  for  an  "  orifice  in 
thin  plate"  this  value  is  reduced  about  3$  by  friction  at  the 
edges,  so  that  for  ordinary  practical  purposes  we  may  write 


vm  =  $  V  fyh  =  0.97  V,     ....     (2) 

in  which  0  is  called  the  coefficient  of  velocity. 

Hence  the  volume  of  flow,  Q,  per  time-unit  will  be 


Q  =  Fmvm  =  CF(f>  Vtofa   on  the  average  =  0.62FV  2gh.   (3) 

It  is  to  be  understood  that  the  flow  is  steady,  and  that  the 
reservoir  surface  (very  large)  and  the  jet  are  both  under  at- 
mospheric pressure.  </>C is  called  the  coefficient  of  'efflux. 

EXAMPLE  1. — Fig.  539.  Eequired  the  velocity  of  efflux, 
^m  ,  at  m,  and  the  volume  of  the  flow  per  second,  Q,  into  the 
air,  if  h  =  21  ft.  6  inches,  the  circular  orifice  being  2  in.  in 
diam. ;  take  0=  0.64.  [Ft.,  lb.,  and  sec.] 

From  eq.  (2), 


vm  =  0.97  t/2  X  32.3  X  21.5  =  36.1  ft.  per  sec.  ; 
hence  the  discharge  is 

Q  =  Fmvm  =  0.64  X  2X  36.1  =  0.504  cub.  ft.  per  second. 


EXAMPLE  2.  —  [Weisbach.]     Under  a  head  of  3.396  metres 
the  velocity  vm  in  the  contracted  section  is  found  by  measure- 


662  MECHANICS    OF   ENGINEERING. 

ments  of  the  jet-curve  to  be  7.98  metres  per  sec;,  and  the  dis- 
charge proves  to  be  0.01825  cub.  metres  per  sec.  Required 
the  coefficient  of  velocity  (0)  and  that  of  contraction  (C\  if 
the  area  of  the  orifice  is  36.3  sq.  centimetres. 

Use  the  metre-kilogram-second  system  of  units,  in  which 
g  =  9.81  met.  per  sq.  second. 

From  eq.  (2), 


V  2gh       V2  X  9.81  X  3.396 
while  from  (3)  we  have 

-01825 


F<f>  V~fyA      Fvn      Tttft.  X  7.98 


_ 

~ 


0  and  (7,  being  abstract  numbers,  are  independent  of  the  sys- 
tem of  concrete  units  adopted. 

NOTE.  —  To  fir.d  the  velocity  vm  of  the  jet  at  the  orifice  by 
measurements  of  the  jet-curve,  as  mentioned  in  Example  2, 
we  may  proceed  as  follows  :  Since  we  cannot  very  readily  as- 
sure ourselves  that  the  direction  of  the  jet  at  the  orifice  is 
horizontal,  we  consider  the  angle  aQ  of  the  parabola  (see  Fig. 
93  and  §  80)  as  unknown,  and  therefore  have  two  unknowns 
to  deal  with,  and  obtain  the  necessary  two  equations  by  meas- 
uring the  a?  and  y  (see  page  84)  of  two  points  of  the  jet,  re- 
membering that  if  we  use  the  equation  (3)  of  page  84  in  its 
present  form  points  of  the  jet  below  the  orifice  will  have  nega- 
tive y's.  The  substitution  of  these  values  ^  ,  »„  ,  y,  ,  and  yz 
in  equation  (3)  furnishes  two  equations  between  constants,  in 
which  only  a0  and  h  are  unknown.  To  eliminate  a9  ,  for 

—  -  —  we  write  1  -f-  tan2  a0  ,  and  taking  a?a  =  2^  for  con*- 

COS    ^-Q 

venience,  we  finally  obtain 


**.«.•+ fly.-y.y,  md ...         /EEWEEZI, 
y.  -  22/.  V     %.  -  ayj 

in  which  yt  and  y2  are  the  vertical  distances  of  the  two  points 


ROUNDED    ORIFICE.  663 

chosen  below  the  orifice  ;  that  is,  we  have  already  made  them 
negative  in  eq.  (3)  of  page  84.  The  h  of  the  preceding  equa- 
tion simply  denotes  vm*  ~  2</,  and  must  not  be  confused  with 
that  of  the  last  two  figures.  For  accuracy  the  second  point 
should  be  as  far  from  the  orifice  along  the  jet  as  possible. 

496.  Orifice  with  Rounded  Approach.*—  -Fig.  541  shows  the 
general  form  and  proportions  of  an  orifice  or  mouth-piece  in 
the  use  of  which  contraction  does  not 
take  place  beyond  the  edges,  the  inner 
surface  being  one  uof  revolution,"  and 
so  shaped  that  the  liquid  filaments  are 

n   i  ^  J 

parallel  on  passing  the  outer  edge  m\ 

hence  the   pressure-head   at   m  is  =  b 

(=  34  ft.  for  water  and  30  inches  for 

mercury)  in    Bernoulli's    Theorem,   if 

efilux  takes   place  into  the    air.     "We  FlG-  541- 

have  also  the  sectional  area  Fm  =  F=  that  of  final  edge  of 

orifice,  i.e.,  the  coefficient  of  contraction,  or  (7,  =  unity  =  1.00, 

so  that  the  discharge  per  time-unit  has  a  volume 


The  tank  being  large,  as  in  Fig.  540,  Bernoulli's  Theorem 
applied  to  m  and  n  will  give,  as  before, 


vm= 
as  a  theoretical  result,  while  practically  we  write 

vm=0V^/i,      .......    (1) 

and  Q  =  F<j>\/~Zgh  ........    (2) 

As  an  average  0  is  found  to  differ  little  from  0.97  with  this 
orifice,  the  same  value  as  for  an  orifice  in  thin  plate  (§  495). 

497.  Problems  in  Efflux  Solved  by  Applying  Bernoulli's 
Theorem.  —  In  the  two  preceding  paragraphs  the  pressure- 
heads  at  sections  m  and  n  were  each  =  pa  ~  y  =  height  of 

*  Smooth  conical  nozzles  for  fire-steams  give  0  =  .  97  with,  h  =  press.- 
heiid  +  veloc.-head  at  base  of  play-pipe  ;  see  p.  833. 


664  MECHANICS   OF   ENGINEEKING. 

the  liquid  barometer  =  b  ;  but  in  the  following  problems  this 
will  not  be  the  case  necessarily.  However,  efflux  is  to  take 
place  through  a  simple  orifice  in  the  side  of  a  large  reservoir, 
whose  upper  surface  (n)  is  very  large,  so  that  vn  may  be  put 
:=  zero. 

Problem  I.  —  Fig.  542.  "What  is  the  velocity  of  efflux,  vm,  at 
the  orifice  m  (i.e.,  at  the  contracted  sec- 
tion, if  it  is  an  orifice  in  thin  plate) 
of  a  jet  of  water  from  a  steam-boiler,  if 
the  free  surface  at  n  is  at  a  height  =  h 
above  m,  and  the  pressure  of  the  steam 
over  the  water  is  pn  ,  the  discharge  tak- 
ing place  into  the  air? 

Applying  Bernoulli's  Theorem  to  sec- 
tion  m  at  the  orifice  [where  the  pres- 
sure-head is  b  and  velocity-head  v^  -H  2^  (unknown)]  and  to 
section  n  at  water-surface  (where  velocity-head  =  0  and  pres- 
sure-head —  pn  -4-  y\  we  have,  taking  m  as  a  datum  for  poten- 
tial heads  so  that  zm  =  0  and  zn  =  A, 


.   „     .     .     .    (1) 


EXAMPLE.  —  Let  the  steam-gauge  read  40  Ibs.  (and  hence 
pn  =  54.7  Ibs.  per  sq.  inch)  and  h  =  2  ft.  4  in.  ;  required  vm. 

Also  if  F—  \  sq.  in.,  in  "  thin  plate,"  required  the  rate  of 
discharge  (volume).  The  temperature  of  saturated  steam  of 
the  given  tension  must  be  286°  Fahr.  [see  foot  of  page  607]. 
The  water  is  practically  at  the  same  temperature  and  hence 
of  a  heaviness,  y,  of  57.7  Ibs.  per  cubic  ft.  (p.  518). 

From  eq.  (1)  above,  then,  with  ft.  Ib.  and  sec.,  noting  that 
for  this  case  b  =  [(14.7  X  144)  -5-  57.7]  feet, 


i       v  -    x  '    x 

* 


=  81.1  ft.  per  sec.,  theoretically;   but*  practically 

*  Another  practical  matter  in  this  case  is  that  some  of  the  hot  water  will 
"  flash  "  into  steam  on  relief  from  the  higher  pressure. 


PROBLEMS   OF   EFFLUX  —  ORIFICES. 


665 


vm  =  0.97  X  81.1  =  78.6  ft.  per  sec.  ; 
so  that  the  discharge  begins  at  the  rate  of 
Q  =  0.64  Fvm  =  0.64  X  i  -  yJr  X  78.6  =  0.174  cub.  ft.  p.  sec. 


Problem  II. — Fig.  543.  With  what  velocity,  vm ,  will  water 
flow  into  the  condenser  C  of  a  steam-engine  where  the  tension 
of  the  vapor  is  pm,  <  one  atmosphere,  if 
h  =  the  head  of  water,  and  the  flow  takes 
place  through  an  orifice  in  thin  plate? 
Taking  position  m  in  the  contracted  section 
where  the  filaments  are  parallel,  and  the 
pressure  therefore  equal  to  that  of  the  sur- 
rounding vapor,  viz.,j?m,  and  position  n  in 
the  (wide)  free  surface  of  the  water  in  the 
tank,  where  (at  surface)  the  pressure  is  one  FIG.  543. 

atmosphere  [and  .-.  ^  =  b  =  34  ft.]  and  velocity  practically 

zero;  we  have,  applying  Bernoulli's  Theorem  to  n  and  m,  tak- 
ing m  as  a  datum  level  for  potential  heads  (so  that  zn  =  h  and 


-,       -      •      *      .      CD 


and 


as  theoretical  results.    But  practically  we  must  write 


(3) 


and 


in  which  F=  area  of  orifice  in  thin  plate,  and  O=  coefficient 
of  contraction  =  about  0.62  approximately  [see  §  495]. 


MECHANICS   OF   ENGINEERING. 


EXAMPLE.  —  If  in  the  condenser  there  is  a  "  vacuum"  of  27£ 
inches  (meaning  that  the  tension  of  the  vapor  would  support 
2J  inches  of  mercury,  in  a  barometer),  so  that 

pm  =  [$$  X  14.7]  Ibs.  per  sq.  inch,  and  h  =  12  feet, 

while  the  orifice  is  £  inch  in  diameter  ;  we  have,  using  the  ft., 
ib.,  and  sec;, 


=  51.1  ft.  per  sec. 
(We  might  also  have  written,  for  brevity, 


=  [2i  :  30]  X  34  =  2.833, 


since  the  pressure  head  for  one  atmos.  =  34  feet,  for  water. 
Hence,  for  a  circular  orifice  in  thin  plate,  we  have  the  volume 
discharged  per  unit  of  time, 

Q  =  CFv  =  0.62  X  f  (]!)' X  51'1  =  0'04:81  cuk  ft<  Per  sec' 

497a.  Efflux  through  an  Orifice  in  Terms  of  the  Internal  and 
External  Pressures. — Fig.  544.  Let  efflux  take  place  through 
a  small  orifice  from  the  plane  side  of  a  large  tank,  in  which  at 
the  level  of  the  orifice  the  hydrostatic  pressure  was  =  p'  be- 
fore the  opening  of  the  orifice,  that  of  the  medium  surround- 
ing the  jet  being  =p".  When  a  steady  flow 
is  established,  after  opening  the  orifice,  the 
pressure  in  the  water  on  a  level  with  the  ori- 
fice will  not  be  materially  changed,  except  in 
the  immediate  neighborhood  of  the  orifice  [see 
§  495] ;  hence,  applying  Bernoulli's  Theorem 
to  m  in  the  jet,  where  the  filaments  are  parallel, 
and  a  point  n,  in  the  body  of  the  liquid  and 
FIO.  544.  at  the  same  level  as  m,  and  where  the  particles 
are  practically  at  rest  [i.e.,  vn  =  0]  (hence  not  too  near  the 


FOECE-PUMP. 


667 


orifice),  we  shall  have,  cancelling  out  the  potential  heads  which 
are  equal, 


«-L     =-.. 

1g        y  '    2</ 


.  ...   (i) 


(In  Fig.  544  p'  would  be  equal  to  pa  +  hy.)    Eq.  (1)  is  con- 
veniently applied  to  the  jet  produced  by 
a  force-pump^  supposing,    for   simplicity, 
the  orifice  to  be  in  the  head  of  the  •pump- 
cylinder,  as  shown  in  Fig.  545.     Let  the  - 
thrust   (force)    exerted    along  the    piston- 
rod  be  =  P,  and   the   area  of    the  piston 
be  =  F' .     Then  the  intensity   of  internal 
pressure    produced   in    the    chamber  AB 
(when  the  piston  moves  uniformly)  is 


FIG.  545. 


while  the  external  pressure  in  the  air  around  the  jet  is  simply 
pa  (one  atmos.). 


.  .  (i)' 


(N.T3.  Of  course,  at  points  near  the  orifice  the  internal 
pressure  is  <  p'\  read  §  495.) 

EXAMPLE.  —Let  the  force,  or  thrust,  P,  [due  tv  steam-pres- 
sure on  a  piston  not  shown  in  figure,]  be  2000  Ibs.,  and  the 
diameter  of  pump-cylinder  be  d  =  9  inches,  the  liquid  being 
salt  water  (so  that  y  —  64  Ibs.  per  cubic  foot). 

Then 


=  0.442  sq.ft., 


And  [ft.,  lb.,  sec.] 


6t)8  MECHANICS    OF    ENGINEERING. 

*,  =  0.97^/2  X  32.2  x  ^^ _  =  65.4  ft.  per  se* 

If  the  orifice  is  well  rounded,  with  a  diameter  of  one  inch, 
the  volume  discharged  per  second  is 

Q  =  Fmvnt  =  Fvm  =  ^(-!-Y  X  65.4  =  0.353  cub.  ft.  per  sec. 


To  maintain  steadily  this  rate  of  discharge,  the  piston  must 
move  at  the  rate  [veloc.  =  v']  of 


vf  =^  Q  --  F'  =  .353  +    j- (—}     =  0.800  ft.  per  sec., 


and  the  force  P  must  exert  a  power  (§  130)  of 

Z  =  Pv'  =  2000  X  0.800  =  1600  ft.  Ibs.  per  sec. 
=  about  3  horse-power  (or  3  H.  P.). 

If  the  water  must  be  forced  from  the  cylinder  through  a 
pipe  or  hose  before  passing  out  of  a  nozzle  into  the  air,  the 
velocity  of  efflux  will  be  smaller,  on  account  of  "fluid  fric- 
tion" in  the  hose,  for  the  same  P\  such  a  problem  will  be 
treated  later  [§513].  Of  course,  in  a  pum ping-engine,  by  the 
use  of  several  pump-cylinders,  and  of  air-chambers,  a  practically 
steady  flow  is  kept  up,  notwithstanding  the  fact  that  the  mo- 
tion of  each  piston  is  not  uniform,  and  must  be  reversed  at  the 
end  of  each  stroke. 

498.  Influence  of  Density  on  the  Velocity  of  Efflux  in  the  Last 
Problem. — From  the  equation 


^ 


of  the  preceding  paragraph,  where  p"  is  the  external  pressure 
around  the  jet,  and  p'  the  internal  pressure  at  the  same  level 
as  the  orifice  but  well  back  of  it,  where  the  liquid  is  sensibly 


RELATION   OF   DENSITY  TO  VELOCITY  OF  EFFLUX.    669 


at  rest,  we  notice  that  for  the  same  difference  of  pressure 
|~j/ — p"~^  the  velocity  of  efflux  is  inversely  proportional  to  the 
square  root  of  the  heaviness  of  the  liquid.  Hence,  for  the 
same  (pf  —  p"\  mercury  would  flow  out  of  the  orifice  with  a 
velocity  only  0.272  of  that  of  water ;  for 


.5_       /  1          272 
£  ~~  V  13.5  ~  1000' 


Again,  assuming  that  the  equation  holds  good  for  the  flow  or 
gases  (as  it  does  approximately  when  p'  does  not  greatly  exceed 
f"\  e.g.,  by  6  or  8  per  cent),  the  velocity  of  efflux  of  atmospheric 
air,  when  at  a  heaviness  of  0.807  Ibs.  per  cub.  foot,  would  be 


times  as  great  as  for  water,  with  the  same  p'  —  p"  .    (See 
§  548,  etc.) 


499.  Efflux  under  Water.    Simple  Orifice.—  Fig.  546. 

and  A3  be  the  depths  of  the  (small)  ori- 
fice below  the  levels  of  the  "  head  "  and 
"  tail  "  waters  respectively.  Then,  using 
the  formula  of  §  49  70,  we  have  for  the 
pressure  at  n  (at  same  level  as  m,  the 
jet) 


FIG.  546. 


Let 


and  for  the  external  pressure,  around 
the  jet  at  m, 


whence,  theoretically, 


(1) 


where  h  =  difference  of  level  between  the  surfaces  of  the  two 

bodies  of  water. 


670 


MECHANICS   OF   ENGINEERING. 


Practically, 


=  0 


but  the  value  of  0  for  efflux  under  water  is  somewhat  uncer- 
tain ;  as  also  that  of  C9  the  coefficient  of  contraction.  "Weis- 
bach  says  that  //,  =  0(7,  is  -^  part  less  than  for  efflux  into  the 
air  ;  others,  that  there  is  no  difference  (Trautwine).  See  also 
p.  389  of  vol.  6,  Jour,  of  'Engin.  Associations,  where  it  is 
stated  that  with  a  circular  mouth-piece  of  0.37  in.  diarn.,  and 
of  "  nearly  the  form  of  the  vena  contracta"  /*  was  found  to  be 
.952  for  discharge  into  the  air,  and  .945  for  submerged  dis- 
charge. 

500.  Efflux  from  a  Small  Orifice  in  a  Vessel  in  Motion. 

CASE  I.  When  the  motion  is  a  vertical  translation  and  uni- 
formly accelerated.  —  Fig.  547.  Suppose  the  vessel  to  move  up- 
ward with  a  constant  acceleration  p. 
(See  §  49a.)  Taking  m  and  n  as  in  the 
two  preceding  paragraphs,  we  know  that 
pm  =p"  =  external  pressure  =  one  at- 

mos.  =pa  (and  .-.  —  =  II).    As  to  the 

internal  pressure  at  n  (same  level  as  m, 
but  well  back  of  oritice),  pn  ,  this  is  not 
equal  to  (o  +  A)/,  because  of  the  acceler- 
ated motion,  but  we  may  determine  it  by  considering  free  the 
vertical  column  or  prism  On  of  liquid,  of  cross-section  =  dF, 
the  vertical  forces  acting  on  which  are  padF,  downward  at  0, 
pndF  upward  at  n,  and  its  weight,  downward,  hdFy.  All 
other  pressures  are  horizontal.  For  a  vertical  upward  acceler- 
ation '=p9  the  algebraic  sum  of  the  vertical  components  of  all 
the  forces  must  =  mass  X  vert.  acceL, 


whence 


Putting  pn  and  pa  equal  to  the  p'  and  p"  ,  respectively,  of 
the  equation,  we  have 


FIG.  547. 


-     ......      (1) 


OKIFICE  IN   MOTION. 


671 


of  §497, 


(2) 


It  must  be  remembered  that  vm  is  the  velocity  of  the  jet  rel- 
atively to  the  orifice,  which  is  itself  in  motion  with  a  variable 
velocity.  The  absolute  velocity  wm  of  the  particles  of  the  jet 
is  found  by  the  construction  in  §  83,  being  represented  graph- 
ically by  the  diagonal  of  a  parallelogram  one  of  whose  sides  is 
vm  ,  and  the  other  the  velocity  c  with  which  the  orifice  itself  is 
moving  at  the  instant,  as  part  of  the  vessel.  The  jet  may 
make  any  angle  with  the  side  of  the  vessel. 

On  account  of  the  flow  the  internal  pressures  of  the  water 
against  the.  vessel  are  no  longer  balanced  horizontally,  and  the 
latter  will  swing  out  of  the  vertical  unless  properly  constrained. 

If  p  —  g  =  ace.  of  gravity,  vm  —  V  2  V  %gh.  If  p  is  nega- 
tive and  =  g,  vm  =  0  ;  i.e.,  there  is  no  flow,  but  both  the  vessel 
and  its  contents  fall  freely,  without  mutual  action. 

CASE  II.  When  the  liquid  and  the  vessel  both  have  a  uni- 
form rotary  motion  about  a  vertical  axis  with  an  angular  veloc- 
ity =  QD  (§  110).  Orifice  small,  so  that  we  may  consider  the 
liquid  inside  (except  near  the  orifice)  to 
be  in  relative  equilibrium.  Suppose  the 
jet  horizontal  at  m,  Fig.  548,  and  the 
radial  distance  of  the  orifice  from  the 
axis  to  be  =  x.  The  external  pressure 
pm  =  pa  ,  and  the  internal  [see  §  428, 

eqs.  (3)  and  (4)]  is 


FIG.  548. 

hence  the  velocity  of  the  jet,  relatively  to  the  orifice,  is  (from 
§  497,  since  pn  and  pm  correspond  to  the  p'  and  p"  of  that 
article), 


672  MECHANICS   OF   ENGINEEKING. 

~T^'; (3; 


in  which  w,  =  GOX,  =  the  (constant)  linear  velocity  of  the  ori- 
fice in  its  circular  path.  The  absolute  velocity  wm  of  the  par- 
tides  in  the  jet  close  to  the  orifice  is  the  diagonal  formed  on 
w  and  vm  (§  83).  Hence  by  properly  placing  the  orifice  in  the 
casing,  wm  may  be  made  small  or  large,  and  thus  the  kinetic 
energy  carried  away  in  the  effluent  water  be  regulated,  within 
certain  limits.  Equation  (3)  will  be  utilized  subsequently  in 
the  theory  of  Barker's  Mill. 

EXAMPLE.  —  Let  the  casing  make  100  revol.  per  min.  (whence 
&  =  [27rlOO  -T-  60]  radians  per  sec.),  &0  ^  12  feet,  and  x  =  2 
ft.  ;  then  (ft.,  lb.,  sec.) 


«,„  -  J 


l  X  32.2  X  12  +  =  34.8  ft.  per  see. 


(while,  if  the  casing  is  not  revolving,  vm  =  V%ghQ  =  only  27.8 
ft.  per  sec.). 

If  the  jet  is  now  directed  horizontally  and  backward,  and 
also  tangentially  to  the  circular  path  of  the  centre  of  the  orifice, 
its  absolute  velocity  (i.e.,  relatively  to  the  earth)  is 

wm  =  vm  —  GDX  —  34.8  —  20.9  =  13.9  ft.  per  sec., 

and  is  also  horizontal  and  backwards.  If  the  volume  of  flow 
is  Q  =  0.25  cub.  feet  per  sec.,  the  kinetic  energy  carried  away 
with  the  water  per  second  (§  133)  is 


ft.  Ibs.  per  second  =  0.085  horse-power. 

501.  Theoretical  Efflux  tlirough  Rectangular  Orifices  of  Con- 
siderable Vertical  Depth,  in  i  Vertical  Plate.—  If  the  orifice  is 
so  large  vertically  that  the  velocities  of  the  different  filaments 
in  a  vertical  plane  of  the  stream  are  theoretically  different,  hav- 
ing different  "  heads  of  water,"  we  proceed  as  follows,  taking 
into  account,  also,  the  velocity  of  approach,  c,  or  mean  velocity 


RECTANGULAR   ORIFICES. 


673 


(if  any  appreciable),  of  the  water  in  the  channel  approaching 
*he  orifice. 

Fig.  549  gives  a  section  of  the  side  of  the  tank  and  orifice. 
Let  b  =  width  of  the  rectangle,  the  sills  of  the  latter  being 
horizontal,  and  a  =  A3  —  A, ,  its  height.  Disregarding  con- 
traction for  the  present,  the  theoretical  volume  of  discharge 
per  unit  of  time  is  equal  to  the 
sum  of  the  volumes  like  vmdF 
(—  vjbdw),  in  which  vm  =  the  ^Ls^rJ 
velocity  of  any  filament,  as  m, 
in  the  jet,  and  bdx  =  cross-sec- 
tion of  the  small  prism  which 
passes  through  any  horizontal 
strip  of  the  area  of  orifice,  in  a 
unit  of  time,  its  altitude  being 
vm.  For  each  strip  there  is  a  FIG  549. 

different  x  or  "  head  of  water,"  and  hence  a  different  velocity. 
Now  the  theoretical  discharge  (volume)  per  unit  of  time  is 

/X- 
V, 


But  from  Bernoulli's  Theorem,  if  k  =  c*  -r-  2g  =  the  velocity, 
head  at  n,  the  surface  of  the  channel  of  approach  nC^b  being 
the  pressure-head  of  n,  and  x  its  potential  head  referred  to  m  as 
datum  (KB.  This  h  =  34  ft.  for  water,  and  must  not  be  con- 
fused with  the  width  b  of  orifice),  we  have  [see  §  492,  eq.  (7)] 


.....    (2)' 

and  since  dx  =  d(x  +  &),  Tc  being  a  constant,  we  have,  from  (1)' 
and  (2);, 

Theoret.  0  =  1  \r\ 


674  MECHANICS   OF   ENGINEERING. 


Theoret.  Q  =  &VTg  [(h,  +  *)•  -  (h,  +  *)•].       .     (1) 

(I  now  denotes  the  width  of  orifice.)     If  c  is  small,  the  chan- 
nel of  approach  being  large,  we  have 


Theoret.  Q  --=  &  Vfy  (hj  -h*)  ....    (2) 

(c  being  —  Q  -r-  area  of  section  of  nC). 

If  ^  =  0,  i.e.,  if  the  orifice  becomes  a  notch  in  the  side,  or 
an  overfall  [see  Fig  550,  which  shows  the*  contraction  which 
actually  occurs  in  all  these  cases],  we  have  for  an  overfall  * 


Theoret.  Q  =  %bv  2g\_(ht  +  fc)l  —  &J (3) 

NOTE. — Both  in  (1)  and  (2)  h,  and  Aa  are  the  vertical  depths 

of  the  respective  sills  of  the  orifice 
from  the  surface  of  the  water 
three  or  four  feet  hack  of  the  plane 
of  the  orifice,  where  the  surface  is 
comparatively  level.  This  must 
be  specially  attended  to  in  deriv- 
Fl°-  55°-  ing  the  actual  discharge  from  the 

theoretical  (see  §  503). 

If  Q  were  the  unknown  quantity  in  eqs.  (1)  and  (3)  it  would 
be  necessary  to  proceed  by  successive  assumptions  and  ap- 
proximations, since  Q  is  really  involved  in  Jc ;  for 

h  =  ^—    and    F0c  —  Q 

(where  F0  is  the  sectional  area  of  the  channel  of  approach  nO). 
With  fc  =  0  (or  c  very  small,  i.e.,  F9  very  large),  eq.  (3)  re- 
duces (for  an  overfall)  to 

Theoret.  Q  =  |M2  V*2gh^,        ..... 


or  $  as  much  as  if  all  parts  of  the  orifice  had  the  same  head  of 
water  —  A2  (as  for  instance  if  the  orifice  were  in  the  horizontal 
bottom  of  a  tank  in  which  the  water  was  A3  deep,  the  orifice 
having  a  width  =  ~b  and  length  =  A3). 

*  The  most  satisfactory  mathematical  treatment  of  the  flow  over  an  overfall 
weir  is  that  of  '.FJamant  (see  p.  96  of  his  Hydraulique,  Paris,  1900,  2d  edition). 
Its  resulting  formula  is  in  remarkable  accord  with  experiment,  but  is  not  con- 
venient for  practical  use. 


TEIANGULAE   ORIFICE. 


675 


502,  Theoretical  Efflux  through  a  Triangular  Orifice  in  a  Thin 
Vertical  Plate  or  Wall.  Base  Horizontal.—  Fig.  551.  Let  the 
channel  of  approach  be  so  large  that  the  velocity  of  approach 
may  be  neglected,  h^  and  A2  =  depths  of  sill  and  vertex, 
which  is  downward.  The  analysis  differs  from  that  of  the 
preceding  article  only  in  having  k  =  0  and  the  length  u,  of  a 
horizontal  strip  of  the  orifice,  variable  ;  b  being  the  length  of 
the  base  of  the  triangle.  From  similar  triangles  we  have 

u      Ao  —  x  b      ,-,         N 

- 


.:  Theoret.  Q  =  fvmdF=fvntudx  =  -=  --  rM1*  —  %)dx\ 
and  finally,  substituting  from  eq.  (2)'  of  §  501,  with  Ic  =  0, 


FIG.  551. 


Tfooret.  Q  = 


FIG.  552. 


For  a  triangular  notch  as  in  Fig.  552,  this  reduces  to 


(4) 


(5) 


i.e.,  ff  of  the  volume  that  "would  be  discharged  per  unit  of 


676 


MECHANICS   OF   ENGINEERING. 


time  if  the  triangular  orifice  witli  base  ~b  and  altitude  A3  were 
cut  in  the  horizontal  bottom  of  a  tank  under  a  head  of  Aa. 
The  measurements  of  A2  and  b  are  made  with  reference  to  the 
level  surface  back  of  the  orifice  (see  figure)  ;  for  the  water- 
surface  in  the  plane  of  the  orifice  is  curved  below  the  level 
surface  in  .the  tank. 

Prof.  Thomson  has  found  by  experiment  that  with 
b  =  2A2  ,  the  actual  discharge  =  theoret.  disch.  X  0.595  ;  and 
with  b  =  4A2  ,  actual  —  theoret.  disch.  X  0.620. 

503.  Actual  Discharge  through  Sharp-edged  Rectangular  Ori- 
fices (sills  horizontal)  in  the  vertical  side  of  a  tank  or  reservoir. 
CASE  I.   Complete  and   Perfect  Contraction.  —  The    actual 
volume  of  water  discharged  per  unit  of  time  is  much  less  than 
the  theoretical  values  derived  in  §  501, 
chiefly  on  account  of  contraction.     By 
complete  contraction  we  mean  that  no 
edge  of  the  orifice  is  flush  with  the 
side  or  bottom  of  the  reservoir  ;  and 
by  perfect  contraction,  that  the  channel 
of  approach,    to    whose    surface    the 
heads  A,  and  A2  are    measured,  is  so 
large  that  the  contraction  is  practically 
the  same  if  the  channel  were  of  infi- 

nite extent  sideways  and   downward 
from  the  orifice. 

For  this  case  (A,  not  zero)  it  is  found  most  convenient  to 
use  the  following  practical  formula  (b  =  width)  : 


FIG.  553. 


Actual      = 


(6) 


in  which  (see  Fig.  553)  a  =  the  height  of  orifice,  Aj  =  the  ver- 
tical depth  of  the  upper  edge  of  the  orifice  below  the  level  of 
the  reservoir  surface,  measured  a  few  feet  back  of  the  plane  of 
the  orifice,  and  //0  is  a  coefficient  of  efflux  (an  abstract  number), 
dependent  on  experiment. 

With  //0.=  0.62  approximate  results  (within  3  or  4  per  cent) 
may  be  obtained  from  eq.  (6)  with  openings  not  more  than 


RECTANGULAR   ORIFICES.  677 

18  inches,  or  less  than  1  inch,  high ;  and  not  less  than  1  inch 

wide;  with  heads  (\  +  -)  from  1  ft.  to  20  or  30  feet. 

V         2/ 

EXAMPLE. — What  is  the  actual  discharge  (volume)  per  min- 
ute through  the  orifice  in  Fig.  553,  14  inches  wide  and  1 
foot  high,  the  upper  sill  being  8  ft.  6  in.  below  the  surface  of 
still  water  ?  Use  eq.  (6)  with  the  ft,  lb.,  and  sec.  as  units,  and 
yw0  =  0.62. 

Solution : 
Q  =  0.62  X  1 X  H  X  1/2  X  32.2[8J+£]  =  17.41  cub. ft. per.sec. 

while  the  flow  of  weight  is 

G=Qy  =  17.41  X  62.5  =  1088  Ibs.  per  second. 

Poncelet  and  Lesbros1  Experiments. — For  comparatively  ac- 
curate results,  values  of  /*0  taken  from  the  following  table 
(computed  from  the  careful  experiments  of  Poncelet  and  Les- 
bros) may  be  used  for  the  sizes  there  given,  and,  where  prac- 
ticable, for  other  sizes  by  interpolation.  To  use  the  table,  the 
values  of  h^ ,  #,  and  b  must  be  reduced  to  metres,  which  can  be 
done  by  the  reduction-table  below ;  but  in  substituting  in  eq. 
(6),  if  the  metre-kilogram-second  system  of  units  be  used  g 
must  be  put  —  9.81  metres  per  square  second  (see  §  51),  and  Q 
will  be  obtained  in  cubic  metres  per  second. 

Since  //„  is  an  abstract  number,  once  obtained  as  indicated 
above,  it  does  not  necessitate  any  particular  system  of  units  in 
making  substitutions  in  eq.  (6).  The  ft.,  lb.,  and  sec.  will  be 
used  in  subsequent  examples. 

TABLE  FOR  REDUCING  FEET  AND  INCHES  TO  METRES. 


1 

foot 

=  0. 

30479 

metre. 

1 

inch 

=  0.0253 

metre. 

2 

feet 

=  0.60959 

a 

2 

inches 

— 

0. 

0507 

u 

3 

u 

=  0.91438 

a 

3 

«      =  0,0761      « 

4 

u 

—  1. 

21918 

metres. 

4 

a 

=  0.1015       " 

5 

a 

=  1.52397 

a 

5 

u 

— 

0. 

1268 

« 

6 

u 

=  1. 

82877 

u 

6 

u 

— 

0. 

1522 

u 

7 

"     =  2.13356       « 

7 

R 

— 

0. 

1776 

" 

8 

u 

=  2. 

43836 

u 

8 

U 

— 

0. 

2030 

u 

9 

" 

—  2. 

74315 

« 

9 

u 

— 

0. 

2283 

u 

10 

"     =  3.04794 

a 

10 

u 

~ 

0. 

2536 

U 

11 

a 

—  : 

0. 

2790 

u 

678  MECHANICS   OF  ENGINEERING. 

TABLE,  FROM  PONCELET  AND  LESBROS. 

VALUES  OF  /<o,  FOB  EQ  (6),  FOB  RECTANGULAR  ORIFICES  IN  THIN  PLATE; 

(Complete  and  perfect  contraction.) 


Value  of  hlt 
Fig.  553  (in 
metres). 

&  =  .20*'- 
a  =  .SO""- 

6  =  .20m 
a  =  .lO""- 

6  =  .20* 
a  =  .05m- 

6  =  .20*»- 
a  =  .03m- 

6  =  .20"" 
a  =  .02"" 

6  =  .20°>- 
a  =  .Ol"1- 

b  =  .60"> 
a=  .20"" 

b  =  .60"». 
a  =  .02"*- 

Mo 

Mo 

^° 

Mo 

Mo 

Mo 

Mo 

Mo 

0.005 

0.705 

.010 

0.607 

0.630 

0.660 

.701 

0.644 

.015 

0.593 

.612 

.632 

.660 

.697 

.644 

.020 

0.572 

.596 

.615 

.634 

.659 

.694 

.643 

.030 

.578 

.600 

.620 

.638 

.659 

.688 

0.593 

.642 

.040 

.582 

.603 

.623 

.640 

.658 

.683 

.595 

.642 

.050 

.585 

.605 

.625 

.640 

.658 

.679 

.597 

.641 

.060 

.587 

.607 

.627 

.640 

.657 

.676 

.599 

.641 

.070 

.588 

.609 

.628 

.639 

.656 

.673 

.600 

.640 

.080 

.589 

.610 

.629 

.638 

.656 

.670 

.601 

.640 

.090 

.591 

.610 

.629 

.637 

.655 

.668 

.601 

.639 

:       .100 

.592 

.611 

.630 

.637 

.654 

.666 

.602 

.639 

.120 

.593 

.612 

.630 

.636 

.653 

.663 

.603 

.638 

.140 

.595 

.613 

.630 

.635 

.651 

.660 

.603 

.637 

.160 

.596 

.614 

.631 

.634 

.650 

.658 

.604 

.637 

;     .iso 

.597 

.615 

.630 

.634 

.649 

.657 

.605 

.636 

.200 

.598 

.615 

.630 

.633 

.648 

.655 

.605 

.635 

.250 

.599 

.616 

.630 

.632 

.646 

.653 

.606 

.634 

.300 

.600 

.616 

.629 

.632 

.644 

.650 

.607 

.633 

.400 

.602 

.617 

.628 

.631 

.642 

.647 

.607 

.631 

.500 

.603 

.617 

.628 

.630 

.640 

.644 

.607 

.630 

.600 

.604 

.617 

.627 

.630 

.638 

.642 

.607 

.629 

.700 

.604 

.616 

.627 

.629 

.637 

.640 

.607 

.628 

.800 

.605 

.616 

.627 

.629 

.636 

.637 

.606 

.628 

.900 

.605 

.615 

.626 

.628  • 

.634 

.635 

.606 

.627 

.000 

.605 

.615 

.626 

.628 

.633 

.632 

.605 

.626 

.100 

.604 

.614 

.625 

.627 

.631 

.629 

.604 

.626 

.200 

.604 

.614 

.624 

.626 

.628 

.626 

.604 

.625 

.300 

.603 

.613 

.622 

.624 

.625 

.622 

.603 

.624 

.400 

.603 

.612 

.621 

.622 

.622 

.618 

.603 

.624 

1.500 

.602 

.611 

.620 

.620 

.619 

.615 

.602 

.623 

1.600 

.602 

.611 

.618 

.618 

.617 

.613 

.602 

.623 

1.700 

.602 

.610 

.617 

.616 

.615 

.612 

.602 

.622 

1.800 

.601 

.609 

.615 

.615 

.614 

.612 

.602 

.621 

1.900 

.601 

.608 

.614 

.613 

.612 

.611 

.602 

.621 

2.000 

.601 

.607 

.613 

.612 

.612 

.611 

.602 

.620 

8.000 

.601 

.603 

.606 

.608 

.610 

.609 

.601 

.615 

EXAMPLE. —With  h,  =  4=  in.  [=0.10  met.],  a  =  S  in. 
[=0.20  met.],  6  =  1  ft.  8  in.  [=0.51  met.],  required  the 
(actual)  volume  discharged  per  second.  See  Fig.  553. 


RECTANGULAR  ORIFICES. 


679 


From  the  foregoing  table, 

for  h,  =  0.10m-,  I  =  0.60rn-  and  a  =  0.20m-,  we  find  //0  =  .602 
«   h,  =  0.10m-,  I  =  0.20m-    «    a=0.20m-,       «        /it  =  .592 

diff.          =  ;oio 

Hence,  by  interpolation, 

for  A,  =  0.10m-  I  =  0.51m-,  and  a  =  0.20m-,  we  have 
K  =  0.602  -  A  [°-602  -  0.592]  =  0.600. 

Hence  [ft.,  lb.,  sec.],  remembering  that  //0  is  an  abstract  num- 
ber, from  eq.  (6), 


X 


X 


=  4.36 


Q  =  0.600  X 

cub.  ft.  per  second. 

CASE  II.  Incomplete  Contraction.  —  This  name  is  given  to 
the  cases,  like  those  shown  in  Fig.  554,  where  one  or  more 
sides  of  the  orifice  have  an  interior  border  flush  with  the  sides 
or  bottom  of  the  (square-cornered)  tank. 

Not  only  is  the  general  direction  of  the  stream  altered,  but 
the  discharge  is  greater,  on  account  of  the  larger  size  of  the 
contracted  section,  since  contraction  is  prevented  on  those  sides 
which  have  a  border.  It  is  assumed  that  the  contraction  which 
does  occur  (on  the  other  edges)  is  perfect  /  i.e.,  the  cross-sec- 
tion of  the  tank  is  large  compared  with  the  orifice.  According 
to  the  experiments  of  Bidone  and 
Weisbach  with  Poncelet's  ori- 
fices (i.e.,  orifices  in  thin  plate 
mentioned  in  the  preceding  table), 
the  actual  volume  discharged  per 
unit  of  time  is 


(7) 


FIG.  554. 


(differing  from  eq.  (6)  only  in 
the  coefficient  of  efflux  //),  in  which  the  abstract  number  ju.  is 
found  thus:  Determine  a  coefficient  of  efflux  //„  as  if  eq.  (6) 
were  to  be  used  in  Case  I ;  i.e.,  as  if  contraction  were  complete 
and  perfect ;  then  write 


680  MECHANICS   OF  ENGINEERING. 


,      .....     (7)' 

where  n  =  the  ratio  of  the  length  of  periphery  of  the  orifice 
with  a  border  to  the  whole  periphery. 
E.g.,  if  the  lower  sill,  only,  has  a  border, 


while  if  the  lower  sill  and  both  sides  have  a  border, 


EXAMPLE.—  If  h,  =  8  ft.  (=  2.43m-),  I  =  2  ft.  (=  0.60m-), 
a  =  4  in.  (=  0.10m-),  and  one  side  is  even  with  the  side  of 
the  tank,  and  the  lower  sill  even  with  the  bottom,  required  the 
volume  discharged  per  second.  (Sharp-edged  orifice,  in  ver- 
tical plane,  etc.) 

Here  for  complete  and  perfect  contraction  we  have,  from 
Poncelet's  tables  (Case  I),  /*6  =  0.608.  Now  ^  =  4;  hence* 
from  eq.  (7)', 

fji  =  0.608  [1  +  0.155  X  i]=  0.6551  ; 
hence,  eq.  (7), 


_ 

Q=  0.655  X  2  X  A  1/2X32.2(8+4. 
=  10.23  cub.  ft.  per  sec. 

CASE  III.  Imperfect  Contraction.  —  If  there  is  a  submerged 
channel  of  approach,  symmetrically 
placed  as  regards  the  orifice,  and  of 
an  area  (cross-section),  =  6>  not 
much  larger  than  that,  =  F,  of  the 
orifice  (see  Fig.  555),  the  contraction 
is  less  than  in  Case  I,  and  is  called 
imperfect  contraction.  Upon  his 
experiments  with  Poncelet's  orifices, 
FIG.  555.  with  imperfect  contraction,  Weisbach 

bases  the  following  formula  for  the  discharge  (volume)  per 

unit  of  time,  viz., 


(8) 


RECTANGULAR  ORIFICES. 


681 


(see  Fig.  553  fei  notation),  with  the  understanding  that  the  co* 
efficient 


where  /«0  fe  the  coefficient  obtained  from  the  tables  of  Case  I 
(as  if  the  contraction  were  perfect  and  complete),  and  ft  an  ab- 
stract number  depending  on  the  ratio  F  :  (?  =  »»,  as  follows: 


0.0760  [91*—  1.00]. 


To  shorten  computation  Weisbach  gives  the  following  table 


for/?: 

EXAMPLE.— Let  h,  =  4'  9J"  (=  1.46 
met.),  the  dimensions  of  the  orifice 

being — 

width  =  b  =  8  in.  (=  0.20m-); 
height  =  a  =  5  in.  (=  0.126m-); 

while  the  channel  of  approach  (CD, 
Fig.  555)  is  one  foot  square.  From 
Case  I,  we  have,  for  the  given  di- 
mensions and  head, 

A*0  =  0.610; 


TABLE  A. 


m. 

ft. 

m. 

0. 

.05 

.009 

.55 

.178 

.10 

.019 

.60 

.308 

.15 

.030 

.65 

.341 

.20 

.043 

.70 

.378 

.35 

.056 

.75 

.319 

.80 

.071 

.80 

.365 

.35 

.088 

.85 

.416 

.40 

.107 

.90 

.473 

.45 

.138 

.95 

.587 

.50 

.153 

1.00 

.608 

F  _  40sq.in. 


We  find  [Table  A] 


ft-  0.062; 


and  hence  p  =  yw0  (1.062),  from  eq.  (8)'.    Therefore,  from  eq. 
(8),  with  ft.,  lb.,  and  sec., 


Q  =  0.610  X  1062  X  A- 
=  3.22  cub.  ft.  per  sec. 


X  32.2  X  5 


CASE  IY.  Head  measured  in  Moving  Water.  —  See  Fig. 
556.  If  the  head  A,  ,  of  the  upper  sill,  cannot  be  measured  to 
the  level  of  still  water,  but  must  be  taken  to  the  surface  of  a 
of  approach,  where  the  velocity  of  approach  is  quite 


682 


MECHANICS   OF  ENGINEERING. 


appreciable,  not  only  is  the  contraction  imperfect,  but 
strictly  we  should  use  eq.  (1)  of  §  501,  in 
which  the  velocity  of  approach  is  considered. 
Let  F  =  area  of  orifice,  and  G  that  of  the 
cross-section  of  the  channel  of  approach; 
then  the  velocity  of  approach  is  c  =  Q  -~  (r, 
and  Tc  (of  above  eq.)  =  0f  ~  2g  =  Q9  -~  2gG*; 
but  Q  itself  being  unknown,  a  substitution  of 
k  in  terms  of  Q  in  eq.  (1),  §  501,  leads  to  an 
FIG.  556.  equation  of  high  degree  with  respect  to  Q. 

Practically,  therefore,  it  is  better  to  write 


and  determine  /«  by  experiment  for  different  values  of  the 
ratio  F-~-  G.,  Accordingly,  Weisbach  found,  for  Poncelet's 
orifices,  that  if  /i0  is  the  coefficient  for  complete  and  perfect 
contraction  from  Case  I,  we  have 

^  =  A(l  +  />0 (9)' 

ft'  being  an  abstract  number,  and  being  thus  related  to^-i-  6r, 


ft'  -  0.641 


(9)" 


A,  was  measured  to  the  surface  one  metre  back  of  the  plane  of 
the  orifice,  and  F :  G  did  not  exceed  0.50. 

Weisbach  gives  the  following  table  computed  from  eq.  (9)" : 


TABLE  B. 


ft. 


in 


EXAMPLE. — A  rectangular  water-trough  4 
wide  is  dammed  up  with  a  vertical  board 
which  is  a  rectangular  orifice,  as  in  Fig.  556,  of 
width  I  —  2  ft.  (=  0.60  met.),  and  height  a  =  6 
in.  (=  0.15  met.) ;  and  when  the  water-level  be- 
hind the  board  has  ceased  rising  (i.e.,  when  the 
flow  has  become  steady),  we  find  that  A,  =  2  ft., 
and  the  depth  behind  in  the  trough  to  be  3  ft. 
Eequired  Q. 

Since  F:  G  =  l  sq.  ft.  ~  12  sq.  ft.  =  .0833, 
we  find  (Table  B)  ft'  =•=  0.005 ;  and  /i0  being  =  0.612  from  Pon, 
celet's  tables,  Case  I,  we  have  finally,  from  eq.  (9), 


F+G. 

0'. 

0.05 

.002 

.10 

.006 

.15 

.014 

.20 

.026 

25 

.040 

.30 

.058 

.85 

.079 

,40 

.103 

.45 

.130 

.50 

.160 

DISCHARGE    OF   OVERFALL-WEIRS. 


683 


Q  =  0.612  (1.005)  2  xi  V2  X  32'.2  X  2.25 
=  7.41  cub.  f-t.  per  second. 

504.  Actual  Discharge    of  Sharp-edged   Overfalls  (Overfall* 
Weirs;  or  Rectangular  Notches  in  a  Thin  Vertical  Plate). 

CASE  I.  Complete  and  Perfect  Contraction  (the  normal 
case),  Fig.  557  ;  i.e.,  no  edge  is  flush 
with  the  side  or  bottom  of  the 
reservoir,  whose  sectional  area  is 
very  large  compared  with  that,  &Aa, 
of  the  notch.  By  deptn,  Aa  ,  of  the 
notch,  we  are  to  understand  the 
depth  of  the  sill  'below  the  surface 
a  few  feet  lack  of  the  notch  where 
it  is  level.  In  the  plane  of  the 
notch  the  vertical  thickness  of  the  stream  is  only  from  £  to  •£$ 
of  A3  .  Putting,  therefore,  the  velocity  of  approach  =  zero, 
and  hence  k  =  0,  in  eq.  (3)  of  §  501,  we  have  for  the 


Fio.  557. 


Actual  Q  = 


(10) 


(5  =  width  of  notch,)  where  /*0  is  a  coefficient  of  efflux  to  be 
determined  by  experiment. 

Experiments  with  overfalls  do  not  agree  as  well  as  might  be 
desired.  Those  of  Poncelet  and  Lesbros  gave  the  results  in 
Table  0. 

EXAMPLE  1.— With 


TABLE  C. 


A,  =  1  ft.  4  in.  (=  .405*), 
I  =  2  ft.  (=  0.60m-), 

we  have,  from  Table  0,  /*.  =  .586, 
and  (ft.,  lb.,  sec.) 


.-.  <?=.586xf  X2xf  I/2X32.2X £ 
=  9.54  cub.  ft.  per  sec. 

EXAMPLE  2.  —  What    width,  £, 
must  be  given  to  a  rectangular  notch,  for  which  Aa  =  10  in. 
(=  0.25m<),  that  the  discharge  may  be  Q  =  6  cub.  feet  per  see.* 


For  6  =  0.20- 

For  6  = 

0.60«. 

metres. 

metres. 

.& 

.636 

.06 

.6?8 

.02 

.620 

.08 

.613 

.03 

.618 

.10 

.609 

.04 

.610 

.12 

.605 

.06 

.601 

.15 

.600 

.08 

.595 

.20 

.592 

.10 

.592 

.30 

.586 

.15 

.589 

.40 

.586 

.20 

.585 

.50 

.586 

.22 

.577 

.60 

.585 

For  approx.  results  /ix0  =  .60 

684 


MECHANICS   OF   ENGINEERING. 


Since  b  is  unknown,  we  cannot  use  the  table  immediately, 
but  take  //0  =  .600  for  a  first  approximation  ;  whence,  eq.  (10), 
(ft.,  lb.,  sec.,) 


6 


0.6  X  |  X  H-  V*  X  32.2  X 


=  2.46  ft. 


Then,  since   this   width   does    not    much  exceed  0.60  metre, 
we  may  take,  in  Table  C,  for  Aa  =  0.25  met.,  /*0  =  .589 ; 


6 


.589  X  I  X  H  V 2  X  32.2  X 


=  2.50  ft. 


CASE  II.  Incomplete  Contraction  ;  i.e.  ,  both  ends  are  flush 
with  the  sides  of  the  tank,  these  being  ~\  to  the  plane  of  the 
notch.  According  to  Weisbach,  we  may  write 


(11) 


in  which  /*  =  1.04:1/<0  ,  /*0  being  obtained  from  Table  C  for  the 
normal  case,  i.e.,  Case  I.     The  section  of  channel  of  approach 
is  large  compared  with  that  of  the  notch  ;  if  not,  see  Case  IV. 
CASE  III.  Imperfect  Contraction  •  i.e.,  the  velocity  of  ap- 
proach is  appreciable  ;  the  sectional  area  G 
of  the  channel  of  approach  not  being  much 
larger  than  that,  F,  =  bh^  =  area  of  notch. 
Fig.  558.     b  =  width,   and    A2  =  depth  of 
notch  (see  Case  I).     Here,  instead  of  using 
a  formula  involving 

lc  =  ca  +  2g  =  [Q  +  Q]*  4-  2(7 


Fl&.  558. 


(see  eq.  (3),  §  501),  it  is  more  convenient  to  put 
as  before,  with 


(12) 
(12)' 


in  which  yuc  is  for  the  normal  case  [Case  I]  ;  and  ft,  according 


OVERFALL- WEIRS. 


685 


to  Weisbach's  experiments,  may  be  obtained  from  the  empiri- 
cal formula 


ft  =  1.718 


(12)" 


TABLE  D. 


[Table  D  is  computed  from  (12)".] 

(The  contraction  is  complete  in  this  case ;  i.e.,  the  ends  are 
not  flush  with  the  sides  of  the  tank.) 

EXAMPLE. — If  the  water  in  the  channel  of  ap- 
proach has  a  vertical  transverse  section  of  G  =  9 
sq.  feet,  while  the  notch  is  2  feet  wide  (i.e., 
b  =  2')  and  1  foot  deep  (A2  —  I7)  (to  level  of 
surface  of  water  3  or  4  ft.  back  of  notch),  we 
have,  from  Table  C,  with  b  =  .60  met.  and 
Aa  =  0.30  met., 

/*.  =  0.586 ; 
while  from  Table  D,  with  F:  G  —  0.222  (or  f), 


F 
& 

A. 

0.05 

.000 

.10 

.000 

.15 

.001 

.20 

.003 

.25 

.007 

.80 

.014 

.35 

.026 

.40 

.044 

.45 

.070 

.50 

.-107 

hence  (ft.-lb.-sec.  system  of  units),  from  eq.  (12), 


Q  =  I  x  0.586  X  1.005  X  2  X  1  X  V64.4  X  1.0 
=  6.30  cub.  ft.  per  second. 

CASE  IY.  Fig.  559.  Imperfect  and  incomplete  contrac- 
tion together ;  both  end-contractions  being  "suppressed"  (by 
making  the  ends  flush  with  the  sides  of  the  reservoir,  these 
sides  being  vertical  and  ~|  to  the  plane  of  the  notch),  and  the 
channel  of  approach  not  being  very  deep,  i.e.,  having  a  sec- 
tional area  G  but  little  larger  than  that,  F^  of  notch.  F=  bht 
as  before. 

Again  we  write 

Q  =  ttMt*Gjfc (13) 

with  ft  computed  from 

/*  =  //„(!  +  /?), (13/ 

/!„  being  obtained  from  Table  C ;  while 


686 


MECHANICS   OF   ENGINEERING. 


ft  =  0.041  +  0.3693  {—)  ,  .     .     .     .     (iav/ 

an  empirical  formula  based  by  Weisbach  on  his  own  experi- 
ments.* To  save  computation,  ft  may  be  found  from  Table  E, 
founded  on  eq.  (13)". 

TABLE  E. 


F 
G~ 

.00 

.05 

.10 

.15 

.20 

.25 

.30 

.35 

.40 

.45 

.50 

3  = 

.041 

.043 

.045 

.049 

.056 

.064 

.074 

.086 

.100 

.116 

.133 

! 

EXAMPLE.— Fig.  559.     With 
ft  =  2  ft.  (=0.60  met.) 

Aa  =  l  ft.  (=0.30  met.), 

we  have,  from  Table  C,  /*0  =  0.586, 
But,  the  ends  being  flush  with  the 
sides  of  the  reservoir  or  channel, 
and  6r  being  =  6  sq.  ft.  (see  figure), 

which  is  not  excessively  large  compared  with  f '=  bht  =  2  sq. 

ft.,  we  have,  from  Table  E,  with  F :  G  =  0.333, 

£  =  .081; 
and  hence  [eq.  (13)  and  (13)'],  //0  being  .586  as  in  last  example, 


FIG.  559. 


Q  =  |  x  0.586  X  (1  +  .081)  X  2  X  1  X  V64.4  X  1.0 
=  6.T8  cub.  ft.  per  sec. 

505.  Francis'  Formula  for  Overfalls  (i.e.,  rectangular  notches), 
— From  extensive  experiments  at  Lowell,  Mass.,  in  1851,  with 
rectangular  overfall-weirs,  Mr.  J.  B.  Francis  deduced  the  fol- 
lowing formula  for  the  volume,  Q,  of  flow  per  second  over 
such  weirs  10  feet  in  width,  and  with  A2  varying  from  0.6  to 
1.6  feet  (from  sill  of  notch  to  level  surface  of  water  a  few  feet 
back); . 

*  Weisbach's  results  in  this  case  differ  considerably  from  those  of  Bazin; 
see  p.  688. 


OVERFALL-WEIES.  687 

Q  =  1x0.6284,  (6-^)4/2^,     .    .    (14) 

in  which  5  =  width. 

This  provides  for  incomplete  contraction,  as  well  as  for  com- 
plete and  perfect  contraction,  by  making 

n  =  2  for  perfect  and  complete  contraction  (Fig.  557) ; 
n  =  1  when  one  end  only  is  flush  with  side  of  channel ; 
n  =  0  when  both  ends  are  flush  with  sides  of  channel. 

The  contraction  was  considered  complete  and  perfect  when 
the  channel  of  approach  was  made  as  wide  as  practicable, 
—  13.96  feet,  the  depth  being  about  5  feet. 

Mr.  Francis  also  experimented  with  submerged  or  "  drowned" 
weirs  in  1883  ;  such  a  weir  being  one  in  which  the  sill  is  be- 
low the  level  of  the  tail-water  (i.e.,  of  receiving  channel). 

508.  Fteley  and  Stearns's  Experiments  at  Boston,  Mass.,  in  1877 
and  1880. — These  may  be  found  in  the  Transactions  of  the 
American  Society  of  Civil  Engineers,  vol.  xn,  and  gave  rise 
to  formulae  differing  slightly  from  those  of  Mr.  Francis  in 
some  particulars.  In  the  case  of  suppressed  end-contractions, 
like  that  in  Fig.  559,  they  propose  formulae  as  follows : 

When  depth  of  notch  is  not  large, 

$  (in  cub.  ft.  per  sec.)  =  3.31  £A2i  + 0.007  5    .    (15) 
(b  and  A3  l}oth  infeet\ 

"  A2 ,  the  depth  on  the  weir,  should  be  measured  from  the  sur- 
face of  the  water  above  the  curvature  of  the  sheet." 

"  Air  should  have  free  access  to  the  space  under  the  sheet." 
The  crest  must  be  horizontal.  The  formula  does  not  apply  to 
Depths  on  the  weir  less  than  0.07  feet. 

When  the  depth  of  notch  is  quite  large,  a  correction  must 
be  made  for  velocity  of  approach,  <?,  thus : 

Q  (in  cub.  ft.  per  sec.)  =  3.31  &[\  + 1.5  ^-T+  0.007  b  (16) 

L  2g  J 

(J,  Aa ,  and  <?*  -f-  2<7,  in  feet). 


688  MECHANICS    OF    ENGINEERING. 

The  channel  should  be  of  uniform  rectangular  section  for 
about  20  ft.  or  more  from  the  weir,  to  make  this  correction 
properly.  If  G  =  the  cross-section,  in  sq.  ft.,  of  the  channel 
of  approach,  c  is  found  approximately  by  dividing  an  approxi- 
mate value  of  Q  by  G  ;  and  so  on  for  closer  results. 

The  weir  may  be  of  any  length,  &,  from  5  to  19  feet. 

506a.  Recent  Experiments  on  Overfall-weirs  in  France.  —  In 

the  Annales  des  Ponts  et  Chaussees  for  October  1888  is  an 
account  of  extensive  and  careful  experiments  conducted  in 
1886  and  1887  by  M.  Bazin  on  the  flow  over  sharp-edged 
overfall-weirs  with  end-contractions  suppressed  ;  i.e.,  like  that 
shown  in  Fig.  559.  The  widths  of  the  weirs  ranged  from 
0.50  to  2.00  metres,  and  the  depths  on  the  weirs  (A2)  from, 
0.05  to  0.60  metre.  With  p  indicating  the  height  of  the  sill 
of  the  weir  from  the  bottom  of  the  channel  of  approach,  M. 
Bazin,  as  a  practical  result  of  the  experiments,  recommends 
the  following  formula  as  giving  a  reasonably  accurate  value 
for  the  volume  of  discharge  per  unit  of  time  : 


.  (IT) 

where  the  coefficient  X  has  a  value 


Eq.  (IT)  is  homogeneous,  i.e.,  admits  of  any  system  of  units. 

Provision  was  made  in  these  experiments  for  the  free  en- 
trance of  air  under  the  sheet  (a  point  of  great  importance), 
while  the  walls  of  the  channel  of  approach  were  continued 
down-stream,  beyond  the  plane  of  the  weir,  to  prevent  any 
lateral  expansion  of  the  sheet.  The  value  of  p  ranged  from 
0.20  to  2.00  metres. 

Herr  Hitter  von  Wex  in  his  "  Hydrodynami'k  "  (Leipsic, 
1888)  derives  formulae  for  weirs,  in  the  establishing  of  which 
some  rather  peculiar  views  in  the  Mechanics  of  Fluids  are 
advanced. 


OVERFALLS. 


689 


Formulae  and  tables  for  discharge  through  orifices  or  over 
weirs  of  some  forms  not  given  here  may  be  found  in  the 
works  of  Weisbach,  Kankine,  and  Trautwfne.* 

Mr.  Hamilton  Smith,  a  noted  American  hydraulic  engineer, 
has  recently  published  "  Hydraulics,"  a  valuable  compilation 
and  resume  of  the  most  trustworthy  experiments  in  all  fields 
of  hydraulics  (New  York,  1886:  John  Wiley  &  Sons). 


f 

•4? 


507.  Efflux  through  Short  Cylindrical  Tubes. — When  efflux 
takes  place  through  a  short  cylindrical  tube,  or  "  short  pipe," 
at  least  2^  times  as  long  as  wide, 
inserted  at  right  angles  in  the 
plane  side  of  a  large  reservoir, 
the  inner  corners  not  rounded 
(see  Fig.  560),  the  jet  issues 
from  the  tube  in  parallel  fila- 
ments and  with  a  sectional  area} 
Fm,  equal  to  that,  F,  of  interior 
of  tube. 

To  attain  this  result,  however,  Fl°-  56°- 

the  tube  must  be  full  of  water  before  the  outer  end  is  un- 
stopped, and  must  not  be  oily ;  nor  must  the  head,  A,  be 
greater  than  about  40  ft.  for  efflux  into  the  air.  Since  at  in 
the  filaments  are  parallel  and  the  pressure-head  therefore  equal 
to  &  (—  34  ft.  of  water,  nearly),  =  that  of  surrounding  medium, 
=  head  due  to  one  atmosphere  in  this  instance ;  an  application 
of  Bernoulli's  Theorem  [eq.  (Y),  §  492]  to  positions  m  and  n 
would  give  (precisely  as  in  §§  454  and  455) 


v    =  veloc.  at  m  = 


as  a  theoretical  result  ;  but  experiment  shows  that  the  actual 
value  of  vm  in  this  case  is 


=  0.815 


(1) 


0.815  being  an  average  value  for  00 ,  the  coefficient  of  velocity,  for 
ordinary  purposes.     It  increases  slightly  as  the  head  decreases, 

*  Mr.  .Rafter's  paper,  in  Vol.  44  (p.  220)  of  the  Trans.  Am.  Soc.  C.  E.,  gives  an 
account  of  Bazin's  experiments  with  weirs  of  irregular  forms ;  as  also  of  similar 
experiments  made  at  the  Hydraulic  Laboratory  of  the  College  of  Civil  Engineer- 
ing at  Cornell  University. 


690 


MECHANICS   OF   ENGINEERING. 


and  is  evidently  much  less  than  the  value  0.97  for  an  orifice  in 
a  thin  pkte,  §  495,  or  for  a  rounded  mouth-piece  as  in  §  496. 

But  as  the  sectional  area  of  the  stream  where  the  filaments 
are  parallel,  at  m,  where  vm  =  0.815  V%gh,  is  also  equal  to  that, 
F,  of  the  tube,  the  coefficient  of  efflux,  /<0  ,  in  the  formula 


is  equal  to  00  ;  i.e.,  there  is  no  contraction,  or  the  coefficient 
of  contraction,  67,  in  this  case  =  1.00. 

Hence,  for  the  volume  of  discharge  per  unit  of  time,  we 
have  practically 

Q  =  <t>,F  Vfyh  =  0.815  TPylp?      ...     (2) 


The  discharge  is  therefore  about  ^  greater  than  through  an 
orifice  of  the  same  diameter  in  a  thin  plate  under  the  same 
head  [compare  eq.  (3),  §  495]  ;  for  although  at  m  the  velocity 
is  less  in  the  present  case,  the  sectional  area  of  the  stream  is 
greater,  there  being  no  contraction. 

This  difference  in  velocity  is  due  principally  to  the  fact  that 
the  entrance  of  the  tube  has  square  edges,  so  that  the  stream 

contracts  (at  m',  Fig.  561)  to  a 
section  smaller  than  that  of  the 
tube,  and  then  re-expands  to  the 
full  section,  F,  of  tube.  The 
eddying  and  accompanying  in- 
ternal friction  caused  by  this  re- 
expansion  (or  "sudden  enlarge- 
ment" of  the  stream)  is  the  prin- 
cipal resistance  which  diminishes 


PIG.  561. 


the  velocity.  It  is  noticeable,  also,  in 
this  case  that  the  jet  is  not  limpid  and 
clear,  as  from  thin  plate,  but  troubled 
and  only  translucent  (like  ground- 
glass).  The  internal  pressure  in  the 
stream  at  m!  is  found  to  be  less  than 
one  atmosphere,  i.e.  less  than  that  at  m, 
as  shown  experimentally  by  the  suck- 
ing in  of  air  when  a  8 ma]!  aperture  is  made  in  the  tube  op 


FIG.  562. 


INCLINED    SHOKT   PIPES.  691 

/ 

posite  mf.  If  the  tube  itself  were  so  formed  internally  as  to 
lit  tins  contracted  vein,  as  in  Fig.  562,  the  eddying  would  be 
diminished  and  the  velocity  at  m  increased,  and  hence  the 
volume  Q  of  efflux  increased  in  the  same  proportion.  (See 
§  509a.) 

If  the  tube  is  less  than  2J-  times  as  long  as  wide,  or  if  the 
interior  is  not  wet  by  the  water  (as  when  greasy),  or  if  the  head 
is  over  40  or  50  ft.  (about),  the  efflux  takes 
place  as  if  the  tube  were  not  there,  Fig.  563, 
and  we  have 


vm  =  0.97  V Zgh,  as  in  §  495. 

EXAMPLE. — The  discharge  through  a  short 
pipe  3  inches  in  diameter,  like  that  in  Fig.  560, 
is  30  cub.  ft.  per  minute,  under  a  head  of 
2'  6",  reservoir  large.  Required  the  coefficient  of  efflux 
;/0 ,  =  00 ,  in  this  case.  For  variety  use  the  inch-pound-min- 
ute system  of  units,  in  which  g  =  32.2  X  12  X  3600  (see  Note, 
§  51).  /e0 ,  being  an  abstract  number,  will  be  the  same  numer- 
ically in  any  system  of  units. 

From  eq.  (2), 

30  X  1728 


x  ga        x  32>2  x  12  x  603  X  30 
=  0.803. 

508.  Inclined  Short  Tubes  (Cylindrical).— Fig.  564.    If  the 
short  tube  is    inclined   at   some    angle 
a  <  90°    to  the  interior    plane   of  the 
reservoir  wall,  the  efflux  i»  smaller  than 
when  the  angle  is  90°,  as  in  §  507. 

We  still  use  the  form  of  equation 


Q  =  fiFYtyh,  =  <t>FVZgh ;-  .    (3) 

but    from    Weisbach's    experiments    /i 
should  be  taken  from  the  following  table: 


692  MECHANICS   OF  ENGINEERING. 

TABLE  F,  COEFFICIENT  OF  EFFLUX  (INCLINED  TUBE). 


Fora  =   90° 
take  ju  =  <p=  .815 

80° 
.799 

70° 

.782 

60° 
.764 

50° 

.747 

40° 
.731 

30° 
.719 

EXAMPLE. — With  h  =  12  ft.,  d  =  diam.  of  tube  —  4  ins., 
and  a  =  46°,  we  have  for  the  volume  discharged  per  sec.  (ft., 
lb.,  and  sec.) 

Q  =  [0.731  +  A  (-016)] ?  j|]f  4/64.4x12=1.79 cub.ft.per  sec, 

The  tube  must  be  at  least  3  times  as  long  as  wide,  to  be 
filled. 

509.  Conical  Diverging,  and  Converging,  Short  Tubes.— With 
conical  convergent  tubes,  as  at  A,  Fig.  565,  with  inner  edges 
not  rounded,  D'Aubuisson  and  Castel  found  by  experiment 
values  of  the  coefficient  of  velocity,  0,  and  of  that  of  efflux,  //, 
[from  which  the  coefficient  of  contraction,  0=  /*  ~-  0,  may  be 


FIG.  565. 

computed,]  for  tubes  1.55  centimeters  wide  at  the  narrow  end, 
and  4.0  centimeters  long,  under  a  head  of  h  =  3  metres,  and 
with  different  angles  of  convergence.  By  angle  of  converg- 
ence is  meant  the  angle  between  the  sides  CE  and  DB,  Fig. 
565.  In  the  following  table  will  be  found  some  values  of  /* 
and  0  founded  on  these  experiments,  for  use  in  the  formulae 


and      Q  = 
in  which  ./^denotes  the  area  of  the  outlet  orifice  EB. 


CONICAL    SHOKT    PIPES.  693 

TABLE  G  (CONICAL  CONVERGING  TUBES). 


Angle  of      )       8o  10, 
convergence  ) 

8° 

10°  20' 

13°  30' 

19°  30' 

30° 

49° 

/*=  .895 

.930 

.938 

.946 

.924 

.895 

.847 

0=  .894 

.932 

.951 

.963 

.970 

.975 

.984 

Evidently  //  is  a  maximum  for 

With  a  conically  divergent  tube  as  at  JOT,  having  the  in- 
ternal diameter  M 0  =  .025  metre,  the  internal  diam.  NP 
=  .032  metre,  and  the  angle  between  JO7' and  PO  —  4°  50', 
"Weisbach  found  that  in  the  equation  Q  =  }*F  V%gh  (where 
F=  area  of  outlet  section  NP)  P  should  be  =  0.553;  the 
great  loss  of  velocity  as  compared  with  V%gh  being  due  to  the 
eddying  in  the  re-expansion  from  the  contracted  section  at  M 
(corners  not  rounded),  as  occurs  also  in  Fig.  549.  The  jet 
was  much  troubled  and  pulsated  violently. 

"When  the  angle  of  divergence  is  too  great,  or  the  head  A 
too  large,  or  if  the  tube  is  not  wet  by  the  water,  efflux  with 
the  tube  filled  cannot  be  maintained,  the  flow  then  taking 
place  as  in  Fig.  563. 

Yenturi  and  Eytelwein  experimented  with  a  conically  di- 
vergent tube  (called  now  "  Ven- 
turis tube"),  with  rounded  en- 
trance to  conform  to  the  shape 
of  the  contracted  vein,  as  in 
Fig.  566,  having  a  diameter  of 
one  inch  at  vnf  (narrowest  part), 
where  the  sectional  area  =  F1 
=  0.7854  sq.  in.,  and  of  1.80 
inches  at  m  (outlet),  where  area  =  F',  the  length  being  8  ins., 
and  the  angle  of  convergence  5°  9'. 

"With  Q  =  ^FVZgh  they  found  A<  =  0.483. 

Hence  2-J-  times  as  much  water  was  discharged  as  would  have 
flowed  out  under  the  same  head  through  an  orifice  in  thin 
plate  with  area  =  F'  =  the  smallest  section  of  the  divergent 
tube,  and  1.9  times  as  much  as  through  a  short  pipe  of  sec- 
tion =  F' .  A  similar  calculation  shows  that  the  velocity  at 
m'  must  have  been  vm>  =1.55  V^gh,  and  hence  that  the  pres- 
sure at  m'  was  much  less  than  one  atmosphere. 


Fra.  566. 


694 


MECHANICS   OF   ENGINEERING. 


Mr.  J.  B.  Francis  also  experimented  with  Yenturi's  tube 
(see  u  Lowell  Hydraulic  Experiments").  See  also  p.  389  of 
vol  6  of  the  Journal  of  Engineering  Societies,  for  experi- 
ments with  diverging  short  tubes  discharging  under  water. 
The  highest  coefficient  (/t)  obtained  by  Mr.  Francis  was  0.782. 

509a.  New  Forms  of  the  Venturi  Tube. — The  statement  made 
in  §  507,  in  connection  with  Fig.  562,  was  based  on  purely 
theoretic  grounds,  but  has  recently  (Dec.  1888)  been  com- 
pletely  verified  by  experiments*  conducted  in  the  hydraulic 
laboratory  of  the  College  of  Civil  Engineering  at  Cornell 
University.  Three  short  tubes  of  circular  section,  each  3  in. 
in  length  and  1  in.  in  internal  diameter  at  both  ends,  were  ex- 
perimented with,  under  heads  of  2  ft.  and  4  ft.  Call  them  A, 
B,  and  C.  A  was  an  ordinary  straight  tube  as  in  Fig.  561; 
the  longitudinal  section  of  B  was  like  that  in  Fig.  562,  the 
narrowest  diameter  being  0.80  in.  [see  §  495;  (0.8)2  =  0.64]; 
while  C  was  somewhat  like  that  in  Fig.  566,  being  formed 
like  B  up  to  the  narrowest  part  (diameter  0.80  in.),  and  then 
made  conically  divergent  to  the  discharging  end.  The  results 
of  the  experiments  are  given  in  the  following  table : 


Name  of 
Tube. 

Head. 

Number 
of  Experi- 
ments. 

Range  of  Values  of  ft. 

Average 
Values  of  /*. 

A 
A 

h  =  2  ft. 
h  =  4  f  t. 

4 
3 

From  0.804  to  0.823 
"    0.819  to  0.823 

0.814 
0.821 

B 
B 

h  =  2  f  t. 
h  =  4  f  t. 

5 

4 

"    0.875  to  0.886 
"    0.881  to  0.902 

0.882 
0.892 

C 
C 

h  =  2  f  t. 
h  =  4  ft. 

5 

4 

"    0.890  to  0.919 
"    0.902  to  0.923 

0.901 
0.914 

The  fact  that  B  discharges  more  than  A  is  very  noticeable, 
•while  the  superiority  of  C  to  B,  though  evident,  is  not  nearly 
jo  great  as  that  of  B  to  A,  showing  that  in  order  to  increase 
the  discharge  of  an  (originally)  straight  tube  (by  encroaching 
on  the  passage-way)  it  is  of  more  importance  to  fill  up  with 
solid  substance  the  space  around  the  contracted  vein  than  to 
make  the  transition  from  the  narrow  section  to  the  discharg- 
ing end  very  gradual. 

*  See  Journal  of  the  Franklin  Inst.,  for  April,  1889. 


"  FLUID   FRICTION."  695 

510.  "  Fluid  Friction." — By  experimenting  with  the  flow  of 
water  in  glass  pipes  inserted  in  the  side  of  a  tank,  Prof.  Rey- 
nolds of  England  has  found  that  the  flow  goes  on  in  parallel 
filaments  for  only  a  few  feet  from  the  entrance  of  the  tube, 
and  that  then  the  liquid  particles  begin'  to  intermingle  and 
cross  each  other's  paths  in  the  most  intricate  manner.  To 
render  this  phenomenon  visible,  he  injected  a  fine  stream  of 
colored  liquid  at  the  inlet  of  the  pipe  and  observed  its  further 
motion,  and  found  that  the  greater  the  velocity  the  nearer  to 
the  inlet  was  the  point  where  the  breaking  up  of  the  parallel- 
ism of  flow  began.  The  hypothesis  of  laminated  flow  is, 
nevertheless,  the  simplest  theoretical  basis  for  establishing 
practical  formulae,  and  the  resistance  offered  by  pipes  to  the 
flow  of  liquids  in  them  will  therefore  be  attributed  to  the  fric- 
tion of  the  edges  of  the  laminae  against  the  inner  surface  of 
the  pipe.* 

The  amount  of  this  resistance  (often  called  skin-friction) 
for  a  given  extent  of  rubbing  surface  is  by  experiment  found — 

1.  To  be  independent  of  the  pressure  between  the  liquid  and 
the  solid ; 

2.  To  vary  nearly  with  the  square  of  the  relative  velocity  / 

3.  To  vary  directly  with  the  amount  of  rubbing  surface; 

4.  To  vary  directly  with  the  heaviness  [y,  §  409]   of  the 
liquid. 

Hence  for  a  given  velocity  v,  a  given  rubbing  surface  of 
area  =  S,  and  a  liquid  of  heaviness  y,  we  may  write 

v* 

Amount  of  friction  (force)  =  fSy  ^— ,       (1) 

Tog 

in  which  f  is  an  abstract  number  called  the  coefficient  of  fluid 
friction,  to  be  determined  by  experiment.  For  a  given  liquid, 
Driven  character  (roughness)  of  surface,  and  small  range  of 
velocities  it  is  approximately  constant.  The  object  of  intro- 

v* 
ducing  the  %g  is  not  only  because  ^—  is  a  familiar  and  useful 

*& 

function  of  v,  but  that  v*  -r-  2^  is  a  height,  or  distance,  and  there- 
fore the  product  of  S  (an  area)  by  v*  -r-  2g  is  a  volume,  and  this 
volume  multiplied  by  y  gives  the  weight  of  an  ideal  prism  of 

*  The  resistance  is  really  due  both  to  the  friction  of  the  water  on  the  sides  of 
the  pipe  and  to  the  friction  of  the  water  particles  on  each  other.  The  assump- 
tion that  it  is  due  to  the  former  action  alone  simply  affects  the  mathematical 
form  of  our  expressions,  without  invalidating  their  accuracy,  since  the  value  of 
/  is  in  any  case  dependent  on  experiment.  See  Engineering  News,  July-Dec. 
1901,  pp.  332  and  476. 


MECHANICS   OF   ENGINEERING. 


the  liquid;  hence  S^-  y  is  &  force  and  ymust  be  cm  abstract 

number  and  therefore  the  same  in  all  systems  of  units,  in  any 
given  case  or  experiment. 

In  his  experiments  at  Torquay,  England,  the  late  Mr.  Froude 
found  the  following  values  for/*,  the  liquid  being  salt  water, 
while  the  rigid  surfaces  were  the  two  sides  of  a  thin  straight 
wooden  board  -fa  of  an  inch  thick  and  19  inches  high,  coated 
or  prepared  in  various  ways",  and  drawn  edgewise  through  the 
water  at  a  constant  velocity,  the  total  resistance  being  measured 
by  a  dynamometer. 

511.  Mr.  Froude's  Results.— (Condensed.)  [The  velocity 
was  the  same  =  10  ft.  per  sec.  in  each  of  the  following  cases. 
For  other  velocities  the  resistance  was  found  to  vary  nearly  as 
the  square  of  the  velocity,  the  index  of  the  power  varying 
from  1.8  to  2.16.] 

TABLE  H. 


Character  of  Surface. 

Value  of/  [from  eq.  (1),  §  510]. 

Varnish  ./— 

2  ft.  long. 

When  the 
8  ft.  long. 

board  was 
20  ft.  long. 

50  ft.  long. 

0.0041 
.0038 
.0030 
.0087 
.0081 
.0090 
.0110 

0.0032 
.0031 
.0028 
.0063 
.0058 
.0062 
.0071 

0.0028 
.0027 
.0026 
.0053 
.0048 
.0953 
.0059 

0.0025 

Parafflne  " 

Tinfoil  

.0025 
.0047 
.0040 
.0049 

Calico  

Fine  Sand  

Medium  Sand  

Coarse  Sand  

N.B.  These  numbers  multiplied  by  100  also  give  the  mean  frictional  resistance  in 
IDS.  per  «q.  foot  of  area  of  surface  in  each  case  (v  =  10'  per  sec.),  considering  the 
heaviness  of  sea  water,  64  Ibs.  per  cubic  foot,  to  cancel  the  2g  =  64.4  ft.  per  sq.  sec.  of 
eq.  (1)  of  the  preceding  paragraph. 

For  use  in  formulae  bearing  on  flow  in  pipes,/*  is  best  deter- 
mined directly  by  experiments  of  that  very  nature,  the  results 
of  which  will  be  given  as  soon  as  the  proper  formulae  have  been 
established. 

512.  Bernoulli's  Theorem  for  Steady  Flow,  with  Friction.— [The 

student  will  now  re-read  the  first  part  of  §  492,  as  far  as  eq. 
(1).]  Considering  free  any  lamina  of  fluid,  Fig.  567,  (according 
to  the  subdivision  of  the  stream  agreed  upon  in  §  492  referred 


BERNOULLI'S  THEOREM  WITH  FRICTION. 


697 


to,)  the  frictioDs  on  tbe  -edges  are  the  only  additional  forces  as 

compared  with  the  system  in  Fig. 

534.     Let  w  denote  the  length 

of  the  wetted  perimeter  of  the 

base  of  this  lamina  (in  case  of  a 

pipe  running  full,  as  we  here 

postulate,  the  wetted   perimeter 

is  of  course  the  whole  perimeter* 

but  in  the  case  of  an  open  chan" 

nel  or  canal,  w  is  only  a  portion 

of  the  whole  perimeter  of  the 

cross-section).     Then,   since  the 

area  of  rubbing  surface  at  the  edge  is  S—  wds',  the  total  fric- 

tion for  the  lamina  is  [by  eq.  (1),  §  510]  =fwy  (v*  ~-  2g)ds'. 

Hence  from  vdv  =  (tan.  accel.)  X  ds,  and  from  (tan.  accel.)  = 

[^"(tang.  compons.  of  acting  forces)]  ~  (mass  of  lamina),  we 

have 

Fp  _  F(p  +  dp)  +  Fyds'  cos  0  —  fwy    -  dsr 


-  :  -  =r^-f 

Fyds'  -r-  g 

As  in  §  492,  so*  here,  considering  the  simultaneous  advance  of 
all  the  laminae  lying  between  any  two  sections  m  and  n  during 
the  small  time  dt,  putting  ds'  =  ds,  and  dsr  cos  0  =  —  dz  (see 
Fig.  568),  we  have,  for  any  one  lamina, 


"Now  conceive  an  infinite  number  of  equations  to  be  formed 
like  eq.  (1),  one  for  each  la- 
mina between  n  and  m,  for  the 
same  dt,  viz.,  a  dt  of  such 
length  that  each  lamina  at  the 
ond  of  dt  will  occupy  the 
same  position,  and  acquire  the 
same  values  of  v,  z,  and  p9 
that  the  lamina  next  in  front 
had  at  the  beginning  of  the  FIO.  568. 

dt  (this  is  the  characteristic  of  a  steady  flow].     Adding  up 


_J DATU"^ 


C98  MECHANICS   OF   ENGINEERING. 

the  corresponding  terras  of  all  these  equations,  we  have  (re« 
membering  that  for  a  liquid  y  is  the  same  in  all  laminae^ 


tfds;  .  (2) 

1 

i.e.?  after  transposition  and  writing  E  for  F~  w,  for  brevity, 

•  » 

This  is  Bernoulli's  Theorem,  for  steady  flow  of  a  liquid  tot, 
a  pipe  of  slightly  varying  sectional  area  F,  and  internal  peri/m- 
eter w,  taking  into  account  no  resistances  or  friction,  except 
the  "  skin-friction,"  or  u  fluid-friction,"  of  the  liquid  and  sides 
of  the  pipe. 

Resistances  due  to  the  internal  friction  of  eddying  occasioned 
by  sudden  enlargements  of  the  cross-section  of  the  pipe,  by 
elbows,  sharp  curves,  valve-gates,  etc.,  will  be  mentioned  later. 
The  negative  term  on  the  right  in  (3)  is  of  course  a  height  or 
head  (one  dimension  of  length),  as  all  the  other  terms  are  such, 
and  since  it  is  the  amount  by  which  the  sum  of  the  three  heads 
(viz.,  velocity-head,  pressure-head,  and  potential  head)  at  mt 
the  down  -stream  position,  lacks  of  being  equal  to  the  sum  of 
the  corresponding  heads  at  n,  the  up-stream  position  or  section, 
we  may  call  it  the  "Loss  of  Head"  due  to  skin-friction  between 
n  and  m\  also  called  friction-head*  or  resistance-heady  or 
height  of  resistance. 

The  quantity  H  =  F  -f-  w  =  sectional-area  -f-  wetted-pe 
rimeter,  is  an  imaginary  distance  or  length  called  the  Hydraur 
lie  Mean  Radius,  or  Hydraulic  Mean  Depth,  or  simply 
hydraulic  radws  of  the  section.  For  a  circular  pipe  of  diam- 
eter =  d~ 


while  for  a  pipe  of  rectangular  section, 

r>_ 

~ 


FRICTION 


PIPES. 


513.  Problems    involving  Friction-heads;    and  Examples  ol 
Bernoulli's  Theorem  with  Friction. 

PROBLEM  I. — Let  the  portion  of  pipe  between  n  and  m  be 
level,  and  of  uniform  cir-        •*&>% 
cular  section  and  diameter   nlSSMt 
=  d.    The  jet  at  m  dis- 
charges into  the  air,  and 
has  the  same  sectional  area, 
F=  J7rd*,as  the  pipe;  then 
the  pressure-head  at  m  is 


34   feet    (for 


m, 


^^ 


FIG.  569. 


water),  and  the  velocity- 
head  at  m  is  =  that  at  n,  since  vm  =  vn.     The  height  of  the 
water  column  in  the  open  piezometer  at  n  is  noted,  and  =  yn 

(so  that  the  pressure-head  at  n  is^  =  yw+5);  while  the 

Y 
length  of  pipe  from  n  to  m  is  =  I. 

Knowing  I,  d,  yn ,  and  having  measured  the  volume  Q,  of 
flow,  per  unit  of  time,  it  is  required  to  jmd  the  form  of  the 
friction-bead  and  the  value  off.  From 

Fmvm  =  Q,    or    %nd*vm  —  Q,  .    .    .    .    (1) 

vm  becomes  known.  Also,  vm  is  known  to  be  =  vn,  and  the 
velocity  at  each  ds  is  v  =  vm ,  since  F  (sectional  area)  is  con- 
stant along  the  pipe,  and  Fv  =  Q.  The  hydraulic  radius  is 

R  =  \d\ (2) 


the  same  for  all  the  ^s  between  n  and  m. 

Substituting  in  eq.  (3)  of  §  512,  with  the  horizontal  axis  of 
the  pipe  as  a  datum  for  potential  heads,  we  have 


i.e.,  since  J^d8  —  l=.  length  of  pipe  from  n  to  m,  the  friction- 
head  for  a  pipe  of  length  =  Z,  and  uniform  circular  section 
</ diameter  =  d,  reduces  to  the  form 


TOO 


MECHANICS   OF  ENGINEERING. 


I        V* 

Friction-head  =  ^f-j  •  ~—  ? 


(4) 


where  v  =  velocity  of  water  in  the  pipe,  being  in  this  case 
also  =  vm  and  =  vn  .  Hence  this  friction-head  varies  directly 
cbs  the  length  and  as  the  square  of  the  velocity,  and  inversely 
as  the  diameter  ;  also  directly  as  the  coefficient  f. 

From  (3),  then,  we  derive  (for  this  particular  problem) 


Piezometer-height  a$n  =  yn  =  4f-^  .  £- 


. 


(5) 


i.e.,  the  open  piezometer-height  atn  is  equal  to  the  loss  of  head 
(all  of  which  is  friction-head  here)  sustained  between  n  and  the 
mouth  of  the  pipe.  (Pipe  horizontal.) 

EXAMPLE.  —  Kequired  the  value  of/,  knowing  that  d  =  3  in., 
yn  (by  observation)  =  10.4  ft.,  and  Q  =  0.1960  cub.  ft.  per 
sec.,  while  I  =  400  ft.  (n  to  m).  From  eq.  (1)  we  find,  in  ft.- 
Ib.-sec.  system,  the  velocity  in  the  pipe  to  be 


then,  using  eq.  (5),  we  determine/  to  be 


400  X 


AA/U?K 


PROBLEM  II.  Hydraulic  Accumulator. — Fig.  570.    JLet  the 
area  Fn  of  the  piston  on  the  left  be  quite  large  compared  with 


FIG.  570. 


that  of  the  pipes  and  nozzle.     The  cylinder  contains  a  friction- 


FRICTION-HEAD   IN   PIPES.  701 

weighted  piston,  producing  (so  long  as  its  downward  slow 
motion  is  uniform)  a  fluid  pressure  on  its  lower  face  of  an 
intensity  pn=  L^  +  ^niV]  -=-  Fn  per  unit  area  (pa  =  one 
atmos.). 
Hence  the  pressure-head  at  n  is 


where  G  =  load  on  piston. 

The  jet  has  a  section  at  m  =  Fm  —  that  of  the  small  straight 
nozzle  (no  contraction).  The  junctions  of  the  pipes  with  each 
other,  and  with  the  cylinder  and  nozzle,  are  all  smoothly 
rounded  ;  hence  the  only  losses  of  head  in  steady  flow  between 
n  and  in  are  the  friction-heads  in  the  two  long  pipes,  neglect. 
ing  that  in  the  short  nozzle.  These  friction-heads  will  be  of 
the  form  in  eq.  (4),  and  will  involve  the  velocities  vl  and  vt 
respectively  in  these  pipes  (supposed  running  full).  vl  and  vt 
may  be  unknown  at  the  outset,  as  here. 

Knowing  G  and  all  dimensions  and  heights,  we  are  required 
to  find  the  velocity  vm  of  the  jet,  flowing  into  the  air,  and  the 
volume  of  flow,  (),  per  unit  of  time,  assuming  f  to  be  known 
and  to  be  the  same  in  both  pipes  (not  strictly  true). 

Let  the  lengths  and  diameters  be  denoted  as  in  Fig.  570, 
their  sectional  areas  Fl  and  F^  the  unknown  velocities  in  them 
t;,  and  vt. 

From  the  equation  of  continuity  [eq.  (3),  §  490],  we  have 


and     V,=          .      ...    (7) 


To  find  vmi  we  apply  Bernoulli's  Theorem  (with  friction), 
eq.  (3),  §  512,  taking  the  down-stream  position  m  in  the  jet 
close  to  the  nozzle,  and  the  up-stream  position  n  just  under  the 
piston  in  the  cylinder  where  the  velocity  vn  is  practically  noth- 
ing. Then  with  m  as  datum  plane  we  have 


.    (8) 


702  MECHANICS    OF   ENGINEERING. 

Apparently  (8)  contains  three  unknown  quantities,  vm9  v  , 
and  va  ;  but  from  eqs.  (7)  vl  and  vt  can  be  expressed  in  terms 
of  vm,  whence  [see  also  eq.  (6)] 


or,  finally, 

\/*(*+£) 

*M  —          /  7    ,-f?    \a  1    tTT    \<^  »  '  ^      ) 

and  hence  we  have  also 

#  =  ^>«,   .......      (11) 

EXAMPLE.  —  If  we  replace  the  force  G  of  this  problem  by 
the  thrust  P  exerted  along  the  pump-pist  n  of  a  steam  fire- 
engine,  we  may  treat  the  foregoing  as  a  close  approximation 
to  the  practical  problem  of  such  an  apparatus,  the  pipes  being 
consecutive  straight  lengths  of  hose,  in  which  (for  the  probable 
values  of  -y,  and  v^)  we  may  take/*=  .0075  (see  "  Fire-streams," 
by  Geo.  Ellis,  Springfield,  Mass.).*  (Strictly,  f  varies  somewhab 
with  the  velocity;  see  §  517.)  Let  P  =  12000  Ibs.,  and  the 
piston-area  atn  =  Fn  =  12  sq.  in.  =  J  sq.  ft.  Also,  let  h  —  20 
ft.,  and  the  dimensions  of  the  hose  be  as  follows  : 

dt  =B  3  in.,    dt  =  2  in.,    dn  (of  nozzle)  =  1  in.; 


With  the  foot-pound-second  system  of  units,  we  DOW  have 
teq.(10)] 


See  §  578. 


FRICTION-HEAD    IN   PIPES.  703 

_      / 

"  V 


_      /2  X  32.2  X  404"  . 


°-59  +  5-62 

vm  =  60.0  ft.  per  sec.     If  this  jet  were  directed  vertically 

v  * 
upward  it  should  theoretically  attain  a  height  =  -—-  =  nearly 

56  feet,  but  the  resistance  of  the  air  would  reduce  this  to  about 
40  or  45  ft. 

We  have  further,  from  eq.  (1), 

Q  =  FmVm  =  f  (4)*X  60'°  =  3-2T  Cllb-  ft  P6r  S6C- 

If  there  were  no  resistance  in  the  hose  we  should  have,  from 
§4970, 


vm  = 


+  A~j  =  1/2  X  32.2  X  404  =  161.3  ft.  per  sec. 


513a.  Influence  of  Changes  of  Temperature.  —  Although  Poi- 
seuille  and  Hagen  found  that  with  glass  tubes  of  very  small 
diameter  the  flow  of  water  was  increased  threefold  by  a  rise  of 
temperature  from  0°  to  45°  Cent.,  it  is  unlikely  that  with  com- 
mon pipes  the  rate  of  flow  is  appreciably  affected  by  the  ordi- 
nary fluctuations  of  temperature  ;  at  any  rate,  experiments  of 
sufficient  precision  are  wanting,  as  regards  such  an  influence. 
See  Mr.  Hamilton  Smith's  "  Hydraulics,"  p.  16,  where  he 
says  :  u  Changes  by  variation  in  T  (temperature)  will  probably 
only  be  appreciable  with  small  orifices,  or  with  very  low  heads 
for  orifices  or  weirs." 

514.  Loss  of  Head  in  Orifices  and  Short  Pipes.  —  So  long  as  the 
steady  flow  between  two  localities  n  and  m  takes  pJace  in  a  pipe 
having  no  abrupt  enlargement  or  diminution  of  section,  nor 
sharp  curves,  bends,  or  elbows,  the  loss  of  head  may  be  ascribed 
solely  to  the  surface  action  (or  "  skin-friction")  between  water. 
and  pipe  ;  but  the  introduction  of  any  of  the  above-mentioned 
features  occasions  eddying  and  internal  disturbance,  and  fric- 
tion  (and  consequent  heat)  ;  thereby  causing  further  deviations 


704  HECHANICS   OF   ENGINEERING. 

from  Bernoulli's  Theorem ;  i.e.,  additional  losses  of  head>  of 
heights  of  resistance. 

From  the  analogy  of  the  form  of  a  friction-head  in  a  long 
pipe  [eq.  (4),  §  513],  we  may  assume  that  any  of  the  above 
heights  of  resistance  is  proportional  to  the  square  of  the  veloc- 
ity, and  may  therefore  always  be  written  in  the  form 

1  j  Loss  of  Head  due  to  any  \  _  ~  tf_  „. 

(  cause  except  skin-friction  j  ~~     2g ' 

in  which  v  is  the  velocity  of  the  water  in  the  pipe  at  the  sec- 
tion where  the  resistance  occurs ;  or  if,  on  account  of  an 
abrupt  enlargement  of  the  stream-section,  there  is  a  correspond- 
ing diminution  of  velocity,  then  v  is  always  to  denote  this 
diminished  velocity  (i.e.,  in  the  down-stream  section).  This 
velocity  v  is  often  an  unknown  at  the  outset. 

C,  corresponding  to  the  abstract  factor  4/  --  in  the  height  of 

Ci 

resistance  due  to  skin-friction  [eq.  (4),  §  513],  is  an  abstract 
number  called  the  Coefficient  of  Resistance,  to  be  determined 
experimentally ;  or  computed  theoretically,  where  possible. 
Roughly  speaking,  it  is  independent  of  the  velocity,  for  a  given 
fitting,  casing,  pipe- joint,  elbow,  bend,  valve-gate  at  a  definite 
opening,  etc.,  etc. 

515.  Heights  of  Resistance  (or  Losses  of  Head)  Occasioned  by 
Short  Cylindrical  Tubes. — When  dealing  with  short  tubes  dis- 
charging into  the  air,  in  §  507,  deviations  from  Bernoulli's 
Theorem  were  made  good  by  using  a  coefficient  of  velocity  0, 
dependent  on  experiment.  This  device  answered  every  pur 
pose  for  the  simple  circumstances  of  the  case,  as  well  as  for 
simple  orifices.  But  the  great  variety  of  possible  designs  of  a 
3ompound  pipe  (with  skin-friction,  bends,  sudden  changes  of 
cross  section,  etc.)  renders  it  almost  impossible,  in  such  a  pipes 
to  provide  for  deviations  from  Bernoulli's  Theorem  by  a  single 
coefficient  of  velocity  (velocity  of  jet,  that  is)  for  the  pipe  as  a 
whole,  since  new  experiments  would  be  needed  for  each  new 
design  of  pipe.  Hence  the  great  utility  of  the  conception  of 
u  loss  of  head,"  one  for  each  source  of  resistance. 


LOSS   OF  HEAD 


A  SHOKT  PIPE. 


705 


If  a  long  pipe  issues  from  the  plane  side  of  a  reservoir  and 
the  corners  of  the  junction  are  not  rounded  [see  Fig.  571],  we 

shall  need  an  expression  for 
"}'  the  loss  of  head  at  the  en- 
7J?  trance,  E,  as  well  as  that 


due  to  the  skin-friction  in  the 

pipe.  But,  whatever  the  velocity,  v,  in  the  pipe  proves  to 
3e,  influenced  as  it  is  both  by  the  entrance  loss  of  head  and 
the  skin-friction  head  (in  applying  Bernoulli's  Theorem),  the 

v* 
loss  of  head  at  E,  viz.,  C,E  — -,  will  be  just  the  same  as  if  efflux 

y 

took  place  through  enough  of  the  pipe  at  E  to  constitute  a 
4  short  pipe,"  discharging  into  the  air,  under  some  head  h 
different  from  h'  of  Fig.  571)  sufficient  to  produce  the  same 

velocity  v.    But  in  that  case  we  should  have 


or 


—  =  0aA 
y 


(1) 


(See  §§  507  and  508,  0  being  the  "  coefficient  of  velocity,"  and 
h  the  head,  in  the  cases  mentioned  in  those  articles.) 

We  therefore  apply  Bernoulli's  Theorem  to  the  cases  of 
those  articles  (see  Figs,  560  and  564)  in  order  to  determine  the 
loss  of  head  due  to  the  short  pipe  and  obtain  (with  m  as  datum 
level  for  potential  heads) 


Now  the  v  of  eq.  (2)  is  equal  to  the  vm  of  the  figures  referred 
to,  and  CE  is  a  coefficient  of  resistance  for  the  ,  short  pipe,  and 
we  now  desire  its  value.  Substituting  for 


706  MECHANICS   OF   ENGINEERING. 

its  value  0aA  from  eq.  (1),  we  have 


Hence  when  a  —  90°  (i.e.,  the  pipe  is  ~]  to  the  inner  reser- 
voir surface),  we  derive 


00S 


(0.815) 


-1  =  0.505;  .     .  l>  =  90°;]  .    . 
8 


and  similarly,  for  other  values  of  a  (taking  0  from  the  table, 
§  508),  we  compute  the  following  values  of  C^  (corners  not 

v9 
rounded)  for  use  in  the  expression  for  "  loss  of  head,"  C#  —  : 


Fora=   90° 
C*=.5Q5 

80° 
.565 

70° 
.635 

60° 
.713 

50° 
.794 

40° 
.870 

30° 

.987 

From  eq.  (4)  we  see  that  the  loss  of  head  at  the  entrance  of 
the  pipe,  corners  not  rounded,  with  a  =  90°,  is  about  one  half 
(.505)  of  the  height  due  to  the  velocity  v  in  that  part  of  the 
pipe  (v  being  the  same  all  along  the  pipe  if  cylindrical).*  The 
value  of  v  itself,  Fig.  571,  depends  on  all  the  features  of  the 
design  from  reservoir  to  nozzle.  See  §  518. 

If  the  corners  at  E  are  properly  rounded,  the  entrance  loss  of 
head  may  practically  be  done  away  with;  still,  if  v  is  quite 
small  (as  it  may  frequently  be,  from  large  losses  of  head 
farther  down-stream),  the  saving  thus  secured,  while  helping 
to  increase  v  slightly  (and  thus  the  saving  itself),  is  insignifi 
cant. 

516.  General  Form  of  Bernoulli's  Theorem,  considering  all 
Losses  of  Head, 

In  view  of  preceding  explanations  and  assumptions,  we  may 
write  in  a  general  and  final  form  Bernoulli's  Theorem  for  a 
steady  flow  from  an  up-stream  position  n  to  a  down-stream 
position  m,  as  follows : 

„.  2  2  (<ill  losses  of  head  } 

— -  -I-  ±-^  -4-  zm  =  —  -\-  ±-2.  -4-  &n —  \  occurring  between  >  .  CZ?f) 

9/1  if  9,n  i/  )  7  I 

*y        f  _!___*_  (          nandm         j 

*  If  the  entrance  of  the  pipe  has  well-rounded  corners  (see  Fig.  541  on  p.  663), 
the  value  of  Cfi  is  very  small ;  viz.,  about  0.05. 


LOSSES   OF  HEAD   IN   GENERAL.  707 

Each  loss  of  head  (or  height  of  resistance)  will  be  of  the  form 
C  —  (except  skin-friction  head  in  long  pipes,  viz.,  4/  —  --J, 

the  v  in  each  case  being  the  velocity,  known  or  unknown,  in 
that  part  of  the  pipe  where  the  resistance  occurs  (and  hence 
is  not  necessarily  equal  to  vm  or  vn). 

517.  The  Coefficient,/,  for  Friction  of  Water  in  Pipes.— See 

eq.  (1),  §  510.  Experiments*  have  been  made  by  Weisbach, 
Eytelwein,  Darcy,  Bossut,  Prony,  Dubuat,  Fanning,  and  oth- 
ers, to  determine  f  in  cylindrical  pipes  of  various  materials 
(tin,  glass,  zinc,  lead,  brass,  cast  and  wrought  iron)  of  diameters 
from  J  inch  up  to  36  inches.  In  general,  the  following  deduc- 
tions may  be  made  from  these  experiments : 

1st.  f  decreases  when  the  velocity  increases ;  e.g.,  in  one 
case  with  the 

same  pipe/  was  =  .0070  for  v  =  2'  per  sec., 
while/  was  =  .0056  for  v  =  20'  per  sec. 

2dly.  /  decreases  slightly  as  the  diameter  increases  (other 
things  being  equal); 

e.g.,  in  one  experiment/ was  =  .0069  in  a  3-in.  pipe, 

while  for  the  same  velocity/ was  =  .0064  in  a  6-in.  pipe. 

3dly.  The  condition  of  the  interior  surface  of  the  pipe 
affects  the  value  of/  which  is  larger  with  increased  roughness 
of  pipe. 

Thus,  Darcy  found,  with  a,  foul  iron  pipe  with  d=  10  in. 
and  veloc.  =  3.67  ft.  per  sec.,  the  value  .0113  for/;  whereas 
Fanning  (see  p.  238  of  his  "  Water-supply  Engineering"),  with 
a  cement-lined  pipe  and  velocity  of  3.74  ft.  per  sec.  and  d  = 
20  inches,  obtained /=  .0052. 

Weisbach,  finding  the  first  relation  very  prominent,  pro- 
posed  the  formula 

/  =0.00359  + 


when  the  velocities  are  great,  while  Darcy,  taking  into  account 
both  the  1  st  and  2d  relations  above,  writes  (see  p.  585,  Ran- 
kine's  Applied  Mechanics) 

*  Among  the  most  important  recent  experiments  are  those  of  Prof.  G.  S.  Wil- 
liams (with  Messrs.  Hubbell  and  Feukell);  and  of  Drs.  Saph  and  Schoder.  See 
Trans.  Am.  Soc.  C.  E.  for  April,  1902,  and  Proc.  Am.  Soc.  C.  E.  for  May  1903. 

Hering  and  Trautwine's  book  on  Kutter's  Formula  contains  many  experi- 
mental data  on  pipes. 


708 


MECHANICS    OF   ENGINEERING. 


/=  .0043  fl  + 1.1     .        I-L 

"9  X  diam.  in  ft.J~ 


.001 


ft.  per  sec. 


fi+ 

L 


18  X  diam.  in  ft. 


For  practical  purposes,  Mr.  J.  T.  Fanning  has  recommended, 
••and  arranged  in  an  extensive  table  (pp.  242-246  of  his  book 
just  mentioned),  values*  of  f 'for  clean  iron  pipe,  of  diameters 
from  £  inch  to  96  inches,  and  for  velocities  of  0.1  ft.  to  20  ft 
per  second.  Of  this  the.  table  opposite  is  an  abridgment,  in- 
serted with  Mr.  Fanning's  permission,  for  use  in  solving  nu- 
merical problems. 

In  obtaining/1  for  slightly  tuberculated  and  for  foul  pipes, 
the  recommendation,0  of  Mr.  Fanning  seem  to  justify  the  fol- 
lowing rules : 


For  slightly  tuberculated  pipes  of  diams.  —  £  ft. 

we  should  add  23$ 
and  for  foul  pipes  of  same  size 


1ft. 


2  ft 

16$ 
38$ 


4ft. 

13$ 
25$ 


of  the/  for  clean  pipes,  to  itself.  For  example,  if/*=  .007 
for  a  certain  %-h.  pipe  when  clean,  with  velocity  =  0.64  ft.  per 
sec.,  we  have/=  .007  X  1.72  =  .01204  when  it  is  foul. 

For  first  approximations  a  mean  value  of  f  =  .006  may  be 
employed,  since  in  some  problems  sufficient  data  may  not  be 
known  in  advance  to  enable  us  to  take /"from  the  table. 

EXAMPLE. — Fig.  572.  In  the  steady  pumping  of  crude 
petroleum  weighing  y  =  55  Ibs.  per  cubic  foot,  through  a  six- 
inch  pipe  30  miles  long, 
to  a  station  700  ft.  higher 
than  the  pump,  it  is 
found  that  the  pressure 
in  the  pump  cylinder  at" 
n,  necessary  to  keep  up 
a  velocity  of  4.4  ft.  per 
sec.  in  the  pipe,  is  1000 

Ibs.  per  sq.  inch.  Kequired  the  coefficient  f  in  the  pipe.  A.S 
all  losses  except  the  friction-head  in  the  pipe  are  insignificant, 
the  latter  only  will  be  considered.  The  velocity-head  at  n  may 

*  Caution.— For  riveted  wrought-iron  pipe,  with  projecting  rivet-teads 
and  lapping  edges,  the  values  of /may  be  nearly  double  those  of  this  table. 
See  Eng.  News,  Dec.  '95,  p.  415  ;  also  in  Jan.,  etc.,  '96,  pp.  59,  74,  193,  393. 


TABLE  OF  VALUES   OF/. 


709 


§  10 

10  II  II 


c^ 

II  II 


11 


o    .a 

a 
O      f 


S-S 


II II 


a  is 

10  ii  if 


II II 


os    co 


TH       CO       O 
CQ      TH      TH 

TrtH        Tt<        ^ 


-I-H     "^     J>     O 

os    QO    £•    £- 

CO     CO      CO     CO 


O 

1C 


Z>  CO  TH  O 
^^  OS  OO  t^* 
CO  IO  IO  IO 


3 


8 


2          8          §          § 

CO     CO     JO     iO 


^COt-C^C^OOlO 
CO      CO     CO     K 


8 


§  22 

co    o 


$  i  s 


8 


t—    THO<MT— loooo-i— los 

lOCOTHCS*>T}<THOSCOt- 
t^     t'*     t^*     CO     CO     CO     CO     iO     iO     iO 


CO  3 


O  00  ^f  C5  O  CO  ^ 
T— I  CO  CO  O  t—  -H/l  i— I 
30  £>  J>  i>  CO  CO  CO 


(M  iO  CO 
CO  OS  1C 


(M 
t» 


CO 


1OOWOOOOOOO 


710 


MECHANICS   OF  ENGINEERING. 


be  pat  =  0  ;  the  jet  at  m  being  of  the  same  size  as  the  pipe, 
the  velocity  in  the  pipe  is  =  vm  ,  and  therefore  vm  —  4.4  ft.  per 
sec.  Notice  that  m,  the  down-stream  section,  is  at  a  higher 
level  than  n. 

From  Bernoulli's  Theorem,  §  516,  we  have,  with  n  as  a 
datum  level, 


Using  the  ft.,  lb.,  and  sec.,  we  have 

h  =  700  ft.,     vm*  -^  fy  =  0.30  ft., 
while 

,       14.7  X  144  ,  pn      1000  X  144 

o  =  -  —  -  =  38.47  ft.,  and  —  =  -  ?5  -  = 
55  y  ™ 

Hence,  in  eq.  (1), 

0.30  +  38.5  +  700  =  2618  -  4/.  ^ 


.. 
ft. 


Solving  for/",  we  have  f  =  .00485  (whereas  for  water,  with 
v  =  4.4  ft.  per  sec.  and  d  =  £  ft.,  the  table,  p.  709,  gives 
f  =  .00601. 
If  the  Y  °^  the  petroleum  had  been  50  Ibs.  per  cubic  foot, 

instead  of  55,  we  should  have  obtained  £*  =  2880  feet  and  f 
=  .0056. 

§18.  Flow  through  a  Long  Straight  Cylindrical  Pipe,  including 
k>th  friction-head  and  entrance  loss  of  head  (corners  not  rounded); 

reservoir  large.  Fig.  573. 
The  jet  issues  directly 
from  the  end  of  the  pipe, 
in  parallel  filaments,  into 
the  air,  and  therefore 
::-v  has  same  section  as  pipe; 

FIG.  573.  ~—  i  t  f   ,1       .   , 

hence,  also,  vm  of  the  jet 
=  v  in  the  pipe  (which  is  assumed  to  be  running  full),  and  i& 


COEFFICIENT   OF   LIQUID   FRICTION.  711 


therefore  the  velocity  to  be  used  in  the  loss  of  head  t,E  -  -  at 


the  entrance  ^(§  515), 

Taking  m  and  n  as  in  figure  and  applying  Bernoulli's 
Theorem  (§  474),  with  m  as  datum  level  for  the  potential  heads 
zm  and  zn  ,  we  have 


Three  different  problems  may  now  be  solved  : 

First,  required  the  head  h  to  keep  up  a  flow  of  given  volume 

=  Q  per  unit  of  time  in  a  pipe  of  given  length  I  and  diameter 

=  d. 
From  the  equation  of  continuity  we  have 


40 

/.  veloc.  of  jet,  which  =  veloc.  in  pipe,  =  vm=  —  ^-.  .     .     (2) 

7t(L 

Having  found  vm  =  v,  from  (2),  we  obtain  from  (1)  the  re- 
quired A,  thus  : 


Now  £E  =  0.505  if  a  =  90°  (see  §  515),  while  /  may  be 
taken  from  the  table,  §  517,  for  the  given  diameter  and  com- 
puted velocity  [ym  =  v,  found  in  (2)],  if  the  pipe  is  clean;  if 
not  clean,  see  end  of  §  517,  for  slightly  tuberculated  and  for 
foul  pipes.* 

Secondly.  Given  the  head  A,  and  the  length  I  and  diameter 
d  of  pipe,  required  the  velocity  in  the  pipe,  viz.,  -y,  =  vm  ,  that 
of  jet;  also  the  volume  delivered  per  unit  of  time,  Q.  Solv- 
ing eq.  (1)  for  vm,  we  have 


vm=          -  -Vtojh;      ...    (4) 


*  "  Hydraulic  Tables,"  for  friction-head  of  water  in  pipes,  by  Prof.  G. 
S.  Williams  and  Mr.  Allen  Hazen  (New  York,  1905,  John  Wiley  &  Sons), 
cover  the  cases  of  pipes  in  various  states  of  tuberculation,  etc. 


712  MECHANICS   OF   ENGINEERING. 

whence  Q  becomes  known,  since 

(5) 


.  —  The  first  radical  in  (4)  might  for  brevity  be  called 
a  coefficient  of  velocity,  0,  for  this  case.  Since  the  jet  has  the 
same  diameter  as  the  pipe,  this  radical  may  also  be  called  a 
coefficient  of  effiux.~\ 

Since  in  (4)  /"depends  on  the  unknown  v  as  well  as  on  the 
known  d,  we  must  first  put/"  =  .006  for  a  first  approximation 
for  vm  ;  then  take  a  corresponding  value  for  f  and  substitute 
again  ;  and  so  on. 

Thirdly,  knowing  the  length  of  pipe  and  the  head  A,  we 
wish  to  find  the  proper  diameter  d  for  the  pipe  to  deliver  a 
given  volume  Q  of  water  per  unit  of  time.  Now 


which  substituted  in  (1)  gives 


that  is, 


.'.  d  =  1 


As  the  radical  contains  d,  we  first  assume  a  value  for  d, 
with/=  .006,  and  substitute  in  (7).  With  the  approximate 
7alue  of  d  thus  obtained,  we  substitute  again  with  a  new  value 
f  or  f  based  on  an  approximate  v  from  eq.  (6)  (with  d  =  its 
first  approximation),  and  thus  a  still  closer  value  for  d  is  de- 
rived ;  and  so  on.  (Trautwine's  Pocket-book  contains  a  table 
of  fifth  roots  and  powers.)  If  I  is  quite  large,  we  may  put 
d  =  0  for  a  first  approximation.  In  connection  with  these 
examples,  see  last  figure. 


LONG   PIPES.  713 

EXAMPLE  1.  —  What  head  h  is  necessary  to  deliver  120  cub. 
ft.  of  water  per  minute  through  a  clean  straight  iron  pipe  140 
ft.  long  and  6  in.  in  diameter  ? 

Prom  eq.  (2),  with  ft.,  lb.,  and  sec.,   we  have 

v  =  vm  =  [4  X  WJ  •*-  <*)3=  10-18  ft-  Per  sec- 

Now  for  v  =  10  ft.  per  sec.  and  d  =  \  ft.,  we  find  (in  table, 
§  517)  /=  .00549  ;  and  hence,  from  eq.  (3), 


of  which  total  head,  as  we  may  call  it,  1.60  ft.  is  used  in  pro- 
ducing the  velocity  10.18  ft.  per  sec.  (i.e.,  vm*  -r-  %g  =  1.60  ft.), 

while  0.808  ft.  [=  C^^j  is  lost  at  the  entrance  ^(with  a  = 

90°),  and  9.82  ft.  (friction-head)  is  lost  in  skin-friction. 

EXAMPLE  2.  —  [Data  from  Weisbach.]  Required  the  de- 
livery, Q,  through  a  straight  clean  iron  pipe  48  ft.  long  and 
2  in.  in  diameter,  with  5  ft.  head  (=  h).  v,  =vm,  being  un- 
known, we  first  take/*=  .006  and  obtain  [eq.  (4)] 


/ 


1  +  .505 


=  6.18  ft.  per  sec. 

From  the  table,  §  517,  for  v  =  6.2  ft.  per  sec.  and  d  =  2  in.f 
/=  .00638,  whence 


.   4  X  .00638  X  48  ^2  x  32'2  x  5 
1  +  .505  -| j 

=  6.04  ft.  per  sec., 
which  is  sufficiently  close.     Then,  for  the  volume  per  second, 

Q  =  ~  d?vm  =  |7r(i-)26.04  —  0.1307  cub.  ft.  per  sec. 


714  MECHANICS   OF   ENGINEERING, 

[Weisbach's  results  in  this  example  are 
vm  =  6.52  ft.  per  sec. 
and  Q  =  0.1420  cub.  ft.  per  sec., 

but  his  values  for/  are  slightly  different.] 

EXAMPLE  3. — [Data  from  Weisbach.]  What  must  be  the 
diameter  of  a  straight  clean  iron  pipe  100  ft.  in  length,  which 
is  to  deliver  Q  —  £  of  a  cubic  foot  of  water  per  second  under 
5ft.  head  (=  A)? 

With/=  .006  (approximately),  we  have  from  eq.  (7),  put- 
ting d  =  0  under  the  radical  for  a  first  trial  (ft,  lb.,  sec.), 


4:0 

whence  v  =  —  ¥-  =  7  ft.  per  sec. 

nd 

For  d  =  3.6  in.  and  v  =  7  ft.  per  sec.,  we  find/=  .00601  ; 
whence,  again, 


-,       s  /L505^X  .30  +4  X  .00601  X  100    (±  x"?V     n  «oA*  . 
=  V~  2  X  32.2  X  5  ""•VirV 

and  the  corresponding  v  =  6.06  ft. 

For  this  d  and  v  we  fiud/=  .00609,  whence,  finally, 


[Weisbach's  result  is  d  =  .318  ft.] 

519.  Ch6zy's  Formula.  —  If,  in  the  problem  of  the  preceding 
paragraph,  the  pipe  is  so  long,  and  therefore  I  :  d  so  great^ 
that  4/7  -r-  d  in  eq.  (3)  is  very  large  compared  with  1  -f  £E) 
we  may  neglect  the  latter  term  without  appreciable  error; 
whence  eq.  (3)  reduces  to 

h  =  4~  -      •  • 


CH^ZY'S  FORMULA.  715 

which  is  known  as  Cheztfs  Formula.    For  example,  if  I  =  1000 

ft.  and  d  =  2  in.  =  -J-  ft.,  and/approx.  —  .006,  we  have  4=f--  = 

cL 

144,  while  1  +  CE  for  square  corners  —  1.505  only. 
If  in  (8)  we  substitute 


(8)  reduces  to 


so  that  for  a  very  long  pipe,  considering  f  as  approximately 
constant,  we  may  say  that  to  deliver  a  volume  =  Q  per  unit 
of  time  through  a  pipe  of  given  length  =  I,  the  necessary  head, 
A,  is  inversely  proportional  to  the  fifth  power  of  the  diameter. 

And  again,  solving  (9)  for  Q,  we  find  that  the  volume  con- 
veyed per  unit  of  time  is  directly  proportional  to  the  fifth  power 
of  the  square  root  of  the  diameter  /  directly  proportional  to 
the  square  root  of  the  head  ;  and  inversely  proportional  to  the 
square  root  of  the  length.  (Not  true  for  short  pipe  ;  see  above 
example.) 

If  we  conceive  of  the  insertion  of  a  great  number  of  piezom- 
eters along  the  long  straight  pipe,  of  uniform  section,  now 
under  consideration,  the  summits  of  the  respective  water 
columns  maintained  in  them  will  lie  in  a  straight  line  joining 
the  discharging  (into  the  air)  end  of  the  pipe  with  a  point  in 
the  reservoir  surface  vertically  over  the  inlet  extremity  (prac- 
tically so),  and  the  "  slope"  of  this  line  (called  the  Hydraulic 
Grade  Line  or  Gradient)^  i.e.,  the  tangent  (or  sine  ;  the  angle 
is  so  small,  generally)  of  the  angle  which  it  makes  with  the 

horizontal  is  =  —  ,  and  may  be  denoted  by  s.     Putting   also 

i 

\d  —  R  =  the  hydraulic  radius  of  the  section  of  the  pipe,  and 
vm  =  v  =  velocity  in  pipe,  we  may  transform  eq.  (8)  into 


or,    v  =  A(£s)*9    .    .    .(10) 


716 


MECHANICS    OF   ENGINE  EKING. 


which  is  the  form  by  which  Mr.  Hamilton  Smith  (see  §  506) 
interprets  all  the  experiments  quoted  by  him  on  long  pipes. 
As  to  notation,  however,  he  uses  n  for  A,  and  r  for  R.  With 
the  foot  and  second  as  units,  the  quantity  A  (not  an  abstract 
number)  varies  approximately  between  60  and  140.  For  a 
given  A  we  easily  tind  the  corresponding  f  from  the  relation 

f=-^.     If  the  pipe  discharges  under  water,  h  =  the  differ- 
A 

ence  of  elevation  of  the  two  reservoirs.  If  the  pipe  is  not 
horizontal,  the  use  of  the  length  of  its  horizontal  projection 

instead  of  its  actual  length  in  the  relation  s  =  -  occasions    an 

error,  but  it  is  in  most  cases  insignificant. 

Similarly,  if  a  steady  flow  is  going  on  in  a  long  pipe  of  uni- 
form section,  at  the  extremities  of  any  portion  of  which  we 
have  measured  the  piezometer  heights  (or  computed  them 
from  the  readings  of  steam  or  pressure  gauges),  we  may  apply 
eq.  (9),  putting  for  A  the  difference  of  level  of  the  piezometer 
summits,  and  for  I  the  length  of  the  pipe  between  them. 

520.  Coefficient  /  in  Fire-engine  Hose.* — Mr.  Geo.  A.  Ellis, 
C.E.,  in  his  little  book  on  "  Fire-streams,"  describing  experi- 
ments made  in  Springfield,  Mass.,  gives  a  graphic  comparison 
(p.  45  of  his  book)  of  the  friction -heads  occurring  in  rubber 
hose,  in  leather  hose,  and  in  clean  iron  pipe,  each  of  2^-  in. 
diameter,  with  various  velocities;  on  which  the  following  state- 
merits  may  be  based  :  That  for  the  given  size  of  hose  and 
pipe  (d  —  2-J  in.)  the  coefficient  f  for  the  leather  and  rubber 
hose  respectively  may  be  obtained  approximately  by  adding  to 
f  for  clean  iron  pipe  (and  a  given  velocity)  the  per  cent  of 
itself  shown  in  the  accompanying  table. 

EXAMPLE. — For  a  clean  iron  pipe 

2-J  in.  diam.,  for  a  velocity  —  10  ft. 

per  sec.,  we  have,  from  §  517,  f  •=. 

.00593.     Hence  for  a  leather  hose  of 

the  same  diameter,  we  have,  for  -y  = 

10  ft.  per  sec., 


Velocity 

Rubber 

Leather 

ft.  per 
sec. 

hose  2$  in. 
diam. 

hose  2£  in. 
diam. 

3.0 

50# 

300$ 

6.5 

20 

80 

10 

16 

43 

13 

12.5 

82 

16 

12 

30 

/=  .00593  +  .43  X  .00593  =  .00848. 


*  For  the  most  recent  and  exhaustive  experiments  in  this  direction  see 
Mr.  J.  R.  Freeman's  "Hydraulics  of  Fire-streams"  in  the  Transac.  Am. 
Soc.  C.  E.  for  1889.  (Also  p.  832  of  this  work.) 


PRESSURE-ENERGY.  717 

621.  Bernoulli's  Theorem  as  an  Expression  of  the  Conservation 
of  Energy  for  the  Liquid  Particles. — In  any  kind  of  flow  with- 
out friction,  steady  or  not,  in  rigid  immovable  vessels,  the 
aggregate  potential  and  kinetic  energy  of  the  whole  mass  of 
liquid  concerned  is  necessarily  a  constant  quantity  (see  §§  148 
and  149),  but  individual  particles  (as  the  particles  in  the  sink- 
ing free  surface  of  water  in  a  vessel  which  is  rapidly  being 
emptied)  may  be  continually  losing  potential  energy,  i.e., 
reaching  lower  and  lower  levels,  without  any  compensating  in- 
crease of  kinetic  energy  or  of  any  other  kind  ;  but  in  a  steady 
flow  without  friction  in  rigid  motionless  vessels,  we  may  state 
that  the  stock  of  energy  of  a  given  particle,  or  small  collection 
of  particles,  is  constant  during  the  flow,  provided  we  recognize 
a  third  kind  of  energy  which  may  be  called  Pressure-energy, 
or  capacity  for  doing  work  by  virtue  of  internal  fluid  pressure  ; 
as  may  be  thus  explained  : 

In  Fig.  574  let  water,  with  a  very  slow  motion  and  under  a 
pressure  p  (due  to  the  reservoir-head  -f-  atmosphere-head  be- 
hind it),  be  admitted  behind  a  pis- 
ton 1;he    space   beyond  which   is 
k.          vacuous.       Let    s  —  length      of 
stroke,  and  F  =  the  area  of  pis- 
ton.     At  the  end  of  the   stroke, 


^  >^—  by  motion  of  proper  valves,  com- 

7/VAC-   munication   with  the  reservoir  is 
Fia  574>  cut  off  on  the  left  of  the  piston 

and  opened  on  the  right,  while  the  watexr  in  the  cylinder  now  on 
the  left  of  the  piston  is  put  in  communication  with  the  vacu- 
ous exhaust-chamber.  As  a  consequence  the  internal  pressure 
of  this  water  falls  to  zero  (height  of  cylinder  small),  and  on 
the  return  stroke  is  simply  conveyed  out  of  the  cylinder, 
neither  helping  nor  hindering  the  motion.  That  is,  in  doing 
the  work  of  one  stroke,  viz., 

W  =  force  X  distance  =  Fp  X  s  =  Fps, 

a  volume  of  water  V=  Fs,  weighing  Fsy  (Ibs.  or  other  unit), 
has  been  used,  and,  in  passing  through  the  motor,  has  experi- 
enced no  appreciable  change  in  velocity  (motion  slow),  and 


718  MECHANICS    OF   ENGINEERING. 

therefore  no  change  in  kinetic  energy,  nor  any  change  of  level, 
and  hence  no  change  in  potential  energy,  ~but  it  has  given  up 
all  its  pressure.  (See  §  409  for  y.) 

Now  TF,  the  work  obtained  by  the  consumption  of  a  weight 
=  G  =  Vy  of  water,  may  be  written 


W=  Fps  =  Fsp  =  Vp  =  Vy.  =  G.     .    .    (1) 

Hence  a  weight  of  water  =  O  is  capable  of  doing  the  work 
G  X  -  =  G-  X  head  due  to  pressure  p,  i.e.,  =  G  X  pressure- 

head^  in  giving  up  all  its  pressure  p  •  or  otherwise,  while  still 
having  a  pressure  p,  a  weight  G-  of  water  possesses  an  amount 
of  energy,  which  we  may  call  pressure-energy,  of  an  amount 

=  G'—  ,  where  y  —  the  heaviness  (§  Y)  of  water,  and  -  =  a 

Y  Y 

height,  or  head,  measuring  the  pressure  p  ;  i.e.,  it  equals  the 
pressure-head. 

We  may  now  state  Bernoulli's  Theorem  without  friction  in 
a  new  form,  as  follows  :  Multiplying  each  term  of  eq.  (7), 
§  451,  by  Qy,  the  weight  of  water  flowing  per  second  (or  other 
time-unit)  in  the  steady  flow,  we  have 

Qr%f+Qr^  +  Qr^=Qr^+Qrf+Qrzn-   (2) 

But  Qy  -2-  =  _M2_-ym2  =  -J-  X  mass  flowing  per  time-unit  X 
2g       2   g 

square  of  the  velocity  =  the  kinetic  energy  inherent  in  the 
volume  Q  of  water  on  passing  the  section  m,  due  to  the  veloc- 

ity at  m.     Also,  Qy  ±—  =  the  pressure-energy  of  the  volume 

Q  at  m,  due  to  the  pressure  at  m  ;  while  Qyzm  =  the  potential 
energy  of  the  volume  Q  at  m  due  to  its  height  zm  above  the 
arbitrary  datum  plane.  Corresponding  statements  may  be 
made  for  the  terms  on  the  right-hand  side  of  (2)  referring  to 
the  other  section,  n,  of  the  pipe.  Hence  (2)  may  be  thus  read  : 
The  aggregate  amount  of  energy  (of  the  three  kinds  mentioned) 
resident  in  the  particles  of  liquid  when  passing  section  m  is 


LOSS   OF   ENERGY. 


719 


equal  to  that  when  passing  any  other  section,  as  n  ;  in  steady 
flow  without  friction  in  rigid  motionless  vessels  y  that  is,  the 
store  of  energy  is  constant. 

522.  Bernoulli's  Theorem  with  Friction,  from  the  Standpoint  of 
Energy. — Multiply  each  term  in  the  equation  of  §  516  by  Qy, 
as  before,  and  denote  a  loss  of  head  or  height  of  resistance  due 
to  any  cause  by  hr ,  and  we  have 


~  +  Or*.  - 


(3) 


Each  term  Qyhr(e.g.^  Qy  4f  — due  to  skin-friction  in  a 

d  2(/ 

long  pipe,  and  Qy  CE-—  due  to  loss  of  head  at  the  reservoir 

entrance  of  a  pipe)  represents  a  loss  of  energy,  occurring  between 
any  position  n  and  any  other  position  m  down-stream  from  n, 
but  is  really  still  in  existence  in  the  form  of  heat  generated  by 
the  friction  of  the  liquid  particles  against  each  other  or  the 
sides  of  the  pipes. 

As  illustrative  of  several  points  in  this  connection,  consider 
two  short  lengths  of  pipe  in 
Fig.  575,  A  and  B,  one  offering 
a  gradual,  the  other  a  sudden, 
enlargement  of  section,  but 
otherwise  identical  in  dimen- 
sions. We  suppose  them  to 
occupy  places  in  separate  lines 
of  pipe  in  each  of  which  a 
steady  flow  with  full  cross-sec- 
tions is  proceeding,  and  so  reg- 
ulated that  the  velocity  and  in- 
ternal pressure  at  n,  in  A,  are 
equal  respectively  to  those  at  n  FWK  575. 

in  B.     Hence,  if  vacuum  piezometers  be  inserted  at  71,  the 


720  MECHANICS    OF    ENGINEERING. 

smaller  section,  the  water  columns  maintained  in  them  by  the 
internal  pressure  will  be  of  the  same  height,  -£?,  for  both  A 

and  B.  Since  at  m,  the  larger  section,  the  sectional  area  is  the 
same  for  both  A  and  B,  and  since  Fn  in  A  =  Fn  in  B,  so  that 
QA  =  QBI  hence  vm  in  A  =  vm  in  B  and  is  less  than  vn . 

Now  in  B  a  loss  of  head  occurs  (and  hence  a  loss  of  energy) 
between  n  and  m,  but  none  in  A  (except  slight  friction-head); 
hence  in  A  we  should  find  as  much  energy  present  at  m  as  ai 
n,  only  differently  distributed  among  the  three  kinds,  while  at 
m  in  B  the  aggregate  energy  is  less  than  that  at  n  in  B. 

As  regards  kinetic  energy,  there  has  been  a  loss  between  n 
and  m  in  both  A  and  B  (and  equal  losses),  for  vm  is  less  than 
vn.  As  to  potential  energy,  there  is  no  change  between  n  and 
m  either  in  A  or  B,  since  n  and  m  are  on  a  level.  Hence  if 
the  loss  of  kinetic  energy  in  B  is  not  compensated  for  by  an 
equal  gain  of  pressure-energy  (as  it  is  in  A),  the  pressure-head 


— )  at  m  in  B  should  be  less  than  that  f  — )    atmin  A.    Ex- 

IB  \y)A 

periment  shows  this  to  be  true,  the  loss  of  head  being  due  to 
the  internal  friction  in  the  eddy  occasioned  by  the  sudden  en- 
largement ;  the  water  column  at  m  in  B  is  found  to  be  of  a 
less  height  than  that  at  m  in  A,  whereas  at  n  they  are  equal. 
(See  p.  467  of  article  "  Hydromechanics"  in  the  Ency.  Bri- 
tannica  for  Mr.  Froude's  experiments.) 

In  brief,  in  A  the  loss  of  kinetic  energy  has  been  made  up 
in  pressure-energy,  with  no  change  of  potential  energy,  but  in 
B  there  is  an  actual  absolute  loss  of  energy  —  Qyhr ,  or 

v  2 
=  QyC,  -^,  suffered  by  the  weight  Qy  of  liquid.     The  value 

of  C  in  this  case  and  others  will  be  considered  in  subsequent 
paragraphs. 

Similarly,  losses  of  head,  and  therefore  losses  of  energy, 
occur  at  elbows,  sharp  bends,  and  obstructions,  causing  eddies 
and  internal  friction,  the  amount  of  each  loss  for  a  given 

weight,  £,  of  water  being  =  Ghr  =  #C--  ;  hr  =  C  —  being 
the  loss  of  head  occasioned  by  the  obstruction  (§  474).  It  is 


SUDDEN   ENLARGEMENTS  IN   PIPES. 

therefore  very  important  in  transmitting  water  through  pipes 
for  purposes  of  power  to  use  all  possible  means  of  preventing 
disturbance  and  eddying  among  the  liquid  particles.  E.g., 
sharp  corners,  turns,  elbows,  abrupt  changes  of  section,  should 
be  avoided  in  the  design  of  the  supply-pipe.  The  amount  of 
the  losses  of  head,  or  heights  of  resistance,  due  to  these  various 
causes  will  now  be  considered  (except  skin -friction,  already 
treated).  Each  such  loss  of  head  will  be  written  in  the  form 

v2 
C  —  ,  and  we  are  principally  concerned  with  the  value  of  the 

2^ 
abstract  number  £,  or  coefficient  of  resistance,  in  each  case. 

The  velocity  v  is  the  velocity,  known  or  unknown,  where  the 
resistance  occurs;  or  if  the  section  of  pipe  changes  at  this 
place,  then  v  —  velocity  on  the  down-stream  section.  The  late 
Professor  "Weisbach,  of  the  mining-school  of  Freiberg,  Saxony, 
was  one  of  the  most  noted  experimenters  in  this  respect,  and 
will  be  frequently  quoted. 

523.  Loss  of  Head  Due  to  Sudden  (i.e.,  Square-edged)  Enlarge- 
ment. Borda's  Formula, — Fig.  576.  An  eddy  is  formed  in  the 
i  angle  with  consequent  loss  of  energy.  Since 
S?  each  particle  of  water  of  weight  =  Gl ,  arriving 
:g  with  the  velocity  vl  in  the  small  pipe,  may  be 
|L^  jfi^lj  considered  to  have  an  impact  against  the  base 
FIO.  576.  of  the  infinitely  great  and  more  slowly  moving 
column  of  water  in  the  large  pipe,  and,  after  the  impact, 
moves  on  with  the  same  velocity,  -ya,  as  that  column,  just  as 
occurs  in  inelastic  direct  central  impact  (§  60),  we  may  find 
the  energy  lost  by  this  particle  on  account  of  the  impact  by 
eq.  (1)  of  §  138,  in  which,  putting  M^  =  Gt  -f-  </,  and  M^  — 
*TI  -T-  g  =  mass  of  infinitely  great  body  of  water  in  the  large 
pipe,  so  that  J/2  =  00,  we  have 

Energy  lost  by  particle  =  6rl  ^  ~  v*>-t      .     .    £r 

and  the  corresponding 

Loss  of  head  =         ^)!, 


722  MECHANICS   OF  ENGINEERING. 

which,  since  F^  =  F^ ,  may  be  written 

['F         "l*^9 
Loss  of  head  in  sudden  enlargement  —     -=f  —  1     ~. 

That  is,  the  coefficient  C  for  a  sudden  enlargement  is 


.     (2) 


FI  and  F9  are  the  respective  sectional  areas  of  the  pipes.     Eq. 
(2)  is  Bordtfs  Formula. 

NOTE. — Practically,  the  flow  cannot  always  be  maintained 
with  full  sections.  In  any  case,  if  we  assume  the  pipes  to  be 
running  full  (once  started  so),  and  on  that  assumption  compute 
the  internal  pressure  at  Fl ,  and  find  jt  to  be  zero  or  negative, 
the  assumption  is  incorrect.  That  is,  unless  there  is  some 
pressure  at  F1  the  water  will  not  swell  out  laterally  to  fill  the 
large  pipe. 

EXAMPLE. — Fig.  577.     In  the  short  tube  AB  containing  a 
sudden  enlargement,  we  have  given  Fy  =  Fm  =  6  sq.  inches, 
F}  —  4  sq.  inches,  and  h  —  9  feet.     Ee- 
quired  the  velocity  of  the  jet  at  m  (in 
the  air,  so  that  j?w  -±-  y  =  ~b  —  34  ft.),  if 
the  only  loss  of  head  considered  is  that 
due  to  the  sudden  enlargement  (skin- 
friction  neglected,  as  the  tube  is  short ; 
the  reservoir  entrance  has  rounded  cor- 
FIG.  577.  ners).     Applying  Bernoulli's  Theorem 

to  m  as  down-stream  section,  and  n  in  reservoir  surface  as  up- 
stream position  (datum  level'at  m),  we  have 


Bat,  here,  va  =  vm  ; 


From  eq.  (3)  we  have 


(5) 


C  =  (!-!)«  =  0.25, 


SUDDEN   ENLARGEMENT  IN  PIPE.  723 

and  finally  (ft.,  lb.,  sec.) 


vm  =  \  V*  X  32.2  X  9  =  0.895  i/2x  32.2x9 

=  21.55  ft.  per  sec. 

(The  factor  0.895  might  be  called  a  coefficient  of  velocity  for 
this  case.)     Hence  the  volume  of  flow  per  second  is 

Q  =  Fmvm  =  ylr  X  21.55  =  0.898  cub.  ft.  per  sec. 


We  have  so  far  assumed  that  the  water  fills  both  parts  of  the 
tube,  i.e.,  that  the  pressure^,  at  Fl  ,  is  greater  than  zero  (see 
foregoing  note).  To  verify  this  assumption,  we  compute  j?, 
by  applying  Bernoulli's  Theorem  to  the  centre  of  Ft  as  down- 
stream position  and  datum  plane,  and  n  as  up-stream  position, 
with  no  loss  of  head  between,  and  obtain 


But  since  Flvl  =  Fj)t  ,  we  have 


and  hence  the  pressure-head  at  Fl  (substituting  from  equations 

above)  is 


and  /.  pl  =  fj  of  14.7  =  11.6  Ibs.  per  sq.  inch,  which  is 
greater  than  zero  ;  hence  efflux  with  the  tube  full  in  both  parts 
can  be  maintained  under  9  ft.  head. 

If,  with  Fl  and  F^  as  before  (and  .*.  £),  we  put  pl  =  0,  and 
solve  for  A,  we  obtain  h  =  42.5  ft.  as  the  maximum  head 
under  which  efflux  with  the  large  portion  full  can  be  secured. 

524.  Short  Pipe,  Square-edged  Internally.  —  This  case,  already 


724 


MECHANICS    OF   ENGINEERING. 


treated  in  §§  507  and  515  (see  Fig,  578 ;  a  repetition  of  560), 
presents  a  loss  of  head  due  to  the  sudden  enlargement  from 
the  contracted  section  at  iri  (whose  sec- 

tional  area  may  be  put  =  CF,  C  being 

an   unknown    coefficient,    or   ratio,  of 
•v          ,x  i      contraction)  to   the   full  section   F  of 
a?_L:r  the  pipe.     From  §  515  we  know  that 


FIG.  578. 


^~Ifr^^^  the  loss  of  head  due  to  the  short  pipe 

^^^.  v     2 

<£m.  is  Ar  =  C*|j  (for  a  =  90°),  in  which 
£E  =  0.505  ;  while  from  Borda's  For- 


[~     ff1  ~~la 

mula,  §  523,  we  have  also  £E  =     -^  —  1     .    Equating  these, 
we  find  the  coefficient  of  internal  contraction  at  m'  to  be 


===  =  0.584, 


or  about  0.60  (compare  with  0=  .64  for  thin-plate  contrac- 
tion, §  495).  It  is  probably  somewhat  larger  than  this  (.584), 
since  a  small  part  of  the  loss  of  head,  Ar,  is  due  to  friction  at 
the  corners  and  against  the  sides  of  the  pipe. 

By  a  method  similar  to  that  pursued  in  the  example  of 
§  523,  we  may  show  that  unless  A  is  leas  than  40  feet,  about, 
the  tube  cannot  be  kept  full,  the  discharge  being  as  in  Fig. 
551.  If  the  efflux  takes  place  into  a  "  partial  vacuum,"  this 
limiting  value  of  h  is  still  smaller.  Weisbach's  experiments 
confirm  these  statements  (but  those  in  the  C.  U.  Hyd.  Lab. 
seem  to  indicate  that  the  limiting  value  for  A  in  the  first  case 
is  about  50  ft.). 


525.  Diaphragm  in  a  Cylindrical  Pipe.—  Fig.  579.  The  dia- 
phragm being  of  "thin  plate," 
let  the  circular  opening  in  it 
(concentric  with  the  pipe)  have 
an  area  =  F,  while  the  sectional 
area  of  pipe  — 


Beyond  F,  the 


FIG.  579. 


stream  contracts  to  a  section  of  area  =  CF  =  Fl ,  in  enlarging 


SHOET   PIPE.      DIAPHRAGM 


PIPE. 


725 


which  to  Jill  the  section  F^ ,  of  pipe,  a  loss  of  head  occurs 
which  by  Borda's  Formula,  §  523,  is 


where  va  is  the  velocity  in  the  pipe  (supposed  full).  Of  course 
Fl  (or  CF)  depends  on  F\  but  since  experiments  are  necessary 
in  any  event,  it  is  just  as  well  to  give  the  values  of  C  itself,  as 
determined  by  Weisbach's  experiments,  viz. : 


For?-  =.10 

-Pa 

.20 

.30 

.40 

.50 

.60 

.70 

.80 

.90 

1.00 

C  =  226. 

48. 

17.5 

7.8 

3.7 

1.8 

.8 

.3 

.06 

0.00 

By  internal  lateral  filling,  Fig.  580,  the  change  of  section 
may  be  made  gradual  and  eddying 
prevented  ;  and  then  but  little  loss 
of  head  (and  therefore  little  loss  of 
energy)  occurs,  besides  the  slight 
amount  due  to  skin-friction  along  FIG.  580. 

this  small  surface.  On  p.  467  of  the  article  Hydromechanics 
in  the  Encyclopaedia  Britannica  may  be  found  an  account  of 
experiments  by  Mr.  Froude,  illustrating  this  fact. 

526.  "The  Venturi  Water-meter."— The  invention  bearing 
this  name  was  made  by  Mr.  Clemens  Herschel  (see  Trans.  Am- 
Soc.  Civ.  Engineers,  for  November  1887),  and  may  be  de- 
scribed as  a  portion  of  pipe  in  which  a  gradual  narrowing  of 
section  is  immediately  succeeded  by  a  more  gradual  enlarge- 
ment, as  in  Fig.  580 ;  but  the  dimensions  are  more  extreme. 
During  the  flow  the  piezometer-heights  are  observed  at  the 
three  positions  r,  n,  and  m  (see  below),  and  the  rate  of  dis- 
charge may  be  computed  as  follows :  Referring  to  Fig.  580, 
let  us  denote  by  r  the  (up-stream)  position  where  the  narrow- 
ing of  the  pipe  begins,  and  by  m  that  where  the  enlargement 
ends,  while  n  refers  to  the  narrowest  section.  Fm  =  Fr. 

Applying  Bernoulli's  Theorem  to  sections  r  and  n,  assuming 


726  MECHANICS    OF   ENGINEERING. 

no  loss  of  head  between,  we  have,  as  the  principle  of  the  ap- 
paratus, 


:  ,         _=n   ,        . 

Y  *~fy  "  r  rV 


whence,  since  Frvr  =  Fnvn  , 


in  which  0  represents  the  first  radical  factor.     0  should  differ 

W 
but  little  from  unity  with  -^  small  (and  such  was  found  to  be 

J^r 

the  case  by  experiment).  Its  theoretical  value  is  constant  and 
greater  than  unity.  In  the  actual  use  of  the  instrument  the 

—  and  —  are  inferred  from  the  observed  piezometer- heights 
Y  Y 

yr  and  yn  (since  —  =  yr  -\-  J,  and  —  =  yn -\-  &,  b  being  =  34  ft.), 

and  then  the  quantity  flowing  per  time-unit  computed,  from 
Q  =  Fnvn ,  vn  having  been  obtained  from  eq.  (2).  This  pro- 
cess gives  a  value  of  Q  about  four  per  cent  in  excess  of  the 
truth,  according  to  the  second  set  of  experiments  mentioned 
below,  if  vn  =35  ft.  per  sec. ;  but  only  one  per  cent  excess  with 
vn  =  5  or  6  ft.  per  sec. 

Experiments  were  made  by  Mr.  Herschel  on  two  meters  of 
this  kind,  in  each  of  which  Fn  was  only  one  ninth  of  Fr ,  a 
ratio  so  extreme  that  the  loss  of  head  due  to  passage  through 
the  instrument  is  considerable.  E.g.,  with  the  smaller  appara- 
tus, in  which  the  diameter  at  n  was  4  in.,  the  loss  of  head  be- 
tween r  and  m  was  10  or  11  ft.,  when  the  velocity  through  n 
was  50  ft.  per  sec.,  those  at  other  velocities  being  roughly  pro- 
portional to  the  square  of  the  velocity.  In  the  larger  instru- 
ment dn  was  3  ft.,  and  the  loss  of  head  between  r  and  m  was 
much  more  nearly  proportional  to  the  square  of  the  velocity 
than  in  the  smaller.  (E.g.,  with  vn  =  34.56  ft.  per  sec.  the 
loss  of  head  was  2.07  ft.,  while  with  vn  =  16.96  ft.  per  sec.  it 


SUDDEN   DIMINUTION   OF   SECTION   OF   PIPE. 


727 


was  0.49  ft.)     The  angle  of  divergence  was  much  smaller  in 
these  meters  than  that  in  Fig.  580. 

527.  Sudden  Diminution  of  Cross-section,  Square  Edges. — Fig. 
581.     Here,  again,  the  resistance  is 


Ua_ 

due  to  the  sudden  enlargement  from  JE:"^  -y_->^ 

the  contracted  section  to  the  full  sec-  -^.-Jsj  5>j  ^jr^dfz^'1'* 

tion  F^  of  the  small  pipe,  so  that  in  ^~~~-~^-^\\ 

the  loss  of  head,  by  Borda's  formula,  FIG.  581. 


-im  •  .  -  .  a) 


the  coefficient 


-1    =   4r-l    .  .     .     (2) 


depends  on  the  coefficient  of  contraction  (7;  but  this  latter  is 
influenced  by  the  ratio  of  F^  to  F9,  the  sectional  area  of  the 
larger  pipe,  C  being  about  .60  when  F^  is  very  large  (i.e., 
when  the  small  pipe  issues  directly  from  a  large  reservoir  so 
that  F^  :  FQ  practically  =  0).  For  other  values  of  this  ratio 
Weisbach  gives  the  following  table  for  O9  from  his  own  ex- 
periments : 


FOT  F*  :  F0  =  .10 

.20 

.30 

.40 

.50 

.60 

.70 

.80 

.90 

1.00 

G=  .624 

.632 

.643 

.659 

.681 

.712 

.755 

.813 

.892 

1.00 

C   being  found,   we  compute    C    from  eq.   (2)  for  use  in 
eq.  (1). 

528.  Elbows, — The  internal  disturbance  caused  by  an  elbow, 
Fig.  582  (pipe  full,  both  sides  of  elbow),  occasions  a  loss  of    • 
head 


FIG.  582.  in  which,  according  to  Weisbach's  experi- 

ments with  tubes  3  centims.,  i.e.  1.2  in.,  in  diameter,  we  may 
put 


728 


MECHANICS    OF   ENGINEERING. 


For  a  =    20° 

40° 

60° 

80° 

90° 

100° 

110° 

120° 

130° 

140° 

C=  .046 

.139 

.364 

.740 

.984 

1.26 

1.556 

1.86 

2.16 

2.43 

computed  from  the  empirical  formula  ; 

C  =  .9457  sin*  %a  +  2.047  sin4  \ot ; 

v  is  the  velocity  in  pipe ;  a  as  in  figure.     For  larger  pipes  C 
would  probably  be  somewhat  smaller ;  and  vice  versa. 

If  the  elbow  is  immediately  succeeded  by  another  in  the  same 
plane  and  turning  the  same  way,  Fig.  583,  the 
loss  of  head  is  not  materially  increased,  since 
the  eddying  takes  place  chiefly       v 
in  the  further  branch  of  the 
second  elbow ;  but  if  it  turns 
in  the  reverse  direction,  Fig. 
584,  but   still    in    the    same 
plane,  the  'total  loss  of  head  fs  double  that  of 
one  elbow ;  while  if  the  plane  of  the  second  is  ~| 
to  that  of  the  first,  the  total  loss  of  head  is  1£ 
times  that  of  one  alone.     (Weisbach.) 


FIG.  583. 


Fio.  584. 


529.  Bends  in  Pipes  of  Circular  Section.— Fig.  585.  Weis- 
bach bases  the  following  empirical 
formula  for  C,  the  coefficient  of  resist- 
ance of  a  quadrant  bend  in  a  pipe 
of  circular  section,  on  his  own  experi- 
Fio.585.  ments*  and  some  of  Dubuat's,  viz. : 


C  =  0.131  +  1.847        , 


for  use  in 


(1) 

(2) 


where  a  =  radius  of  pipe,  r  =  radius  of  bend  (to  centre  of 
pipe),  and  v  —  velocity  in  pipe ;  hr  =  loss  of  head  due  to 
bend. 

*  Weisbach's  experiments  were  probably  made  with  small  pipes.  Those 
made  at  Detroit  in  1893  to  1898  on  large  water  mains  (diain.  =  12  to  30  ins.) 
gare  different  results.  (See  Trans.  Am.  Soc.  C.  E.  for  April,  1902.) 


ELBOWS  AND   BENDS   IN   PIPES. 


729 


.  It  is  understood  that  the  portion  I>C  of  the  pipe  is  keptfuH 
by  the  flow ;  which,  however,  may  not  be  practicable  unless 
JBCis  more  than  three  or  four  times  as  long  as 
wide,  and  is  full  at  the  outset.  A  semicircular 
bend  occasions  about  the  same  loss  of  head  as  a 
quadrant  bend,  but  two  quadrants  forming  a  re- 
verse curve  in  the  same  plane,  Fig.  586,  occasion  a 
double  loss.  By  enlarging  the  pipe  at  the  bend, 
or  providing  internal  thin  partitions  parallel  to  the 
sides,  the  loss  of  head  may  be  considerably  dimin- 
ished. Weisbach  gives  the  following  table  com- 
puted from  eq.  (1),  but  does  not  state  the  absolute  size  of  the 
pipes. 


For  ^-=.10 

.20 

.30 

.40 

.50 

.60 

.70 

.80 

.90 

1.0 

£=.131 

.138 

.158 

.206 

.294 

.440 

.661 

.977 

1.40 

1.98 

Accounts  of  many  of  Weisbach's  hydraulic  experiments  are 
contained  in  the  Cimlingenieur^  vols.  ix,  x,  and  xi. 

529a.  Common  Pipe-elbows. — Prof.  L.  F.  Bellinger  of  Nor- 
wich University,  Vermont,  conducted  a  set  of  experiments  in 
1887,  when  a  student  at  Cornell,  on  the 
loss  of  head  occasioned  by  a  common  el- 
bow (for  wrought-iron  pipe),  whose  longi- 
tudinal section  is  shown  in  Fig.  586a. 
The  elbow  served  to  connect  at  right 
angles  two  wrought-iron  pipes  having  an 
internal  diameter  of  0.482  in. 

The  internal  diameter  of  the  short  bend 
or  elbow  was  £  in.,  and  the  radius  of  its  curved  circular  axis 
(a  quadrant)  was  £  in.  Its  internal  surface  was  that  of  an 
ordinary  rough  casting. 

A  straight  pipe  of  the  same  character  and  size  and  14  feet 
long  was  first  used,  and  the  loss  of  head  due  to  skin-friction 
(the  only  loss  of  head  in  that  case)  carefully  determined  for  a 
range  of  velocities  from  2  to  20  ft.  per  sec. 

Two  lengths  of  similar  pipe  were  then  joined  by  the  elbow 


FIQ.  586a. 


730 


MECHANICS   OF   ENGINEERING. 


mentioned,  forming  a  total  length  of  14  feet,  and  tbe  total  loss 
of  head  again  determined  through  the  same  range  of  velocities. 
By  subtraction,  the  loss  of  head  due  to  the  elbow  was  then 
easily  found  for  each  velocity,  and  assuming  the  form 


for  the  loss  of  head,  C  was  computed  in  each  case. 

From  Fig.  586a  it  is  seen  that  the  stream  meets  with  a  sud- 
den enlargement  and  a  sudden  diminution,  of  section,  as  well 
as  with  the  short  bend  ;  so  that  the  disturbance  is  of  a  rather 
complex  nature. 

The  principal  results  of  Prof.  Bellinger's  experiments,  after 
the  adjustment  of  the  observed  quantities  by  "  least  squares," 
were  found  capable  of  being  represented  fairly  well  by  the 
formula 

C  =  0.621  +  [2n  -  1]  X  0.03T6,       ...     (2) 

where  n  =  [veloc.  in  pipe  in  ft.  per  sec.]  -+-  5.     The  following 
table  was  computed  from  eq.  (2)  (where  v  is  in  ft.  per  second) : 


0  = 

2 

4 

6 

8 

10 

12 

14 

16 

18 

20 

c= 

.633 

.649 

.670 

.697 

.734 

.782 

.845 

.929 

1.039 

1.185 

530.  Valve-gates*  and  Throttle-valves  in  Cylindrical  Pipes. — 
Adopting,  as  usual,  the  form 


(1) 


for  the  loss  of  head  due  to  a  valve-gatet  Fig.  587,  or  for  a 

L        .     , 


FIG.  587.  PIG.  588. 

throtile-vafoe,  Fig.  588,  each  in  a  definite  position,  Weisbach's 

*  Mr.  Kuichling's  experiments  on  a  24-inch  stop-valve  are  described  in 
the  Trans.  Am.  8.  C.  E.  for  May  1892,  p.  439.  See  also  Eng.  News,  Aug. 
1892,  p.  117. 


VALVE-GATE.      THROTTLE-VALVE. 


731 


experiments  furnish  us  with  a  range  of  values  of  C  in  the  ease 
of  these  obstacles  in  a  cylindrical  pipe  1.6  inches  in  diameter, 
as  follows  (for  meaning  of  s,  d,  and  <*,  see  figures,  v  is 
the  velocity  in  the  full  section  of  pipe,  running  full  on  both 
sides.) 


Valve-gate. 

Throttle-valve. 

8 
d 

c 

a 

C 

5° 

.24 

1.0 

.00 

10° 

.52 

* 

.07 

15° 

20° 

.90 
1.54 

I 

.26 

25° 
30° 

2.51 
3.91 

§ 

.81 

35° 

6.22 

40° 

10.8 

i 

2.06 

45° 

18.7 

i 

5.52 

50° 
55° 

32.6 

58.8 

i 

17.00 

60° 
65° 

118.0 
256.0 

i 

97.8 

70" 

751. 

631.  Examples  involving  Divers  Losses  of  Head. — We  here 
suppose,  as  before,  that  the  pipes  are  full  during  the  flow. 
Practically,  provision  must  be  made  for  the  escape  of  the  air 
which  collects  at  the  high  points.  If  this  air  is  at  a  tension 
greater  than  one  atmosphere,  automatic  air-valves  will  serve  to 
provide  for  its  escape ;  if  less  than  one  atmosphere,  an  air. 
pump  can  be  used,  as  in  the  case  of  a  siphon  used  at  the 
Kansas  City  "Water  Works.  (See  p.  346  of  the  Engineering 
ffews  for  November  1887.) 


EXAMPLE  1. — Fig.  589.    What  head,  =  A,  will  be  required 
to  deliver  i  U.  S.  gallon  (i.e.  231  cubic  inches)  per  second 


732  MECHANICS   OF  ENGINEERING. 

through  the  continuous  line  of  pipe  in  the  figure,  containing  twr» 
sizes  of  cylindrical  pipe  (d0  =  3  in.,  and  d9  =  1  in.),  and  two 
90°  elbows  in  the  larger.  The  flow  is  into  the  air  at  m,  the 
jet  being  1  in.  in  diameter,  like  the  pipe.  At  E,  a  =  90°,  and 
the  corners  are  hot  rounded  ;  at  K,  also,  corners  not  rounded. 
Use  the  ft.-lb.-sec.  system  of  units  in  which  g  =  32.2. 

Since  Q  =  i  gal.  =  }  -  fffa  =  .0668  cub.  ft.  per  sec.,  and 
therefore  the  velocity  of  the  jet 

vm  =  v,  =  Q  +  JarOjV)'  =  12.25  ft.  per  sec.; 

hence  the  velocity  in  the  large  pipe  is  to  be  v0  =  Q-)X  =  1.36 
ft.  per  sec.  From  Bernoulli's  Theorem,  we  have,  with  m  as 
datum  plane, 


involving  six  separate  losses  of  head,  for  each  of  which  there 
is  no  difficulty  in  finding  the  proper  C  or/,  since  the  velocities 
and  dimensions  are  all  known,  by  consulting  preceding  para- 
graphs. (Clean  iron  pipe.) 

From  §  515,  table,  for  a  =  90°  we  have  .     .     .     C*  =  0.505 
«     §  517,  for  d.  =  3  in.,  and  v0  =1.36  ft.  per  sec.,/0  =   .00725 
«          "      "  d,  =  l  in.,  and  v,-=  12.25  "    "      /,=   -00613 
«     §  528  (elbows),  for  a  —  90°      ....     Ce*.  =0.984 
"     §  527,  for  sudden  diminution  at  K  we  have 
[since  F%  -s-  F.  =  T  -5-  3a  =  0.111,  /.  G  =0.625] 


<* =(.«-•)•=»• 


0.360. 


Solving  the  above  equation  for  A,  then,  and  substituting 
above  numerical  values  (in  ft.-lb.-scc.-system),  we  have  (noting 
that  vm  =  v»  and  v0  = 


X  984) 


.360+4x-00-613x20l; 

TT5"  -J 


EXAMPLES;    WITH  LOSSES  OF  HEAD. 


733 


Asa  (12.257  ri  ,  /Q0623 +.07160 +  .0243)+(.360 + 5.8848  I; 
64.4     L  J 

.-.  h  =  2.323  X  7.3469  =  17.09  ft.— Ans.      ' 

It  is  here  noticeable  how  small  are  the  losses  of  head  in  the 
large  pipe,  the  principal  reason  of  this  being  that  the  velocity 
in  it  is  so  small  (VQ  =  only  1.36  ft.  per  sec.),  and  that  in  gen- 
eral losses  of  head  depend  on  the  square  of  the  velocity 
(nearly). 

In  other  words,  the  large  pipe  approximates  to  being  a  reser- 
voir in  itself. 

With  no  resistances  a  head  h  =  vm8  -r-  2g  =  2.32  ft.  would  be 
sufficient. 

EXAMPLE  2.— Fig.  590.  With  the  valve-gate  Fhalf  raised 
(i.e.,  s  =  \d  in  Fig.  587),  required  the  volume  delivered  per 
second  through  the  clean  pipe  here  shown.  The  jet  issues 


> fc-80-' * 


FIG.  59p. 


from  a  short  straight  pipe,  or  nozzle  (of  diameter  d^  =  1£  in.) 
inserted  in  the  end  of  the  larger  pipe,  with  the  inner  corners 
not  rounded.  Dimensions  as  in  figure.  Radius  of  each  bend 
=  r  =  2  in.  The  velocity  vm  of  the  jet  in  the  air  =  velocity 
»t  in  the  small  pipe  ;  hence  the  loss  of  head  at 


Now  vm  is  unknown,  as  yet  ;  but  VQ  ,  the  velocity  in  the  large 

r(-Y~i 

pipe,  is  =vm  \~t    ;  Le.,v0  =  ^$vm.    From  Bernoulli's  The- 


734  MECHANICS   OF   ENGINEERING. 

orem  (m  as  datum  level)  we  obtain,  after  transposition, 


Of  the  coefficients  concerned,  f0  alone  depends  on  the  un- 
known velocity  v0.    For  the  present  [first  approximation], 
put   ................   /0  =  .006 

From  §  515,  with  a  =  00°,  ......  '  .     .    £E  =  .505 

From  §  517,  valve-gate  with  s  =  £d,  .     ,     .     .     .    £F  =  2.06 
From  §  529,  with  a  :  r  =  0.5,    .......    £B  =  0.294 

While  at  K,  from  §  527,  having 


we  find  from  table,      .....    .....     C  =  0.700 

and/.  6r=^-  l]'  =  (0.428)'.    .     .    .    i.e.,  C*=  0.183 

Substituting  in  eq.  (1)  above,  with  v*  =  (^-)2  vm*,  we  have 


in  which  the  first  radical,  an  abstract  number,  might  be  called 
a  coefficient  of  velocity,  0,  for  the  whole  delivery  pipe ;  and 
also,  since  in   this   case   Q,  =  Fmvm  =  F^ ,  may  be  written 
Q  =  nFt  Vfyh,  it  may  be  named  a  coefficient  of  efflux,  p. 
Hence 


r505  +  2.06,Sx.294  +  iJL^^l+.1 

.%  vm=  0.421  Vfyh  =  0.421  1/2x32.2x25  =  16.89  ft.  per  seu 

(The  .421  might  be  called  a  coefficient  of  velocity.)    The 
volume  delivered  per  second  is 

Q  =  ±*d;  vm  =  t?r(/T)*  16.89  =  .207  cub.  ft.  per  sec. 


(As  the  section  of  the  jet  Fm  =  F^  ,  that  of  the  short  pipe  or 
nozzle,  we  might  also  say  that  .421  =  //  —  coefficient  of  efflux, 
for  we  may  write  Q  =  pF^  V%gh,  whence  ^  =  .421.) 


FLOW   THROUGH   SIPHONS.  735 

532,  Siphons.— In  Fig.  532,  §  490,  the  portion  HN%0  is 
above  the  level,  BCy  of  the  surface  of  the  water  in  the  head 
reservoir  BL,  and  yet  under  proper  conditions  a  steady  flow 
can  be  maintained  with  all  parts  of  the  pipe  full  of  water,  in- 
cluding HN^O.  If  the  atmosphere  exerted  no  pressure,  this 
would  be  impossible ;  but  its  average  tension  of  14.7  Ibs.  per 
sq.  inch  is  equivalent  to  an  additional  depth  of  nearly  34  feet 
of  water  placed  upon  BC.  With  no  flow,  or  a  very  small 
velocity,  the  pipe  may  be  kept  full  if  JV9  is  not  more  than 
33  or  34  feet  above  BG\  but  the  greater  -y,,  the  velocity  of 
flow  at  N^ ,  and  the  greater  and  more  -numerous  the  losses 
of  head  between  Z  and  N^  the  less  must  be  the  height  of  N% 
above  BC  for  a  steady  flow. 

The  analytical  criterion  as  to  whether  a  flow  can  be  main- 
tained or  not,  supposing  the  pipe  completely  filled  at  the  out- 
set, is  that  the  internal  pressure  must  be  >  0  at  all  parts  of 
the  pipe.  If  on  the  supposition  of  a  flow  through  a  pipe  of 
given  design  the  pressure^?  is  found  <  0,  i.e.  negative,  at  any 
point  \Nt  being  the  important  section  for  test]  the  supposition 
is  inadmissible,  and  the  design  must  be  altered. 

For  example,  Fig.  532,  suppose  LN^N^  to  be  a  long  pipe  of 
uniform  section  (diameter  =  d,  and  length  =  Z),  and  that  under 
the  assumption  of  filled  sections  we  have  computed  v4,  the 
velocity  of  the  jet  at  ^ ;  i.e., 


/ 
y 


.   .    . 


(1) 


To  test  the  supposition,  apply  Bernoulli's  Theorem  to  the 
surface  BC  and  the  point  Nt  where  the  pressure  is  p99  velocity 
«?,(=  vt  ,  since  we  have  supposed  a  uniform  section  for  whole 
pipe),  and  height  above  BC=h^.  Also,  let  length  of  pipe 
•  Whence  we  have 


[B  C  being  datum  plane."| 


736  MECHANICS   OF  ENGINEERING. 


Solving  for    %  we  have 


We  note,  then,  that  for^>8  to  be  >  0, 


In  the  practical  working  of  a  siphon  it  is  found  that  atmos- 
pheric air,  previously  dissolved  in  the  water,  gradually  collects 
at  N't  ,  the  highest  point,  during  the  flow  and  finally,  if  not  re- 
moved, causes  the  latter  to  cease.  See  reference  below. 

One  device  for  removing  the  air  consists  in  first  allowing  it 
to  collect  in  a  chamber  in  communication  with  the  pipe  be- 
neath. This  communication  is  closed  by  a  stop-cock  after  the 
water  in  it  has  been  completely  displaced  by  air.  Another 
stop-cock,  above,  being  now  opened,  water  is  poured  in  to  re- 
place the  air,  which  now  escapes.  Then  the  upper  stop-cock  is 
shut  and  the  lower  one  opened.  The  same  operation  is  again 
necessary,  after  some  hours. 

On  p.  346  of  the  Engineering  News  of  November  1887  may 
be  found  an  account  of  a  siphon  which  has  been  employed  since 
1875  in  connection  with  the  water-works  at  Kansas  City.  It 
is  1350  ft.  long,  and  transmits  water  from  the  river  to  the 
artificial  "  well  "  from  which  the  pumping  engines  draw  their 
supply.  At  the  highest  point,  which  is  16  ft.  above  low-water 
level  of  the  river,  is  placed  a  "  vacuum  chamber  "  in  which  the 
air  collects  under  a  low  tension  corresponding  to  the  height, 
and  a  pump  is  kept  constantly  at  work  to  remove  the  air  and 
prevent  the  "  breaking"  of  the  (partial)  vacuum.  The  diam- 
eter of  the  pipe  is  24  in.,  and  the  extremity  in  the  "  well  "  dips 
5  ft.  below  the  level  of  low  water.  See  Trautwine's  Pocket- 
book,  for  an  account  of  Maj.  Orozet's  Siphon. 

532a.  Branching  Pipes.*  —  If  the  flow  of  water  in  a  pipe  is 
caused  to  divide  and  pass  into  two  others  having  a  common 

*  Problems  of  this  kind  are  best  solved  by  tables  or  diagrams.  Mr.  Cof- 
fin's book  "  Graphical  Solution  of  Hydraulic  Problems"  is  useful  for  this 
purpose. 


BRANCHING  PIPES.  737 

junction  with  the  first,  or  vice  versa  ;  or  if  lateral  pipes  lead 
out  of  a  main  pipe,  the  problem  presented  may  be  very  com- 
plicated. As  a  comparatively  simple  instance,  let  us  suppose 
that  a  pipe  of  diameter  d  and  length  I  leads  out  of  a  reservoir, 
and  at  its  extremity  is  joined  to  two  others  of  diameters  dt  and 
d^  and  lengths  lt  and  19  respectively,  aod  that  the  further  extrem- 
ities of  the  latter  discharge  into  the  air  without  nozzles  under 
heads  hl  and  A,  below  the  reservoir  surface.  Cal»  these  two 
pipes  Nos.  1  and  2.  That  is,  the  system  forms  a  Y  in  plan. 

Assuming  that  all  entrances  and  junctions  are  smoothly 
rounded,  so  that  all  loss  of  head  is  due  to  skin-friction,  it  is  re- 
quired to  lind  the  three  velocities  of  flow,  v9  v1  ,  and  t>a,  in  the 
respective  pipes.  First  applying  Bernoulli's  Theorem  to  a 
stream-line  from  the  reservoir  surface  through  the  main  pipe 
to  the  jet  at  the  discharging  end  of  pipe  No.  1,  we  have 


and  similarly,  dealing  with  a  stream-line  through  the  main 
pipe  and  No.  2, 


while  the  equation  of  continuity  for  this  case  is 

.      ....    (3) 


From  these  three  equations,  assuming/  the  same  in  all  pipes 
as  a  first  approximation,  we  can  find  the  three  velocities  (best 
by  numerical  trial,  perhaps)  ;  and  then  the  volume  of  discharge 
of  the  system  per  unit  of  time 


(4) 


533.  Time  of  Emptying  Tertical  Prismatic  Vessels  (or  Inclined 
Prisms  if  Bottom  is  Horizontal)  under  Variable  Head. 

CASE  I.  Through  an  orifice  or  short  pipe  in  the  bottom  and 
opening  into  the  air.  —  Fig.  591.  As  the  upper  free  surface,, 


738  MECHANICS   OF   ENGINEERING. 

of  area  =  Ff,  sinks,  F'  remains  constant.  Let  z  =  head  of 
water  at  any  stage  of  the  emptying  ;  it  =  s0  at  the  outset,  and 
=  0  when  the  vessel  is  empty.  At  any 
instant,  Q,  the  rate  of  discharge  (=  vol- 
ume per  time-unit)  depends  on  z  and  is 

Q  =  »FVfyS;  .    .    .    (1) 

where  /*  =  coefficient  of  efflux  =  <pC  = 
coefficient  of  velocity  X  coefficient  of  con- 
traction [see  §  495,  eq,,  (3)].  We  here  suppose  F'  so  large 
compared  with  F,  the  area  of  the  oritice,  that  the  free  surface 
of  the  water  in  the  vessel  does  not  acquire  any  notable  velocity 
at  any  stage,  and  that  hence  the  rate  of  efflux  is  the  same  at 
any  instant,  as  for  a  steady  flow  with  head  =  z  and  a  zero 
velocity  in  the  free  surface.  ^  is  considered  constant.  From 
(1)  we  have 


dV=  (vol.  discharged  in  time  dt)  =  Qdt  =  ^FVfyz  dt.  .  (2 

But  this  is  also  equal  to  the  volume  of  the  horizontal  lamina, 
F'dZ)  through  which  the  free  surface  has  sunk  in  the  same 
time  dt.  - 

.%  dt= 

We  have  written  minus  F'dz  because,  dt  being  an  increment, 
dz  is  a  decrement.  To  reduce  the  depth  from  z^  (at  the  start, 
time  ==  t  =  zero)  to  zn  ,  demands  a  time 


W 

>>=--4-=    \ 
>jiFVZz 


whence,  by  putting  zn  =  0,  we  have  the  time  necessary  to 
empty  the  whole  prism 

Ct__    %F'z*          iF'z.  2  X  volume  of  vessel  .  „. 

"  nFVfy    :  pFV2gz0  ""  initial  rate  of  discharge  ' 


TIME   OF   EMPTYING   VESSELS. 


739 


that  is,  to  empty  the  vessel  requires  double  the  time  of  dis- 
charging the  same  amount  of  water  if  the  vessel  had  been  kept 
full  (at  constant  head  =  ZQ .=  altitude  of  prism). 

To  Jill  the  same  vessel  through  an  orifice  in  the  bottom,  the 
flow  through  which  is  supplied  from  a 
body  of  water  of  infinite  extent  hori- 
zontally, as  with  the  (single)  canal  lock 
of  Fig.  592,  will  obviously  require  the 
same  time  as  given  in  eq.  (5)  above, 
since  the  effective  head  z  diminishes 
from  ZQ  to  0,  according  to  the  same  law. 

EXAMPLE. — What  time  will  be  needed 
to  empty  a  parallelopipedical  tank  (Fig.  591)  4  ft.  by  5  ft.  in 
horizontal  section  and  6  ft.  deep,  through  a  stop-cock  in  the 
bottom  whose  coefficient  of  efflux  when  fully  open  is  known 
to  be  /*  =  0.640,  and  whose  section  of  discharge  is  a  circle  of 
diameter  —  \  in.  ?  From  given  dimensions  F'  =  4  X  5  =  20 
sq.  ft.,  while  z0  =  6  ft.  Hence  from  eq.  (5)  (ft.-lb.-sec.) 


time  of  \  = 
emptying  j       0.64  X 


2  X  20  X 


_  j      13980  seconds 


V2  X  32.! 


CASE  II.  Two  communicating  prismatic  veuels.     Required 
the  time  for  the  water  to  come  to  a  common  level  ON,  Fig. 

T 593,  efflux  taking  place  through  a  small 

orifice,  of  area  =  F,  u-ider  water.     At 
any  instant  the  rate  of  discharge  is 


• 


FIG.  593. 


as  before,    z  =  difference  of  level.  Now 
if  F'  and^7/  are  the  horizontal  sectional 
areas  of  the  two  prismatic  vessels  (axes 
vertical)  we  have  F'x  =  F"y^  and  hence  z>  which  =  x 


,     , 
-,    and   dx  = 


740  MECHANICS   OF  ENGINEEKING. 

As  before,  we  have 


F'F"      z  ' 


or     <    =    ~ 


Hence,  integrating,  the  time  for  the  difference  of  level  to 
change  from  20  to  zn 


F'+F 


and  by  making  zn  =  0  in  (6),  we  have  the 

%F'F"      1       l~z^ 
time  of  coming  to  a  common  level  =  -p,,-^,,  •  ~W/  ~^f   CO 


ALGEBRAIC  EXAMPLE. — In  the  double  lock  in  Fig.  594,  let 
L'  be  full,  while  in  L"  the  water  stands  at  a  level  MN  the 

same  as  that  of  the  tail- 
water.  F'  and  F"  are  the 
horizontal  sectional  areas  of 
the  prismatic  locks.  Let 
the  orifice,  0,  between 
them,  be  at  a  depth  =  hl 
below  the  initial  level  KE 
of  L ',  and  a  height  =  Aa 
above  that,  MN,  of  L". 
The  orifice  at  0,  area  =  F,  being  opened,  efflux  from  L'  be- 
gins into  the  air,  and  the  level  of  L"  is  gradually  raised  from 
MN  to  OD,  while  that  of  L'  sinks  from  KE  to  AE  a  distance 
=  a,  computed  from  the  relation  vol.  F'a  =  vol.  F"Ji^^  and 
the  time  occupied  is  [eq.  (4)] 


FIG.  594. 


As  soon  as  0  is  submerged,  efflux  takes  place  un^er  water,  and 
we  have  an  instance  of  Case  II.  Hence  the  time  of  reaching 
a  common  level  (after  submersion  of  0)  (see  eq.  7)  is 


and  the  total  time  is  =  t,  +  4 ,  with  a  =  F"\  -r-  F'. 


TIME   OF   EMPTYING  VESSELS. 


741 


CASE  III.  Emptying  a  vertical  prismatic  vessel  through 
rectangular  "notch"  in  the  side,  or  over- 
fall.— Fig.  595.  As  before,  let  even  the 
initial  area  (=  zjb)  of  the  notch  be  small 
compared  with  the  horizontal  area  F'  of 
tank.  Let  z  =  depth  of  lower  sill  of  notch 
below  level  of  tank  surface  at  any  instant, 
and  b  =  width  of  notch.  Then,  at  any  in- 
stant  (see  eq.  10,  §  504),  Flo.  595. 


Bate  ofdisch.  (vol.)  =  <>  = 

.-.  vol.  of  disch.  in  dt  =  \)& 


and  putting  this  =  —  F'dz  =  vol.  of  water  lost  by  the  tank 
in  time  dt,  we  have 


whence 


3    F 


3    F' 


c- 


3F 


r  1    _ 


(10) 


as  the  time  in  which  the  tank  surface  sinks  from  a  height  z^ 
above  sill  to  a  height  zn  above  sill.  If  we  inquire  the  time  tf 
for  the  water  to  sink  to  the  level  of  the  sill  of  the  notch  we 
put  zn  —  zero,  whence  t'  =  infinity.  As  explanatory  of  this 
result,  note  that  as  z  diminishes  not  only  does  the  velocity  of 
flow  diminish,  but  the  available  area  of  efflux  (=  zb)  also  grows 
less,  whereas  in  Cases  I  and  II  the  orifice  of  efflux  remained 
of  constant  area  =  F. 

Eq.  (10)  is  applicable  to  the  waste- weir  of  a  large  reservoir 
or  pond. 

534.  Time  of  Emptying  Vessels  of  Variable  Horizontal  Sec- 
tions.— Considering  regular  geometrical  forms  first,  let  us  take 


742 


MECHANICS   OF  ENGINEERING. 


CASE  I.    Wedge-shaped  vessel,  edge  horizontal  and  under- 

neath,  orifice  F  in  the  edge,  so  that 
2,  the  variable  head,  is  always  the 
altitude  of  a  triangle  similar  to  the 
section  ABC  of  the  body  of  water 
when  efflux  begins.  At  any  instant 
during  the  efflux  the  area,  $,  of 
the  free  surface,  variable  here, 
takes  the  place  of  F'  in  eq.  (3)  of 
§  533,  whence  we  have, 

far  any  case  of  variable  free  surface,  dt  =  — ^  /__  .    .    (11) 


FIG.  596. 


In  the  present  case  S  =  ul,  and  from  similar  triangles 

u  :  z  ::  b  :  z^\ 
whence 

S=llz  +  z9, 
and 


and  hence  the  time  of  emptying  the  whole  wedge,  putting 
^  =  0,  is 


4       $blz0       _  4  Vol.  of  wedge 

%'  }iF  V%gz<>       3  initial  rate  of  discharge 


(13) 


i.e.,  f  as  long  as  to  discharge  the  same  volume  of  water  under 
a  constant  head  =  z0 .     This  is  equally  true  if  the  ends  of  the 
wedge  are  oblique,  so  long  as  they  are  parallel. 
CASE  II.    Right  segment  of  paraboloid  of 
revolution. — Fig.    597.     Axis   vertical.     Ori- 
fice at  vertex.     Here  the  variable  free  surface 
has  at  any  instant  an  area,  =  S,=  nu*,  u  be- 
ing  the  radius  of    the  circle  and  variable. 
From  a  property  of  the  parabola 

u* :  5* : :  z  :  z. ;    .*.   S  =  ntfz  -?-  z. ,  FIG.  597. 


TIME   OF   EMPTYING  VESSELS. 

sod  hence,  from  eq.  (11), 


743 


whence,  putting  zn  =  0,  we  have  the  time  of  emptying  the 
whole  vessel 

4    7ttf^z0      _  4  total  vol.  „ .. 

°       3  jjiFV%gzQ       3  initial  rate  of  disck? 

same  result  as  for  the  wedge,  in  Case  I ;  in  fact,  it  applies  to 
any  vessel  in  which  the  areas  of  horizontal  sections  vary 
directly  with  their  heights  above  the  orifice. 

CASE  III.  Any  pyramid  or  cone  ;  vertex  down  ;  small  ori- 
fice in  vertex. — Fig.  598.     Let  area  of  the 
base  =  S9 ,  at  upper  edge  of  vessel.     At 
any  stage  of  the  flow  S  =  area  of  base  of 
pyramid  of  water.    From  similar  pyra-    &^^^r  \         4o 
mids 


and[eq.(ll)] 


So 


whence  (sn  =  0)  the  time  of  emptying  the  whole  vessel  is 

2      Sz* 


or, 


6          Total  volume 
tf  =-.-r-—  -  —  -  .....    (15) 

5  ^n^t^al  rate  of  d^sch. 


CASE  IY.  Sphere.  —  Similarly,  we  may  show  that  to  empty 


744 


MECHANICS   OF   ENGINEERING. 


a  sphere,  of  radius  =  r,  through  a  small  orifice,  of  area  =  F, 
in  lowest  part,  the  necessary  time  is 


I6 

15 


8 


Vol. 


init.  rate  of  disch. 


535.  Time  of  Emptying  an  Obelisk-shaped  Vessel. — (An  obe- 
lisk may  be  defined  as  a  'solid  of  six  plane  faces,  two  of  which 
are  rectangles  in  parallel  planes  and  with  sides  respectively 
parallel,  the  others  trapezoids;  a  frustum  of  a  pyramid  is  a 
particular  case.) 

A  volume  of  this  shape  is  of  common  occurrence ;  see  Fig. 
599.  Let  the  altitude  =  A,  the  two  rectangular  faces  being 
horizontal,  with  dimensions  as  in  figure.  By  drawing  through 

F,  G,  and  II  right  lines  par- 
allel to  EC,  to  cut  the  upper 
base,  we  form  a  rectangle 
KLMC  equal  to  the  lower 
base.  Produce  ML  to  P  and 
KL  to  N,  and  join  PG  and 
NG-.  We  have  now  sub- 
divided the  solid  into  a  paral- 
lelopiped  KLMC  -  EHGF, 
a  pyramid  PBNL  -  G,  and 
two  wedges,  viz.  APLR-HG  and  LNDM-FG,  with 
their  edges  horizontal ;  and  may  obtain  the  time  necessary  to 
empty  the  whole  obelisk-volume  by  adding  the  times  which 
would  be  necessary  to  empty  the  individual  component  vol- 
umes, separately,  through  the  same  orifice  or  pipe  in  the  bot- 
tom plane  EG.  These  have  been  already  determined  in  the 
preceding  paragraphs.  The  dimensions  of  each  component 
volume  may  be  expressed  in  terms  of  those  of  the  obelisk,  and 
all  have  a  common  altitude  =  h. 

Assuming  the  orifice  to  be  in  the  bottom,  or  else  that  the 
discharging  end  of  the  pipe,  if  such  is  used,  is  in  the  plane  of 
the  bottom  EG,  we  have  as  follows,  F  being  the  area  of  dis- 
charge : 


FIG.  599. 


TIME   OF   EMPTYING    KESEKV01ES.  745 

Time  to  empty  the  parallelepiped]        ^  _     ^A       ,7-    ,.., 
separately  would  be  (Case  /,  §  533)  j  *       l      uF  Vty 

Time  to  empty  the  two  )  0^/7      / 

wedges      separately  K  =  I  •  ^~ 
(6W/,§534)       *)'       3 

For  the  pyramid  \          /  __  2   (I  —  I,)  (b  —  b,) 
(<7^  ///,  §  534)  f  '      -*8-5*       ^  <fij       V/l' 

Hence  to  empty  the  whole  reservoir  we  have 


_ 


(4) 


EXAMPLE.  —  Let  a  reservoir  of  above  form,  and  with  b  =  50  ft.> 
I  =  60  ft.,  b,  =  10  ft.,  Zt  =  20  ft.,  and  depth  of  water  h  =  16 
ft.,  be  emptied  through  a  straight  iron  pipe,  horizontal,  and 
leaving  the  side  of  the  reservoir  close  to  the  bottom,  at  an  angle 
a  =  36°  with  the  inner  plane  of  side.  The  pipe  is  80  ft.  long 
and  4  inches  in  internal  diameter  ;  and  of  clean  surface.  The 
jet  issues  directly  from  this  pipe  into  the  air,  and  hence 
F=ln($f  sq.  feet.  To  find  /*,  the  "coefficient  of  efflux" 
(=0,  the  coefficient  of  velocity  in  this  case,  since  there  is  no 
contraction  at  discharge  orifice),  we  use  eq.  (4)  (the  first  radical) 
of  §  518,  withyapprox.  =  .006,  and  obtain 


i  -4X.006X80 

—  T— 


°'36L 


(N".B.    Since  the  velocity  in  the  pipe  diminishes  from  a 
value 

v  =  .361  V2g  X  16  =  11.6  ft.  per  sec. 

at  the  beginning  of  the  flow  to  v  —  zero  at  the  close,  f  =  .006 
is  a  reasonably  approximate  average  with  which  to  compute 
the  average  0  above  ;  see  §  517. 


746  MECHANICS   OF   ENGINEERING. 

Hence  from  eq.  (4)  of  this  paragraph  (ft.-lb.-sec.  system) 
[3  X  50  X  60  +  8  X  10 X 20+2(50 X 20+10 X 60)]2  VlS 


15  X  0.361  X  |(|)  V2X  32.2 


nt\i  ^  A  01,      *      •      (  Probably  within  2  or  3$  of 

=  29110  sec.  =  8  hrs.  5  mm.  j  '  the  truth> 

536.  Time  of  Emptying  Reservoirs  of  Irregular  Shape.  Simp- 
son's Rule.  —  From  eq.  (11),  §  534,  we  have,  for  the  time  in 
which  the  free  surface  of  water  in  a  vessel  of  any  shape  what- 
ever sinks  through  a  vertical  distance  =dz, 

~~       ,  whence  [time  =  -  1-=-_    fte-We,  ,  .  (1) 
L= 


where  8  is  the  variable  area  of  the  free  surface  at  any  in- 
stant, and  z  the  head  of  water  at  the  same  instant,  efflux 
proceeding  through  a  small  orifice  (or  extremity  of  pipe)  of 
area  =  F.  If  8  can  be  expressed  in  terms  of  z,  we  can  in- 
tegrate eq.  (1)  (i.e.,  provided  that  S&-*  has  a  known  anti- 
derivative)  ;  but  if  not,  the  vessel  or  reservoir  being  irregular 
in  form,  as  in  Fig.  600  (which  shows  a  pond  whose  bottom 
has  been  accurately  surveyed,  so  that  we  know  the  value  of  S 
for  any  stage  of  the  emptying),  we  can  still  get  an  approximate 

solution  by  using  Simpson's 
Rule  for  approximate  inte- 
gration. Accordingly,  if  we 
inquire  the  time  in  which 
the  surface  will  sink  from  0 
to  the  entrance  JKof  the  pipe 
in  Fig.  600  (any  point  n  ;  at 
E.  or  short  of  that),  we 
FlG-600-  divide  the  vertical  distance 

from  0  to  n  (4  in  this  figure)  into  an  even  number  of  equal 
parts,  and  from  the  known  form  of  the  pond  compute  the  area 
8  corresponding  to  each  point  of  division,  calling  them  #,,#,, 
etc.  Then  the  required  time  is  approximately 


r 


TIME   OF   EMPTYING   POND.  747 

-^+4(3,  A_ 
*        -« 


EXAMPLE.  —  Fig.  600.  Suppose  we  have  a  pipe  Em  of  the 
same  design  as  in  the  example  of  §  535,  and  an  initial  head  of 
z^  =  16  ft.,  so  that  the  same  value  of  /*,  =  .361,  may  be  used. 
Let  zn  —  z0  —  8  feet,  and  divide  this  interval  (of  8  ft.)  into 
four  equal  vertical  spaces  of  2  ft.  each.  If  at  the  respective 
points  of  division  we  find  from  a  previous  survey  that  S0  = 
400000  sq.  ft.,  S,  =  320000  sq.  ft,  S,  =  270000  sq.  ft.,  £3  = 
210000  sq.  ft.,  and  £4  =  180000  sq.  ft.  ;  while  n  •  =  4,  p  =  .361, 
and  the  area  F—  \n($f  —  .0873  sq.  ft.,  we  obtain  (ft.,lb.,  sec.) 


G= 


16  —  8  r4QOOOO       £X  320000 

7==- 


0.361  X. 0873  1/2  X  32.2  X  3  X  4  L    i/16 

_  2  X  270000   ,    4  X  210000      ISOOOpn  =  2444000  sec. 

V12~  1/10  V8"  J=  28d-  6b'  53m-  20S- 


volume  discharged,  V,  may  also  be  found  by  Simpson's 
Rule,  thus  :  Since  each  infinitely  small  horizontal  lamina  has 
a  volume 


or,  approximately, 


F=  ^^-S.  +  45,  +  2$  +  45,  +  .  .  .  +  & 

0  071         L 

Hence  with  w  =  4  we  have  (ft.,  lb.,  sec.) 


+  180000~]  =  2,160,000  cub.  ft. 


748 


MECHANICS   OF   ENGINEERING. 


537.  Volume  of  Irregular  Reservoir  Determined  by  Observing 
Progress  of  Emptying. — Transforming  eq.  (11),  §  534,  we  have 


But  Sdz  is  the  infinitely  small  volume  d  V  of  water  lost  by 
the  reservoir  in  the  time  dt,  so  that  the  volume  of  the  reser- 
voir between  the  initial  and  final  (0  and  n)  positions  of  the 
horizontal  free  surface  (at  beginning  and  end  of  the  time  tn) 
may  be  written 

A* 
'    \  z*dt. (1) 

Jo 

This  can  be  integrated  approximately  by  Simpson's  Rule,  if 
the  whole  time  of  emptying,  =  tn,  be  divided  into  an  even 
..  .•...-.-..•:.-.:,-...-.->.v^  ,.-,:-„.:...  number     of     equal 

parts,  and  the  values 
20 ,  0j ,  02 ,  etc.,  of  the 
head  of  water  noted 
at  these  equal  inter- 
vals of  time  (not  of 
vertical  height).  The 
corresponding  sur- 
face planes  will  not 


FIG.  601. 


be  equidistant,  in  general, 
when  n  =  4  (see  Fig.  601) 


Whence  for  the  particular  case 


CHAPTER  VII. 

HYDRODYNAMICS  (Continued)— STEADY  FLOW  OF  WATER  IN 
OPEN  CHANNELS. 

538.  Nomenclature. — Fig.  602.  When  water  flows  in  an 
open  channel,  as.  in  rivers,  canals,  mill-races,  water-courses, 
ditches,  etc.,  the  bed 
and  banks  being  rigid, 
the  upper  surface  is 
free  to  conform  in 
shape  to  the  dynamic 
conditions  of  each  case, 
which  therefore  regu- 
late to  that  extent  the 
shape  of  the  cross-sec- 
tion- 

In  the  vertical  trans-  Fl°- 602- 

verse  section  AC  in  figure,  the  line  AC  is  called  the  air-profile 
(usually  to  be  considered  horizontal  and  straight),  while  the 
line  ABC,  or  profile  of  the  bed  and  banks,  is  called  the  wetted 
perimeter.  It  is  evident  that  the  ratio  of  the  wetted  perimeter 
to  the  whole  perimeter,  though  never  <  -J,  varies  with  the 
form  of  the  transverse  section. 

In  a  longitudinal  section  of  the  stream,  EFGH,  the  angle 
made  by  a  surface  filament  EF  with  the  horizontal  is  called 
the  slope,  and  is  measured  by  the  ratio  s  =  h  :  I,  where  I  is  the 
length  of  a  portion  of  the  filament  and  h  =  the  fall,  or  vertical 
distc^nce  between  the  two  ends  of  that  length.  The  angle  be- 
tween the  horizontal  and  the  line  HG  along  the  bottom  is  not 
necessarily  equal  to  that  of  the  surface,  unless  the  portion  of 
the  stream  forms  a  prism ;  i.e.,  the  slope  of  the  bed  does  not 
necessarily  =  s  =  that  of  surface. 

EXAMPLES. — The  old  Croton  Aqueduct  has  a  slope  of  1.10 
ft.  per  mile ;  i.e.,  s  =  .000208.  The  new  aqueduct  (for  New 

710 


750  MECHANICS    OF    ENGINEEKING. 

York)  has  a  slope  $  =  .000132,  with  a  larger  transverse  section. 
For  large  sluggish  rivers  s  is  much  smaller. 

539.  Velocity  Measurements.  —  Yarious  instruments  and 
methods  may  be  employed  for  this  object,  some  of  which  are 
the  following : 

Surface-floats  are  small  balls,  or  pieces  of  wood,  etc.,  so 
colored  and  weighted  as  to  be  readily  seen,  and  still  but  little 
affected  by  the  wind.  These  are  allowed  to  float  with  the  cur- 
rent in  different  parts  of  the  width  of  the  stream,  and  the  sur- 
face velocity  c  in  each  experiment  computed  from  c=.l-r-t, 
where  I  is  the  distance  described  between  parallel  transverse 
alignments  (or  actual  ropes  where  possible),  whose  distance 
apart  is  measured  on  the  bank,  and  t  =  the  time  occupied. 

Double-floats.  Two  balls  (or  small  kegs)  of  same  bulk  and 
condition  of  surface,  one  lighter,  the  other  heavier  than  water, 

are  united  by  a  slender  chain,  their 
weights  being  so  adjusted  that  the 
light  ball,  without  projecting  notably 
above  the  surface,  buoys  the  other 
ball  at  any  assigned  depth.  Fig.  603. 
It  ia  assumed  that  the  combination 
moves  with  a  velocity  c',  equal  to  the 
arithmetic  mean  of  the  surface  veloc- 
ity c0  of  the  stream  and  that,  c,  of  the  water  filaments  at  the 
depth  of  the  lower  ball,  which  latter,  <?,  is  generally  less  than 
c0 .  That  is,  we  have 

C'=%(GO  +  C)    and    /.   c  =  2c'-c0.  .    .     .    (1) 

Hence,  c0  having  been  previously  obtained,  eq.  (1)  gives  the 
velocity  c  at  any  depth  of  the  lower  ball,  c'  being  observed. 

The  floating  staff  is  a  hollow  cylindrical  rod,  of  adjustable 
length,  weighted  to  float  upright  with  the  top  just  visible.  Its 
observed  velocity  is  assumed  to  be  an  average  of  the  velocities 
of  all  the  filaments  lying  between  the  ends  of  the  rod. 

Woltmanrfs  Mill ;  or  Tachometer  •  or  Current-meter,  Fig. 
604,  consists  of  a  small  wheel  with  inclined  floats  (or  of  a  small 


CURRENT-METERS,    ETC. 


751 


FIG.  604. 


"  ecrew-propeller"  wheel  /S)  held  with  its  plane  1  to  the  cur 

rent,  which  causes  it  to  re- 
volve at  a  speed  nearly  pro- 
portional to  the  velocity,  c, 
of  the  water  passing  it. 
By  a  screw-gearing  W  on 
the  shaft,  connection  is 
made  with  a  counting  ap- 
paratus to  record  the  num- 
ber of  revolutions.  Some- 
times a  vane  B  is  attached, 
to  compel  the  wheel  to  face 
the  current.  It  is  either 
held  at  the  extremity  of  a  pole  or,  by  being  adjustable  along 
a  vertical  staff  fixed  in  the  bed,  may  be  set  at  any  desired  depth 
below  the  surface.  It  is  usually  so  designed  as  to  be  thrown 
in  and  out  of  gear  by  a  cord  and  spring,  that  the  time  of  mak- 
ing the  indicated  number  of  revolutions  may  be  exactly  noted. 
By  experiments  in  currents  of  known  velocities  a  table  or 
formula  can  be  constructed  by  which  to  interpret  the  indica- 
tions of  any  one  instrument ;  i.e.,  to  find  the  velocity  c  of  the 
current  corresponding  to  an  observed  number  of  revolutions 
per  minute. 

A  peculiar  form  of  this  instrument  has  been  recently  in- 
vented, called  the  Ititchie-HasJcell  Direction-current  Meter, 
for  which  the  following  is  claimed :  u  This  meter  registers 
electrically  on  dials  in  boat 
from  which  used,  the  direction 
and  velocity,  simultaneously, 
of  any  current.  Can  be  used 
in  river,  harbor,  or  ocean  cur- 
rents." 

Pitotfs  Tube  consists  in  prin-  r 
ciple  of  a  vertical  tube  open    =£ 
above,  while  its  lower  end,  also    "^ 
open,  is  bent  horizontally  up-    2 
stream;  see  A  in  figure.   After  ~  FIG.  eosT" 

the  oscillations  have   ceased,  the  water  in  the  tube  remains 


MECHANICS    OF   ENGINEERING. 

stationary  with  its  free  surface  a  height,  7i',  above  that  of  the 
stream,  on  account  of  the  continuous  impact  of  tbe  current 
against  the  lower  end  of  the  column.  By  the  addition  of 
another  vertical  tube  (see  B  in  figure)  with  the  face  of  its 
lower  (open)  end  parallel  to  the  current  (so  that  the  water- 
level  in  it  is  the  same  as  that  of  the  current),  both  tubes  being 
provided,  with  stop-cocks,  we  may,  after  closing  the  stop- 
cocks, lift  the  apparatus  into  a  boat  and  read  off  the  height  hf 
at  leisure.  We  may  also  cause  both  columns  of  water  to 
mount,  through  flexible  tubes,  into  convenient  tubes  in  the 
boat  by  putting  the  upper  ends  of  both  tubes  in  communica- 
tion with  a  receiver  of  rarefied  air,  and  thus  watch  the  oscilla- 
tions and  obtain  a  more  accurate  value  of  h'.  [See  Van  JS~os- 
trand's  Mag.  for  Mar.  '78,  p.  255.]  Theoretically  *  (see  §  565), 
the  thickness  of  the  walls  of  the  tube  at  the  lower  end  being 
considerable,  we  have 

.     .     .....     (1) 


as  a  relation  between  c,  the  velocity  of  the  particles  impinging 
on  the  lower  end,  and  the  static  height  Ti'  (§  565).  Eq.  (1)  is 
verified  fairly  well  by  Weisbach's  experiments  with  fine  in- 
struments, used  with  velocities  of  from  0.32  to  1.24  meters 
per  second.  Weisbach  found 

c  =    3.54  Vh'  (in  meters)   met.  per  sec., 
whereas  eq.  (1)  gives 

c=    3.133  Vhf  (in  meters)   met.  per  sec. 

In  the  instruments  used  by  Weisbach  the  end  of  the  tube 
turning  up-stream  was  probably  straight  ;  i.e.,  neither  flaring 
nor  conically  convergent.  A  change  in  this  respect  alters  the 
relation  between  c  and  ti  ';  see  §  565  for  Pilot's  and  Darcy's 
results. 

Pilot's  Tube,  though  simple,  is  not  so  accurate  as  the  ta- 
chometer. 

*  In  the  foot-note  of  p.  803,  if  a  be  assumed  =  60°,  instead  of  90°,  with  .Fas 
before,  we  obtain 

c  =  Vlgh'. 


CURRENT-METERS. 


753 


The  Hydrometric  Pendulum,  a  rather  uncertain  instrument, 
is  readily  understood  from  Fig.  606.  The  side  AJ3,  of  the 
quadrant  ABC,  being  held  vertical,  the 
plane  of  the  quadrant  is  made  parallel  to 
the  current.  The  angle  9  between  the 
cord  and  the  vertical  depends  on  G,  the 
effective  weight  (i.e.,  actual  weight  dimin- 
ished by  the  buoyant  effort)  of  the  ball 
(heavier  than  water),  and  the  amount  of  P, 
the  impulse  or  horizontal  pressure  of  the 
current  against  the  latter,  since  the  cord 
will  take  the  direction  of  the  resultant  It, 
for  equilibrium. 

Now  P  (see  §  5Y2)  for  a  ball  of  given  size  and  character  of 
surface  varies  (nearly)  as  the  square  of  the  velocity ;  i.e.,  if  P' 
is  the  impulse  on  a  given  stationary  ball,  when  the  velocity  of 
the  current  =  c',  then  for  any  other  velocity  c  we  have 


P' 

P  =  impulse  =  —  c* 


(2) 


From  this  and  the  relation  tan  6  =  —^  we  derive 


With  a  given  instrument  and  a  specified  system  of  units,  the 
numerical  value  of  the  first  radical  may  be  determined  as  a 
single  quantity,  by  experimenting  with  a  known  velocity  and 
the  value  of  6  then  indicated,  and  may  then,  as  a  constant  fac- 
tor, be  employed  in  (3)  for  finding  the  value  of  c  for  any  ob- 
served value  of  6  ;  but  the  same  units  must  be  used  as  before 


540.  Velocities  in  Different  Parts  of  a  Transverse  Section,  — 
The  results  of  velocity-measurements  made  by  many  experi- 
menters do  not  agree  in  supporting  any  very  definite  relation 
.between  the  greatest  surface  velocity  (c0  max.)  of  a  transverse 


754  MECHANICS   OF   ENGINEERING. 

section  and  the  velocities  at  other  points  of  the  section,  but 
establish  a  few  general  propositions  : 

1st.  In  any  vertical  line  the  velocity  is  a  maximum  quite 
near  the  surface,  and  diminishes  from  that  point  both  toward 
the  bottom  and  toward  the  surface. 

2d.  In  any  transverse  horizontal  line  the  velocity  is  a  maxi- 
mum near  the  middle  of  the  stream,  diminishing  toward  the 
banks. 

3d.  The  mean  velocity  =  v,  of  the  whole  transverse  section, 
i.e.,  the  velocity  which  must  be  multiplied  by  the  area,  F,  of 
the  section,  to  obtain  the  volume  delivered  per  unit  of  time, 


(1) 


Is  about  83  per  cent  of  the  maximum  surface  velocity  (<?0  max.) 
observed  when  the  air  is  still  ;  i.e., 

v  =  0.83  X  (o9  max.)  .......     (2) 

Of  eight  experimenters  cited  by  Prof.  Bowser,  only  one  gives 
a  value  (=  0.62)  differing  more  than  .05  from  .83,  while  others 
obtained  the  values  .82,  .78,  .82,  .80,  .82,  .83. 

In  the  survey  of  the  Mississippi  Eiver  by  Humphreys  and 
Abbot,  1861,  it  was  found  that  the  law  of  variation  of  the 
velocity  in  any  given  vertical  line  could  be  fairly  well  repre- 
sented by  the  ordinates  of  a  parabola  (Fig.  607)  with  its  axis 
horizontal  and  its  vertex  at  a  distance  d, 
j£fjEf^'j—  :-=  below  the  surface  according  to  the  follow- 
~=  ing  relation,  f"  being  a  number  dependent 
T^.  on  the  force  of  the  wind  (from  0  for  no 


_ 

~~'  wind  to  10  for  a  hurricane)  : 


(3) 


^-+?'__~=-  where  d  is  the  total  depth,  and  the  double 
sign  is  to  be  taken  +  for  an  up-stream,  - 
FK>.  GOT.  for  a   down-stream,  wind.     The  following 

relations  were  also  based  on  the  results  of  the  survey : 


(putting,  for  brevity,  B  =  1.69  -f-  Vd  +  1.5,)      .     (4) 


VELOCITIES   IN   OPEtf   CHANNEL.  755 

......     (5) 


jKH-ta),   ....     (6) 
and 

c*  =  cin  +  3FtfBv.      .....    (7) 

(These  equations  are  not  of  homogeneous  form,  but  call  for 
the  foot  and  second  as  units.) 
In  (4),  (5),  and  (6), 

c  =  velocity  at  any  depth  z  below  the  surface  ; 
cm  =  mean  velocity  in  the  vertical  curve  ; 
cdl  =  max.         «  "  "          " 

qdl  =  "at  mid-depth  ; 

cd  =  velocity  at  bottom  ; 
v  =  mean  velocity  of  the  whole  transverse  section. 

It  was  also  found  that  the  parameter  of  the  parabola  varied 
inversely  as  the  square  root  of  the  mean  velocity  cm  of  curve. 

In  general  the  bottom  velocity  (cd)  is  somewhat  more  than  -J- 
the  maximum  velocity  (cdl)  in  the  same  vertical.  In  the  Mis- 
sissippi the  velocity  at  mid-depth  in  any  vertical  was  found  to 
be  very  nearly  .96  of  the  surface  velocity  in  the  same  vertical; 
this  fact  is  important,  as  it  simplifies  the  approximate  gauging 
of  a  stream. 

541.  Gauging  a  Stream  or  River,  —  Where  the  relation  (eq.  (2), 
§  540)  v  =  .83  (c0  max.)  is  not  considered  accurate  enough  for 
substitution  in  Q  =  Fv  to  obtain  the  volume  of  discharge  (or 
delivery)  Q  of  a  stream  per  time-unit,  the  transverse  section 
may  be  divided  into  a  number  of  subdivisions  as  in  Fig.  608, 
of  widths  al  ,  «Q,  etc.,  and 
mean  depths  dl  ,  d^  ,  etc., 
and  the  respective  mean 
velocities,  cl  ,  £,  ,  etc.,  com- 
puted from  measurements  FIG.  eos. 

with  current-meters  ;  whence  we  may  write 

...     (7) 


756' 


MECHANICS   OF   ENGINEEBING. 


With  a  small  stream  or  ditch,  however,  we  may  erect  a  ver- 
tical boarding,  and  allow 
the  water  to  flow  through  a 
rectangular  notch  or  over- 
fall, Fig.  609,  and  after  the 
head  surface 
permanent, 


has  become 
measure  A2 
(depth  of  sill  below  the 
level  surface  somewhat 
back  of  boards),  and  b 
(width)  and  use  the  formu- 
lae of  §  504;  see  examples 


FIG.  609. 


in  that  article. 


542.  Uniform  Motion  in  an  Open  Channel.  —  We  shall  tiow 
consider  a  straight  stream  of  indefinite  length  in  which  the 
flow  is  steady,  i.e.,  a  state  of  permanency  exists,  as  distin- 
guished from  a  freshet  or  a  wave.  That  is,  the  flow  is  steady 
when  the  water  assumes  fixed  values  of  mean  velocity  v,  and 
sectional  area  F,  on  passing  a  given  point  of  the  bed  or  bank  ; 
and  the 


Eq.  of  continuity  .  .  Q  =  Fv  =  FQvQ  =  F^vl  =  constant  .  .  (1) 

holds  good  whether  those  sections  are  equal  or  not. 

By  uniform  motion  is  meant  that  (the  section  of  the  bed 
and  banks  being  of  constant  size  and  shape)  the  slope  of  the 
bed,  the  quantity  of  water  (volume  =  Q)  flowing  per  time- 
unit,  and  the  extent  of  the  wetted  perimeter,  are  so  adjusted 
to  each  other  that  the  mean  velocity  of  flow  is  the  same  in  all 
transverse  sections,  and  consequently  the  area  and  shape  of  the 
transverse  section  is  the  same  at  all  points  ;  and  the  slope  of 
the  surface  =  that  of  the  bed.  We  may  therefore  consider, 
for  simplicity,  that  we  have  to  deal  with  a  prism  of  water  of 
indefinite  length  sliding  down  an  inclined  rough  bed  of  con- 
stant slope  and  moving  with  uniform  velocity  (viz.,  the  mean 
velocity  v  common  to  all  the  sections)  ;  that  is,  there  is  no  ac- 
celeration. Let  Fig.  610  show,  free,  a  portion  of  this  prism, 
of  length  =  Z,  and  having  its  bases  ~\  to  the  bed  and  surface. 


UNIFORM   MOTION.      OPEN   CHANNEL.  757 

The  hydrostatic  pressures  at  the  two  ends  balance  each  other 
from  the  identity  of  conditions.     The  only  other  forces  having 


FIG.  610. 

Components  parallel  to  the  bed  and  surface  are  the  weight 
&  =  Fly  of  the  prism  (where  y  =  heaviness  of  water)  making 
an  angle  =  s  (=  slope)  with  a  normal  to  the  surface,  and  the 
friction  between  the  water  and  the  bed  which  is  parallel  to  the 
surface.  The  amount  of  this  friction  for  the  prism  in  question 
may  be  expressed  as  in  §  510,  viz.: 

P=fric.=fSr^=fwly?-,     ...    (2) 

in  which  S  =  wl  =  rubbing  surface  (area)  =  wetted  perimeter, 
w,  X  length  (see  §  538),  and  /"an  abstract  number.  Since  the 
mass  of  water  in  Fig.  610  is  supposed  to  be  in  relative  equili- 
brium, we  may  apply  to  it  the  laws  of  motion  of  a  rigid  body, 
and  since  the  motion  is  a  uniform  translation  (§  109)  the  com- 
ponents, parallel  to  the  surface,  of  all  the  forces  must  balance. 


.*.  Q-  sin  s  must  =  P  =  fric, ;    .*.   Fly  •=-  =fwly  — ; 

(/  2i(j 

thence 

»=/$•$' w 

or 

*^y>» (3) 

in  which  F—w  is  called  7?,  the  hydraulic  mean  depth,  or 
hydraulic  radius.     (3)  is  sometimes  expressed  by  saying  that 


758 


MECHANICS    OF  ENGINEERING. 


the  whole  fall,  or  head,  A,  is  (in  uniform  motion)  absorbed  in 
friction-head.     Also,  since  the  slope  s  =  h  —  I,  we  have 


which  is  of  the  same  form  as  Chezy's  formula  in  §  519  for  a 
very  long  straight  pipe  (the  slope  s  of  the  actual  surface  in  this 
case  corresponding  to  the  slope  along  piezometer-summits  in 
that  of  a  closed  pipe).  In  (4)  the  coefficient  A  =  V%g  H-/  is 
not,  like/",  an  abstract  number,  but  its  numerical  value  depends 
on  the  system  of  units  employed. 

542a.  Experiments  on  the  Flow  of  Water  in  Open  Channels. — 
Those  of  Darcy  and  Bazin,  begun  in  1855  and  published  in 
1865  ("  Kecherches  Hydrauliques"),  were  ^ery  carefully  con- 
ducted with  open  conduits  of  a  variety  of  shapes,  sizes,  slopes, 
and  character  of  surface.  In  most  of  these  i  uniform  flow  was 
secured  before  the  taking  of  measurements.  The  velocities 
ranged  between  from  about  0.5  to  8  or  10  ft.  per  second,  the 
hydraulic  radii  from  0.03  to  3.0  ft.,  with  deliveries  as  high  as 
182  cub.  ft.  per  second.  For  example,  the  following  results 
were  obtained  in  the  canals  of  Marseilles  and  Craponne,  the 
quantity  A  being  for  the  foot  and  second.  The  sections  were 
nearly  all  rectangular.  See  eq.  (4)  above. 


No. 

Q- 
(cub.  ft. 
per  sec.) 

R. 
(ft.) 

s. 
abs. 
numb. 

V. 

(ft.  per 
sec.) 

A. 

(foot  and 
sec.) 

Character  of  the  masonry 
surface. 

1 
2 
3 
4 
5 
6 
7 

182.73 
143.74 
43.93 
43.93 
43.93 
43.93 
167.68 

1.504 

1.774 
.708 
.615 
.881 
.835 
2.871 

.0037 
.00084 
.029 
.060 
.0121 
.014 
.00043 

10.26 
5.55 
11.23 
13.93 

7.58 
8.36 
2.54 

137.1 
125. 
78.4 
72.5 
73.5 
77.3 
72.2 

Very  smooth. 
Quite      " 

Hammered  stone. 
Rather  rough. 

Mud  and  vegetation. 

[In  Experiment  No.  7  the  flow  had  not  fully  reached  a  state 
of  permanency.] 

Fteleyand  Stearns's  experiments  on  the  Sudbnry  conduit  at 
Boston,  Mass.  (Trans.  A.S.  C.  E,  '83),  from  1878  to  1880,  are 
also  valuable.  This  open  channel  was  of  brick  masonry  with 


UNIFOEM   MOTION.       KUTTEE'S   FOKMULA.  759 

good  mortar  joints,  and  about  9  ft.  wide ;  the  depths  of  water 
ranging  from  1.5  to  4.5  ft.  With  plaster  of  pure  cement  on 
the  bed  in  one  of  the  experiments  the  high  value  of  A  =  153.6 
was  reached  (foot  and  second),  with  v  =  2.805  ft.  per  second, 
R  =  2.111  ft.,  s  =  .0001580,  and  Q  =  87.17  cu.  ft.  per  second. 

Captain  Cunningham,  in  his  experiments  on  the  Ganges 
Canal  at  Koorkee,  India,  in  1881,  found  A  to  range  from  48 
to  130  (foot  and  second). 

Humphreys  and  Abbot's  experiments  on  the  Mississippi 
River  and  branches  (see  §  540),  with  values  of  R  —  from  2  or 
3  ft.  to  72  ft.,  furnish  values  of  A  —  from  53  to  167  (foot  and 
second). 

542b.  Kutter's  Formula. — The  experiments  upon  which 
Weisbach  based  his  deductions  for/,  the  coefficient  of  fluid 
friction,  were  scanty  and  on  too  small  a  scale  to  warrant  gener- 
al conclusions.  That  author  considered  that/  depended  only 
on  the  velocity,  disregarding  altogether  the  degree  of  rough- 
ness of  the  bed,  and  gave  a  table  of  values  in  accordance  with 
that  view,  these  values  ranging  from  .0075  for  15  ft.  per  sec. 
to  .0109  for  0.4  ft.  per  sec.;  but  in  1869  Messrs.  Kutter  and 
Ganguillet,  having  a  much  wider  range  of  experimental  data 
at  command,  including  those  of  Darcy  and  Bazin,  and  those 
obtained  on  the  Mississippi  River,  evolved  a  formula,  known 
as  Kutter }s  Formula,  for  the  uniform  motion  of  water  in  open 
channels,  which  is  claimed  to  harmonize  in  a  fairly  satisfactory 
manner  the  chief  results  of  the  best  experiments  in  that  direc- 
tion. They  make  the  coefficient  A  in  eq.  (4)  (or  rather  the 

factor  -—  contained  in  A)  a  function  of  R,  s,  and  also  n  an 

J 

abstract  number,  or  coefficient  of  roughness,  depending  on  the 
nature  of  the  surface  of  the  bed  and  banks  ;  viz., 


1.811       .00281 


41.6  +  ±l±±  + 


v  in 

per 
sec. 


which  is  Kutter's  Formula.* 


*  A  book  of  "  Diagrams  of  Mean  Velocity  based  on  Kutter's  Formula,"  by 
the  present  writer  (New  York,  J.  Wiley  &  Sons,  1902),  obviates  the  necessity 
of  numerical  substitution  in  Kutter's  formula  for  all  practical  purposes. 


760  MECHANICS   OF   ENGINEERING. 

That  is,  comparing  (5)  with  (4),  we  have  f  a  function  of 
JR,  and  s,  as  follows : 


/= 


n 


.     .    .    (6) 


From  (6)  it  appears  that  /  decreases  with  an  increasing  R, 
ts  has  been  also  noted  in  the  case  of  closed  pipes  (§  517) ;  that 
it  increases  with  increasing  roughness  of  surface ;  and  that  it 
is  somewhat  dependent  on  the  slope.  The  makers  of  the 
formula  give  the  following  values  for  n. 

Values  of  n.    n  = 

,009  for  well-planed  timber  bed  ; 

.010  for  plaster  in  pure  cement ; 

.011  for  plaster  in  cement  with  \  sand  ; 

.012  for  unplaned  timber ; 

.013  for  ashlar  and  brickwork ;  * 

.015  for  canvas  lining  on  frames; 

.017  for  rubble ; 

.020  for  canals  in  very  firm  gravel ; 

.025  for  rivers  and  canals  in  perfect  order  and  regimen,  and 
perfectly  free  from  stones  and  weeds ; 

.030  for  rivers  and  canals  in  moderately  good  order  and  regi- 
men, having  stones  and  weeds  occasionally ; 

.035  for  rivers  and  canals  in  bad  order  and  regimen,  overgrown 
with-  vegetation  and  strewn  with  stones  or  detritus  of 
any  sort. 

Kutter's  Formula  is  claimed  to  apply  to  all  kinds  and  sizes  of 
watercourses,  from  large  rivers  to  sewers  and  ditches  ;  for  uni- 
form motion.  If  VR  is  the  unknown  quantity,  Kutter's  For- 
mula leads  to  a  quadratic  equation  ;  if  s  the  slope,  to  a  cubic. 
Hence,  to  save  computation,  tables  have  been  prepared,  some 
of  which  will  be  found  in  vol.  28  of  Yan  Nostrand's  Magazine 

*  For  ordinary  brick  sewers  Mr.  R.  F.  Hartford  claims  that  n  =  .014 
gives  good  results.  See  Jour.  Eng.  Societies  for  '84- '85,  p.  220. 


UNIFORM    MOTION.      OPEN   CHANNEL. 


761 


(pp.  135  and  393)  (sewers),  and  in  Jackson's  works  on  Hydrau- 
lics (rivers).  See  foot-note  on  p.  759. 

The  following  table  will  give  the  student  an  idea  of  the 


variation   of   the   coefficient  A,  = 


3 ,  of  eq.  (4),  or  large 
J 


bracket  of  eq.  (5),  with  different  hydraulic  radii,  slopes,  and 
values  of  7i,  according  to  Kutter's  Formula ;  from  R  —  \  ft., 
for  a  small  ditch  or  sluice-way  (or  a  wide  and  shallow  stream), 
to  R  =  15  ft.,  for  a  river  or  canal  of  considerable  size.  Under 
each  value  of  R  are  given  two  values  of  A  ;  one  for  a  slope  of 
3'  =  .001,  and  the  other  for  s"  =  .00005.  All  these  values  of 
A  imply  the  use  of  the  foot  and  second. 

These  values  of  A  have  been  scaled  by  the  writer  from  a 
diagram  given  in  Jackson's  translation  of  Kutter's  "  Hydraulic 
Tables,"  and  are  therefore  only  approximate.  The  corre- 
sponding values  of  /",  the  coefficient  of  fluid  friction,  can  be 

2(7 

computed  from  f=—^. 


n 

R  =  $  f  t. 

R  =  1  ft. 

R  =  3  f  t. 

R  =  6  ft. 

R  =  15  ft. 

for          for 

s'            s" 

for           for 

s'             s" 

for          for 

s'            s" 

for          for 

s'            s" 

for      for 

s'        s" 

0.010 
0.015 
0.020 
0.025 
0.030 
0  035 

133        114 
83          68 
58          49 
45          38 
36          31 
28          25 

149        137 
96          87 
70          63 
54          48 
43         40 
38          34 

174        174 
118        118 
87          87 
70          70 
58          58 
50          50 

187        196 
128        137 
98        106 
80         85 

66          72 
58          64 

199    222 
138    158 
110    126 
90    106 
77      90 
68      81 

The  formula  used  in  designing  the  New  Aqueduct  for  New 
York  City,  in  1885,  by  Mr.  Fteley,  consulting  engineer,  was 

[see  (4)]  

v  (ft.  per  sec.)  =  142  VR(inft.)  X  <<?,...     (7) 

whereas  Kutter's  Formula  gives  for  the  same  case  (a  circular* 
section  of  14  ft.  diameter,  and  slope  of  0.7  ft.  to  the  mile), 
with  n  =  0.013, 


v  (ft.  per  sec)  =  140.7  VR(inft)  X  *.  .    .     .     (8) 

*  The  aqueduct  has  this  circular  form  for  a  small  portion,  only,  of  its 
length;  a  "horseshoe"  section  of  very  nearly  the  same  flowing  capacity 
being  given  to  the  greater  portion  of  the  remainder. 


762  MECHANICS   OF   ENGINEEKING. 

To  quote  from  a  letter  of  Mr.  I.  A.  Shaler  of  the  Aqueduct 
Corps  of  Engineers,  "  Mr.  Fteley  states  that  the  cleanliness  of 
the  conduit  (Sudbury)  had  much  to  do  in  affecting  the  flow. 
He  found  the  flow  to  be  increased  by  7  or  8  per  cent  in  a  por- 
tion which  had  been  washed  with  a  thin  wash  of  Portland 
cement." 

EXAMPLE  1.  —  A  canal  1000  ft.  long  of  the  trapezoidal  sec- 
tion in  Fig.  611  is  required  to  deliver  300  cubic  ft.  of  water 
per  second  with  the  water  8  ft.  deep  at  all 
sections  (i.e.,  with  uniform  motion),  the 
slope  of  the  bank  being  such  that  for  a  depth 
of  8  ft.  the  width  of  the  water  surface  (or 
length  of  air-profile)  will  be  20  ft.;  and  the 
coefficient  for  roughness  being  n  =  .020.  What  is  the  neces- 
sary slope  to  be  given  to  the  bed  (slope  of  bed  =  that  of  sur- 
face, here)  (ft.,  lb.,  sec.)  ? 
The  mean  velocity 

v  =  Q  +  jr—  300  -r-  i  (20  +  8)  8  =  2.67  ft.  per  sec. 

[So  that  the  surface  velocity  of  mid-channel  in  any  section 
would  probably  be  (<?omax)  =  v  -f-  0.83  —  3.21  ft.  per  sec.  (eq. 
(2),  §  540).] 

The  wetted  perimeter 

w  =  8  +  2  1/8'  +  6'  =  28  ft., 
and  therefore  the  mean  hydraulic  depth 

=  E  =  F+  w  =  112  -7-  28  =  4  ft. 

To  obtain  a  first  approximation  for  the  slope,  we  may  use 
the  value  /'  =  .00795  given  by  Weisbach  for  a  velocity  of  2.67 
ft.  per  sec.,  and  obtain,  from  (3), 


A==.  00795  X  1000  X          .        =  ()221f 
112  X  2  X  32.2 


i.e.  s  =  h  +  l  =  .000221. 


UNIFOKM   MOTION   IN   OPEN   CHANNEL. 


763 


With  this  value  for  the  slope  and  E  =  4  ft.  (see  above),  we 
then  have,  from  eq.  (6)  (putting  n  =  .020), 


.00281  \   .020 


K 
' 


.2256        .00035 


=  .0071, 


with  which  value  of  f  we  now  obtain 

h  =  0.200  feet ;     i.e.,    slope  =  s  =  .00020. 

EXAMPLE  2. — If  the  bed  of  a  creek  falls  20  inches  every 
1500  ft.  of  length,  what  volume  of  water  must  be  flowing  to 
maintain  a  uniform  mean  depth  of  4£  ft.,  the  corresponding 
surface- width  being  40  ft.,  and  wetted  perimeter  46  ft.  ?  The 
bed  is  "  in  moderately  good  order  and  regimen  ;"  use  Kutter's 
Formula,  putting  n  =  0.030  (ft.  and  sec.). 

First  we  have 


=  y/(40X  44)  -  (46  X  q~?)  =  .066, 


while  VrR~(K)  =1.98,  and  the  slope  =  s  =  ff  -r-  1500=.  00111  ; 
hence 


= 


104.43  X  .066 
1.6685       ' 


1.98 
v  =  4.13  ft.  per  sec. 


or 

Hence,  also, 

Q  =  Fv  =  40  X  4J  X  4.13  =  Y43.4  cub.  ft.  per  sec. 


[N.B.  Weisbach  works  this  same  example  by  eq.  (3)  with  a 
value  off  taken  from  his  own  table,  his  result  being  v  =  6.1 


764  MECHANICS    OF   ENGINEERING. 

ft.  per  sec.,  which  would  probably  be  attained  in  practice  only 
by  making  the  bed  and  banks  smoother  than  as  given.] 

EXAMPLE  3. — The  desired  transverse  water-section  of  a  canal 
is  given  in  Fig.  612.  The  slope  is  to  be 
3  ft.  in  1600 ;  i.e.,  s  =  3  -r-  1600 ;  or,  for 
I  —  1600  ft.,  h  =  3  ft.  What  must  be  the 
velocity  (mean)  of  each  section,  for  a  uni- 
form motion,  the  corresponding  volume 
delivered  per  sec.,  Q,  =  Ft),  —  ? ;  assuming  that  the  character 
of  the  surface  warrants  the  value  n  =  .030  ? 

Knowing  the  slope  s,  =  3  ~-  1600  ;  and  the  hydraulic  radius 
R,  =  F+-w,  =  79.28  sq.  ft.  -r-  24.67  ft.,  =  3.215  feet ;  with 
w=  .030  we  substitute  directly  in  eq.  (5),  obtaining  v  =  4.6T 
ft.  per  sec. ;  whence  Q  =  Fv  —  370  cub.  ft.  per  sec. 

543.  Hydraulic  Mean  Depth  for  a  Minimum  Frictional  Resist- 
ance.— We  note,  from  eq.  (3),  §  542,  that  if  an  open  channel 
of  given  length  I  and  sectional  area  F  is  to  deliver  a  given 
volume,  Q,  per  time- unit  with  uniform  motion,  so  that  the 
common  mean  velocity  v  of  all  sections  (—  Q  -f-  F)  is  also  a 
given  quantity,  the  necessary  fall  =  A,  or  slope  s  —  h  -=-  /,  is 
seen  to  be  inversely  proportional  to  7?,  the  hydraulic  mean 
depth  of  the  section,  =  (F  -r-  w\  =  sectional  area  -=-  wetted 
perimeter. 

For  h  to  be  as  small  as  possible,  we  may  design  the  form  of 
transverse  section,  so  as  to  make  R  as  large  as  possible ;  i.e., 
to  make  the  wetted  perimeter  a  minimum  for  a  given  F\  for 
in  this  way  a  minimum  of  frictional  contact,  or  area  of  rub- 
bing surface,  is  obtained  for  a  prism  of  water  of  given  sectional 
area  T^and  given  length  I. 

In  a  closed  pipe  running  full  the  wetted  perimeter  is  the 
whole  perimeter;  and  if  the  given  sectional  area  is  shaped  in 
the  form  of  a  circle,  the  wetted  perimeter,  =  w,  is  a  minimum 
(and  R  a  maximum).  If  the  full  pipe  must  have  a  polygonal 
shape  of  n  sides,  then  the  regular  polygon  of  n  sides  will  pro- 
vide a  minimum  w. 

Whence  it  follows  that  if  the  pipe  or  channel  is  running 


UNIFORM   MOTION   IN    OPEN   CHANNEL. 


765 


half  full,  and  thus  becomes  an  open  channel,  the  semicircle, 

of  all  curvilinear  water  pro- 
files, gives  a  minimum  w. 
Also,  of  all  trapezoidal  pro- 
files with  banks  at  60°  with 
the  horizontal  the  half  of  a 
regular  hexagon  gives  a 
minimum  w.  Among  all 
rectangular  sections  the  half 
square  gives  a  minimum  w  ; 

and  of  all  half  octagons  the  half  of  a  regular  octagon  gives  a 
minimum  w  (and  max.  ^)  for  a  given  F.  See  Fig.  613  for 
fill  these. 

The  egg-shaped  outline,  Fig.  614,  small  end  down,  is  fre- 
quently given  to  sewers  in  which  it  is  important  that  the 
different  velocities  of  the  water  at  dif- 
ferent stages  (depths)  of  flow  (depend- 
ing on  the  volume  of  liquid  passing  per 
unit-time)  should  not  vary  widely  from 
each  other.  The  lower  portion  ABC, 
providing  for  the  lowest  stage  of  flow 
AB,  is  nearly  semicircular,  and  thus  in- 
duces a  velocity  of  flow  (the  slope  being 
constant  at  all  stages)  which  does  not 
differ  extremely  from  that  occurring 
when  the  water  flows  at  its  highest 
stage  DE,  although  this  latter  velocity  is  the  greater;  the 
reason  being  that  ABC  from  its  advantageous  form  has  a 
hydraulic  radius,  R,  larger  in  proportion  to  its  sectional  area, 
F\  than  DCE. 

That  is,  F  -=-  w  for  ABC  is  more  nearly  equal  to  F -^  w  for 
DEC  than  if  DEC  were  a  semicircle,  and  the  velocity  at  the 
lowest  stage  may  still  be  sufficiently  great  to  prevent  the  de- 
posit of  sediment.  See  §  575. 

544.  Trapezoid  of  Fixed  Side-slope. — For  large  artificial  water- 
courses and  canals  the  trapezoid,  or  three-sided  water-profile 
(symmetrical),  is  customary,  and  the  inclination  of  the  bank, 


FIG.  614. 


766 


MECHANICS    OF   ENGINEERING. 


or  angle  6  with  the  horizontal,  Fig.  615,  is  often  determined 

by  the  nature  of  the  material 
composing  it,  to  guard  against 
washouts,  caving  in,  etc.  We 
are  therefore  concerned  with  the 
following  problem  :  (riven  the 
area,  F,  of  the  transverse  section, 
and  the  angle  0,  required  the  value  of  the  depth  x  (or  of  upper 
width  z,  or  of  lower  width  y,  both  of  which  are  functions  of  x) 
to  make  the  hydraulic  mean  depth,  R  ==  F  -f-  w,  a  maximum, 
or  w  -r-  F  a  minimum.  F\s>  constant. 
From  the  figure  we  have 


FlG-  615- 


and 
whence 


0, 


.    (1) 


3cot.  6; 
=  i.(^-a?3cot.  8),     ........    (2) 


substituting  which  in    (1)  and  dividing  by  F9  noting  that 
2  cosec.  6  —  cot.  0  =  —  ^  —  —  »  we  have 


sin  0 


w 


For  a  minimum  w  we  put 


1    .  2  —  cos0 


dx 


=  0; 


.'.  a?  (for  max.  or  min.  w)  =  ±  A  /=  —        ^. 

"  y  2  —  cos  6 

The  +  sign  makes  the  second  derivative  positive,  and  hence 
for  a  min.  w  or  max.  12  we  have 

VF*™6 


x  (call  it  a/)  =x'= 


—  cos 


TRAPEZOID   FOB  MINIMUM    WETTED   PERIMETEE.     767 

while  the  corresponding  values  for  the  other  dimensions  are 

y'  =  -^—  x'  cot.  0  ...    ......     (5) 

and 

z'=y'  +  2x'cot.0=^-  +  a'cot.0.    ...     (6) 
OK 

For  the  corresponding  hydraulic  mean  depth  R!  [see  (3)], 
i.e.9  the  max.  R,  we  have 


_1_       1        2  —  cosfl    ,__2_ 
'~    '  "  *     =    ' 


™      j/      1  . 

=  ..... 


Equations  (4),  (5),  ...  (8)  hold  good,  then,  for  the  trapezoi- 
dal section  of  least  frictional  resistance  for  a  given  angle  0. 

PROBLEM.  —  Required  the  dimensions  of  the  trapezoidal  sec- 
tion of  minimum  frictional  resistance  for  6  =  45°,  which  with 
h  =  6  inches  fall  in  every  1200  feet  (=  I)  is  required  to  de- 
liver Q  =  360  cub.  ft.  of  water  per  minute  with  uniform 
motion. 

Here  we  have  given,  with  uniform  motion,  A,  Z,  and  Q, 
with  the  requirement  that  the  section  shall  be  trapezoidal,  with 
0  =  450,  and  of  minimum  frictional  resistance.  The  following 
equations  are  available  : 

Eq.  of  continuity  .  .  .  Q  =  Fv,  ......    (!') 


Eq.  (8)  preceding,  for  con- )       r>/  __  _!      /     8ip  # 
dition  of  least  resistance   )  '          "  2   \  2  —  cos  0 


There  are  three  unknown  quantities,  v,  F,  and  It'.     Solve 


768 


MECHANICS   OF   ENGINEERING. 


(!')  for  v;  solve (2')  for  R'\  substitute  their  values  in  (3')$ 
whence 


-D 


._r2/7V2^o^#n* 

SghVsin  V      J 


Since/  cannot  be  exactly  computed  in  advance,  for  want  of 
knowing  the  value  of  It,  we  calculate  it  approximately  [eq.  (6)> 
§  542  b]  for  an  assumed  value  of  R,  insert  it  in  the  above 
equation  (4'),  and  thus  find  an  approximate  value  of  F\  and 
then,  from  (8),  a  corresponding  value  of  R,  from  which  a  new 
value  of  /can  be  computed.  Thus  after  one  or  two  trials  a 
satisfactory  adjustment  of  dimensions  can  be  secured. 

545.  Variable  Motion.  —  If  a  steady  flow  of  water  of  a  de- 
livery  Q,  =  Fv,  =  constant,  takes  place  in  a  straight  open 
channel  the  slope  of  whose  bed  has  not  the  proper  value  to 
maintain  a  "  uniform  motion"  then  "  variable  motion"  ensues 
(the  flow  is  still  steady,  however);  i.e.,  although  the  mean 
velocity  in  any  one  transverse  section  remains  fixed  (with  lapse 
of  time),  this  velocity  has  different  values  for  different  sections  ; 
but  as  the  eq.  of  continuity, 


etc., 


still  holds  (since   the  flow  is  steady),  the  different  sections 

have  different  areas.  If, 
Fig.  616,  a  stream  of 
water  flows  down  an 
inclined  trough  without 
friction,  the  relation 


between  the   velocities 
v0  and   vl  at   any  two 


FIG.  616.  sections  0  and  1  will  be 


the  same  as  for  a  material  point  sliding  down  a  guide  without 
friction  (see  §  79,  latter  part),  viz.  : 


v,9  _  v0* 


VARIABLE  MOTION.       OPEN   CHANNEL.  769 

an  equation  of  heads  (really  a  case  of  Bernoulli's  Theorem, 
§  492).  But,  considering  friction  on  the  bed,  we  must  sub- 

tract the  mean  friction-head  f  -=>-  •—  [see  eqs.  (3)  and  (3'), 
§  542]  lost  between  0  and  1  ;  this  friction-head  may  also  bet 
written  thus:  f^-L  •  and  therefore  eq.  (1)  becomes 


which  is  the  formula  for  variable  motion /  and  in  it  I  is  the 
length  of  the  section  considered,  which  should  be  taken  short 
enough  to  consider  the  surface  straight  between  the  end-sec- 
tions, and  the  latter  should  differ  but  slightly  in  area.  The 
subscript  m  may  be  taken  as  referring  to  the  section  midway 
between  the  ends,  so  that  vm*  =  i(^02  +  v,2).  The  wetted  pe- 
rimeter wm  =  J(w0  -f-  w,)9  and  Fm  =  %(F0  +  F^.  Hence  eq. 
(2)  becomes 

and  again,  by  putting  VQ  =  Q  ~-  F0 ,  v,  =  Q  -r-  Fl ,  we  may 
write 

A=r|-l         1,1  f  %».  +  »>)  (I     ,    1 
whence 


From  eq.  (4),  having  given  the  desired  shapes,  areas,  etc.,  of 
the  end-sections  and  the  volume  of  water,  Q,  to  be  carried  per 
unit  of  time,  we  may  compute  the  necessary  fall,  A,  of  the  sur- 
face, in  length  =  I  ;  while  from  eq.  (5),  having  observed  in  an 
actual  water-course  the  values  of  the  sectional  areas  F9  and  F.  $ 
the  wetted  perimeters  w9  and  wl  ,  the  length,  =  I,  of  the  pen- 


770 


MECHANICS   OF 


tion  considered,  we  may  calculate  Q  and  thus  gauge  the  stream 
approximately,  without  making  any  velocity  measurements. 

As  to  the  value  of/1,  we  compute  it  from  eq.  (6),  §  542b, 
using  for  R  a  mean  between  the  values  of  the  hydraulic  radii 
of  the  end-sections. 

546.  Bends  in  an  Open  Channel. — According  to  Humphreys 
and  Abbot's  researches  on  the  Mississippi  Kiver  the  loss  of 
head  due  to  a  bend  may  be  put 


536  n  '• 


(i) 


i     A. 


in  which  v  must  be  in  ft.  per  sec.,  and  tf,  the  angle  ABC,  Fig. 
617,  must  be  in  7r-measure,  i.e.  in  radians. 
The  section  F  must  be  greater  than  100 
sq.  ft.,  and  the  slope  s  less  than  .0008.  v 
is  the  mean  velocity  of  the  water.  Hence 
if  a  bend  occurred  in  a  portion  of  a 
stream  of  length  Z,  eq.  (3)  of  §  542  be- 

comes 

v*^ 

Fro.  617. 


h=      " 


while  eq.  (2)  of  §  545  for  variable  motion  would  then  become 


(o  and  S  as  above.)    (For  "  radian"  see  p.  544.) 

547.  Equations  for  Variable  Motion,  introducing  the  Depths. 
--— Fig.  618.     The  slope  of  the  bed  being  sin  a  (or  simply  a, 
arnicas.),  while  that  of  the  surface  is 
different,  viz.,  ""l/T"^ 

sin/?  =  5  =  A-5-Z, 
we  may  write 


FIG.  6ia 


VAKIABLE  MOTION.      OPEN   CHANNEL.  771 

in  which  d0  and  dt  are  the  depths  at  the  end-sections  of  the 
portion  considered  (steady  flow  with  variable  motion).  With 
these  substitutions  in  eq.  (4),  §  545,  we  have,  solving  for  I, 

Q* 


From  which,  knowing  the  slope  of  the  bed  and  the  shape 
and  size  of  the  end-sections,  also  the  discharge  Q,  we  may 
compute  the  length  or  distance,  £,  between  two  sections  whose 
depths  differ  by  an  assigned  amount  (d0  —  d^).  But  we  can- 
not compute  the  change  of  depth  for  an  assigned  length  I  from 
(6).  However,  if  the  width  b  of  the  stream  is  constant,  and 
the  same  at  all  depths  ;  i.e.,  if  all  sections  are  rectangles  hav- 
ing a  common  width  ;  eq.  (6)  may  be  much  simplified  by  intro- 
ducing some  approximations,  as  follows  :  We  may  put 


F; 


approx'  = 


<0  —  <,    v 


and,  similarly, 


which  approx.  =  2fe-  —  . 
djb  2g 

Hence  by  substitution  in  eq.  (6)  we  have 


--  sin 
djb  2g 


547a.  Backwater. — Let  us  suppose  that   a  steady  flow  "has 
been  proceeding  with  uniform  motion  (i.e.,  the  surface  parallel 


772 


MECHANICS    OF    ENGINEERING. 


to  the  bed)  in  an  open  channel  of  indefinite  extent,  and  that  a 
vertical  wall  is  now  set  up  across  the  stream.  The  water  rises 
and  flows  over  the  edge  of  the  wall,  or  weir,  and  after  a  time 
a  steady  flow  is  again  established.  The  depth,  y0  ,  of  the  water 
close  to  the  weir  on  the  up-stream  side  is  greater  than  d0  ,  the 
original  depth.  We  now  have  "  variable  motion  "  above  the 
weir,  and  at  any  distance  x  up-stream  from  the  weir  the  new 
depth  y  is  greater  than  d0  .  This  increase  of  depth  is  called 
backwater,  and,  though  decreasing  up-stream,  may  be  percep- 
tible several  miles  above  the  weir.  Let  s  be  the  slope  of  the 
original  uniform  motion  (and  also  of  present  bed),  and  v  the 

v* 
velocity   of  the   original   uniform   motion,  and   let    ~k  —  —  . 

y 

Then,  if  the  section  of  the  stream  is  a  shallow  rectangle  of 
constant  width,  we  have  the  following  relation  (Rankine)  : 


(1) 


»  =  l[y.  -  y  +  (d.  -  2*X0  -  < 
where  0  is  a  function  of  ^-,  as  per  following  table  : 


For  —  =  1.0 

1.10 

1.20 

1.30 

1.40 

1.50 

1.60 

1.70 

do 

<p  =  oo 

.680 

.480 

.376 

.304 

.255 

.218 

.189 

•For|-  =1.80 

1.90 

2.00 

2.20 

2.40 

2.60 

2.80 

3.0 

0=  .166 

.147 

.132 

.107 

.089 

.076 

.065 

.056 

</>0  is  found  from  ^-,  precisely  as  0  from  ^-,  by  use  of  the  table. 

a0.  a0 

With  this  table  and  eq.  (1),  therefore,  we  can  find  #,  the  dis- 
tance (u  amplitude  of  backwater")  from  the  weir  of  the  point 
where  any  assigned  depth  y  (or  "  height  of  backwater,"  y  —  d0) 
will  be  found. 

For  example,  Prof.  Bowser  cites  the  case  from  D'Aubuis- 
son's  Hydraulics  of  the  river  Weser  in  Germany,  where  the 
erection  of  a  weir  increased  the  depth  at  the  weir  from  2.5  ft. 
to  10  ft,  the  flow  having  been  originally  "uniform"  for  10 
miles.  Three  miles  above  the  dam  the  increase  (>j  —  dQ)  of 
depth  was  1.25  ft.,  and  even  at  four  miles  it  was  0  75  ft. 


CHAPTEK  VIII. 

DYNAMICS  OF  GASEOUS  FLUIDS. 

548,  Steady  Flow  of  a  Gas. — [N.B.  The  student  should  now 
review  §  492  up  to  eq.  (5).]     The  differential  equation  from 
which  Bernoulli's  Theorem  was  derived  for  any  liquid,  with- 
out friction,  was  [eq.  (5),  §  492] 

-vdv  +  dz  +  -dj>  =  0, (A) 

y  / 

and  is  equally  applicable  to  the  steady  flow  of  a  gaseous  fluid, 
but  with  this  difference  in  subsequent  work,  that  the  heaviness, 
y  (§  '"0?  °f  the  gas  passing  different  sections  of  the  pipe  or 
stream-line  is,  or  may  be,  different  (though  always  the  same  at 
a  given  point  or  section,  since  the  flow  is  steady).  For  the 
present  we  neglect  friction  and  consider  the  flow  from  a  large 
receiver,  where  the  great  body  of  the  gas  is  practically  at  rest, 
through  an  orifice  in  a  thin  plate,  or  a  short  nozzle  with  a 
rounded  entrance. 

In  the  steady  flow  of  a  gas,  since  y  is  different  at  different 
points,  the  equation  of  continuity  takes  the  form 

Flow  of  weightier  time-unit  =  F^vlyl  =  F^v^y^  =  etc. ;  .  (a) 

i.e.,  the  weight  of  gas  passing  any  section,  of  area  F,  per  unit 
of  time,  is  the  same  as  for  any  other  section,  or  Fvy  =  con- 
stant, y  being  the  heaviness  at  the  section,  and  v  the  velocity. 

549.  Flow  through  an   Orifice — Remarks. — In    Fig.  619  we 
have  a  large  rigid  receiver  containing  gas  at  some  tension, pn, 
higher  than  that,  pm,  of  the  (still)  outside  air  (or  gas),  and  at 
some  absolute  temperature  Tn ,  and  of  some  heaviness  yn ;  that 
is,  in  a  state  n.    The  small  orifice  of  area  F  being  opened,  the 
gas  begins  to  escape,  and  if  the  receiver  is  very  large,  or  if  the 
supply  is  continually  kept  up  (by  a  blowing-engine,  e.g.),  after 

773 


774 


MECHANICS   OF   ENGINEERING. 


FIG.  619. 


a  very  short  time  the  flow  becomes  steady.  Let  nm  represent 
any  stream-line  (§  495)  of  the  flow.  According  to  the  ideal 
subdivision  of  this  stream-line  into 
laminae  of  equal  mass  or  weight  (not 
equal  volume,  necessarily)  in  estab- 
lishing eq.  (A)  for  any  one  lamina, 
each  lamina  in  the  lapse  of  time  dt 
moves  into  the  position  just  vacated 
by  the  lamina  next  in  front,  and 
assumes  precisely  the  same  velocity, 
pressure,  and  volume  (and  there- 
fore heaviness}  as  that  front  one  had  at  the  beginning  of  the 
dt.  In  its  progress  toward  the  orifice  it  expands  in  volume, 
its  tension  diminishes,  while  its  velocity,  insensible  at  n,  is 
gradually  accelerated  on  account  of  the  pressure  from  behind 
always  being  greater  than  that  in  front,  until  at  m,  in  the 
"  throat"  of  the  jet,  the  velocity  has  become  vm,  the  pressure 
(i.e.,  tension)  has  fallen  to  a  value  pm ,  and  the  heaviness  has 
changed  to  ym.  The  temperature  Tm  (absolute)  is  less  than 
Tn ,  since  the  expansion  has  been  rapid,  and  does  not  depend 
on  the  temperature  of  the  outside  air  or  gas  into  which  efflux 
takes  place,  though,  of  course,  after  the  effluent  gas  is  once 
free  from  the  orifice  it  may  change  its  temperature  in  time. 

We  assume  the  pressure^  (in  throat  of  jet)  to  be  equal  to 
that  of  the  outside  medium  (as  was  done  with  flow  of  water), 
so  long  as  that  outside  tension  is  greater  than  .527 pn ;  but  if  it 
is  less  than  .527  pn  and  is  even  zero  (a  vacuum),  experiment 
seems  to  show  that^>m  remains  equal  to  0.527  of  the  interior 
tension  pn:  probably  on  account  of  the  expansion  of  the 
effluent  gas  beyond  the  throat,  Fig.  620,  so 
that  although  the  tension  in  the  outer  edge, 
at  a,  of  the  jet  is  equal  to  that  of  the  outside 
medium,  the  tension  at  m  is  greater  because 
of  the  centripetal  and  centrifugal  forces  devel- 
oped in  the  curved  filaments  between  a  and 
m.  (See  §  553.) 

550.  Flow  through  an  Orifice;   Heaviness  assumed  Constant 
during  Flow.  The  Water  Formula. — If  the  inner  tension  pn  ex- 


FIG.  620. 


STEADY   FLOW   OF   GASES.  775 

ceeds  cne  outer,  pm  ,  but  slightly,  we  may  assume  that,  like 
water,  the  gas  remains  of  the  same  heaviness  during  flow. 
Then,  for  the  simultaneous  advance  made  by  all  the  laminae  or 
a  stream-line,  Fig.  619,  in  the  time  dt,  we  may  conceive  an 
equation  like  eq.  (A)  written  out  for  each  lamina  between  n 
and  m.)  and  corresponding  terms  added  ;  i.e., 

/     f*m  f*m  f*m  rfn 

(For  orifices)  ....    i/>»  +  .)(>+./„    ££  =  <>.    .  (Ej 

In  general,  y  is  different  in  the  different  laminae,  but  in  the 
present  case  it  is  assumed  to  be  the  same  in  all;  hence,  with 
m  as  rktura  level  and  h  =  vertical  distance  from  n  to  m,  we 
have,  />om  eq. 


-       =  0.  (1) 

2?      fy  y       y 

But  we  may  put  vn  =  0  ;  while  A,  even  if  several  feet,  is 
small  compared  with  -^  —  ^  .  E.g.,  with  pm  =  15  Ibs.  per 

sq.  in.  and  pn  —  16  Ibs.  per  sq.  in.,  we  have  for  atmospheric 
air  at  freezing  temperature 

^  _  ^  -  1638  feet. 

y      y 

Hence,  putting  vn  —  0  and  h  —  0  in  eq.  (1),  we  have 

^»2  _  Pn  —Pm  (   Water  formula  •  for  small  \  /%\ 

2p  yn  \  difference  of  pressures,  only,  f 

The  interior  absolute  temperature  Tn  being  known,  the  yn 
(interior  heaviness)  may  be  obtained  from  yn  =pny0T0  -f-  Tnp9 
(§  472),  and  the  volume  of  flow  per  unit  of  time  then  obtained 
(fir.«t  solving  (2)  for  vm)  is 


(3) 


where  Fm  is  the  sectional  area  of  the  jet  at  m.     If  the  mouthr 
piece  or  orifice  has  well-rounded  interior  edges,  as  in  Fig.  541, 


776  MECHANICS   OF   ENGINEERING. 

its  sectional  area  F  may  be  taken  as  the  area  Fm  .  But  if  it  is 
an  orifice  in  "thin  plate,"  putting  the  coefficient  of  contraction 
=-.  C'=  0.60,  we  have 

Fm=CF=Q.WF\    and     <?*  =  0.60^w.    .      (4) 

This  volume,  Qm  ,  is  that  occupied  by  the  flow  per  time-unit 
when  in  state  ra,  and  we  have  assumed  that  ym  =  yn  ;  hence 
tfco  weight  of  flow  per  time-unit  is 

G  =  Qmym  =  FmvmYn  =  FmvmYn.    ...    (5) 

EXAMPLE.  —  In  the  testing  of  a  blowing-engine  it  is  found 
capable  of  maintaining  a  pressure  of  18  Ibs.  per  sq.  inch  in  a 
large  receiver,  from  whose  side  a  blast  is  steadily  escaping 
through  a  "  thin  plate"  orifice  (circular)  having  an  area  F  =  4 
sq.  inches.  The  interior  temperature  is  30°  Cent,  and  the  out- 
side tension  15  Ibs.  per  sq.  in. 

Required  the  discharge  of  air  per  second,  both  volume  and 
weight.  The  data  are  :  pn  —  18  Ibs.  per  sq.  in.,  Tn  =  303° 
Abs.  Cent.,  F=  4  sq.  inches,  and^?m  =  15  Ibs.  per  sq.  in.  Use 
ft.-lb.-sec.  system. 

First,  the  heaviness  in  the  receiver  is 

"  =  t  '  Tu  r<  ^  55  '  m  x  -°807  =  -°89  lb8'  per  cub"  ft 

Then,  from  eq.  (2), 


/^Pn-pm_     / 

vm  —  \/    *y  ~\/ 

V  Yn  V 


_    555.3 

feet 
per  sec. 


(97  per  cent  of  this  would  be  more  correct  on  account  of  fric- 
tion.) 

•'•  Qm=Fmvm=.6Fvm  =  TV-rh-X  555.3  =  9.24  cub.  ft.  per  sec. 

at  a  tension  of  15  Ibs.  per  sq.  in.,  and  of  heaviness  (by  hypoth- 

esis) =  .089  Ibs.  per  cub.  ft.     Hence  weight 

=  G  =  9.24  X  .089  =  .82  Ibs.  per  sec. 


FLOW    OF   GASES   BY   MAKIOTTE'S   LAW.  777 

The  theoretical  power  of  the  air-compressor  or  bio  wing-  en- 
gine to  maintain  this  steady  flow  can  be  computed  as  in  Exam- 
ple 3,  §  483. 

551,  Flow  through  an  Orifice  on  the  Basis  of  Mariotte's  Law; 
or  Isothermal  Efflux.  —  Since  in  reality  the  gas  expands  during 
flow  through  an  orifice,  and  hence  changes  its  heaviness  (Fig. 
619),  we  approximate  more  nearly  to  the  truth  in  assuming 
this  change  of  density  to  follow  Mariotte's  law,  i.e0?  that  the 
heaviness  varies  directly  as  the  pressure,  and  thus  imply  that 
the  temperature  remains  unchanged  during  the  flow.  We 
again  integrate  the  terms  of  eq.  (B\  but  take  care  to  note  that, 
noiv,  y  is  variable  (i.e.,  different  in  different  laminae  at  the 
same  instant),  and  hence  express  it  in  terms  of  the  variable  p 
(froineq.  (2),  §475),  thus: 


/»W  ^ny 

Therefore  the  term  /     -^-  of  eq.  (B\  becomes 

Un        y 

=  _P_n      Pn  (1) 

n  Yn         'W 

and,  integrating  all  the  terms  of  eq.  (B\  neglecting  A,  and  call 
ing  vn  zero,  we  have 

v_^_  _  Pn  i        Pn  (  efflux  ly  Mariotte's  \  /2\ 

2<7        yn          pm  '      '  \  Law  through  orifice  \  ' 


T      r> 

As  before,  yn  =  -—-  -  —  y0  ,  and  the  flow  of  volume  per  time- 

%  ^n    Po 

unit  at  ?n  is 

Q.  =  F~v.',    .......    (3) 

while  if  the  orifice  is  in  thin  plate,  Fm  may  be  put  =  .60  F, 
and  the 

weight  of  the  flow  per  time-unit  —  G  —  Fmvmym.  .    (4) 

If  the  mouth-  piece  is  rounded,  Fm  —  F=  area  of  exit  orifice 
of  mouth-  piece. 


778  MECHANICS   OF   ENGINEERING. 

EXAMPLE. — Applying  eq.  (2)  to  the  data  of  the  example  in 
§  550,  where  yn  was  found  to  be  .089  Ibs.  per  cub.  ft.,  we  have 
("ft.,  lb.,  sec.] 


vm  = 


.'•  &  =  ^>m  =  0.60  X  y^f  X  584.7  =  9.745  cub.  ft.  per  sec. 
Since  the  heaviness  at  m  is,  from  Mariotte's  law, 

ym  =  S^  yn  =  f|  of  .089,    i.e.,     ym  =  .0741  Ibs.  per  cub.  ft., 

Pn 

hence  the  weight  of  the  discharge  is 

G  =  QmYm  =  9.745  X  .0741  =  0.722  Ibs.  per  sec., 

or  about  12  per  cent  less  than  that  given  by  the  "  water  for- 
mula." If  the  difference  between  the  inner  and  outer  tensions 
had  been  less,  the  discrepancy  between  the  results  of  the  two 
methods  would  not  have  been  so  marked. 

552.  Adiabatic  Efflux  from  an  Orifice. — It  is  most  logical  to 
assume  that  the  expansion  of  the  gas  approaching  the  orifice, 
being  rapid,  is  adiabatic  (§  478).  Hence  (especially  when  the 
difference  between  the  inner  and  outer  tensions  is  considerable) 
it  is  more  accurate  to  assume  y  as  varying  according*  to  Pois- 
son's  Law,  eq.  (1),  §  478 ;  i.e.,  y  =  [yn  -t-prf]p%,  in  integrat- 
ing eq.  (B).  Then  the  term 

/*m/7r>  m  § 

L  f  ^= 


*  Bead  the  foot-note  on  p.  623. 


ADIABATIC   FLOW   OF   GASES   THROUGH   ORIFICES.      779 


ana  eq.  (2?),  neglecting  h  as  before,  and  with  vn  =  0,  becomes 
(See  Fig.  619) 


-  .  (Adiabaticflow;  orifice.)  .  (1) 


Having  observed  pn  and  7^  in  the  reservoir,  we  compute 

rri 

yn  =  P^°    °  (from  §  472).     The  gas  at  m,  just  leaving  the 

•LnPo 

orifice,  having  expanded  adiabaticallj  from  the  state  n  to  the 
state  m,  has  cooled  to  a  temperature  Tm  (absolute)  found  thus 
(8  478), 

Tm=Tn(^J,     ......    (2) 

and  is  of  a  heaviness 


and  the  flow  per  second  occupies  a  volume  (immediately  on 
exit) 

Qm  =  Fmvn, (4) 

and  weighs 

0  =  Fmvmym (5) 

EXAMPLE  1. — Let  the  interior  conditions  in  the  large  reser- 
voir of  Fig.  619  be  as  follows  (state  n) :  pn  =  22£  Ibs.  per  sq. 
in.,  and  Tn  =  294°  Abs.  Cent,  (i.e.,  21°  Cent.) ;  while  ex- 
ternally the  tension  is  15  Ibs.  per  sq.  inch,  which  may  be  taken 
as  being  =  pm  =  tension  at  m,  the  throat  of  jet.  The  opening 
is  a  circular  orifice  in  "  thin  plate"  and  of  one  inch  diameter. 
Required  the  weight  of  the  discharge  per  second  [ft.,  lb.,  sec.; 
g  =  32.2]. 

22  5  V  144    273 
FirSt>  r"  =  14.7x144 '  294X  -0807=  °-116  lbs-  Per  c«l>.  ft. 

Then,  from  (1), 


—  •   72X 32.2X3X22.5X144          8        __ 

0.116  "  L         Mr]  ~  '  per 


780  MECHANICS   OF   ENGINEERING. 

Xow  F  =  i?r(TV)2  =  .00546  sq.  ft. 

- .-.  Qm  =  OFvm  =  .WFvm  =  0.60  X  .00546  x  825  =  2.702 

cub.  ft.  per  sec. ,  at  a  temperature  of 

Tm  =  294  ff  =  257°  Abs.  Cent.  =  -  16°  Cent.,* 
and  of  a  heaviness 

ym  =  0.116  V(f)'  =  0.0885  Ibs.  per  cub.  ft., 
so  that  the  weight  of  flow  per  sec. 

=  a  =  Qmym  =  2.702  X  .0885  =  .239  Ibs.  persec. 

EXAMPLE  2. — Let  us  treat  the  example  already  solved  by  the 
two  preceding  approximate  methods  (§§  550  and  551)  by  the 
present  more  accurate  equation  of  adiabatic  flow,  eq.  (1). 

The  data  were'(Fig.  619): 

pn  —  18  Ibs.  per  sq.  in. ;  Tn  =  303°  Abs.  Cent. ; 
pm  =  15       "      "     "     ;  and  F  =  4  sq.  inches 

[F being  the  area  of  orifice].     yn  was  found  =  .089  Ibs.  per 
cub.  ft.  in  §  550 ;  hence,  from  eq.  (1), 


From  (4), 

Qm=Fmvm=.6Fvm=.Qx^fX 576.2  =  9.603  cub.  ft.  per  sec.; 

and  since  at  m  it  is  of  a  heaviness 

ym  =  .089  V(ffy3  =  .0788  Ibs.  per  cub.  ft., 
we  have  weight  of  flow  per  sec. 

=  G  =  Qmym  =  9.603  x  .0788  =  0.756  Ibs.  per  see. 

*  By  the  impact  of  the  effluent  air  on  the  outside  air,  with  extinction  of 
velocity,  the  temperature  rises  again. 


THEORETICAL   MAXIMUM   FLOW   OF    WEIGHT   OF   GAS.      781 

Comparing  the  three  methods  for  this  problem,  we  see  that 

By  the  "  water  formula?  .  .  .   £  =:  0.82    Ibs.  per  sec. 
"      isothermal  formula,  .  .  G  —  0.722      "      " 
"      adiabatic  formula,    .  .   G  —  0.756      "      " 

553.  Practical  Notes.     Theoretical  Maximum  Flow  of  Weight. 

*—  If  in  the  equations  of  §  552  we  write  for  brevity  ^>TO  -;-£>„:=  a* 
we  derive,  by  substitution  from  (1)  and  (3)  in  (5), 


This  function  of  x  is  of  such  a  form  as  to  be  a  maximum  for 

»  =  0>«-5-^.)=(t)1=.5l2;    .    /  .    .    (2) 

i.e.,  theoretically,  if  the  state  n  inside  the  reservoir  remains 
the  same,  while  the  outside  tension  (considered  =pm  of  jet, 
Fig.  619)  is  made  to  assume  lower  and  lower  values  (so  that 
#,  =-pm  -r-  pni  diminishes  in  the  same  ratio),  the  maximum  flow 
of  weight  per  unit  of  time  will  occur  when  pm  =  .512  pnj  a 
little  more  than  half  the  inside  tension.  (With  the  more  ac- 
curate value  1.41  (1.408),  instead  of  f,  see  §  478,  we  should 
obtain  .527  instead  of  .512  for  dry  air;  see  §  549.) 

Prof.  Cotterill  says  (p.  544  of  his  "  Applied  Mechanics")  : 
"The  diminution  of  the  theoretical  discharge  on  diminution 
of  the  external  pressure  below  the  limit  just  now  given  is  an 
anomaly  which  had  always  been  considered  as  requiring  ex- 
planation, and  M.  St.  Tenant  had  already  suggested  that  it 
could  not  actually  occur.  In  1866  Mr.  R.  D.  Napier  showed 
by  experiment  that  the  weight  of  steam  of  given  pressure  dis- 
charged from  an  orifice  really  is  independent  of  the  pressure 
of  the  medium  into  which  efflux  takes  place*;  and  in  1872 
Mr.  Wilson  confirmed  this  result  by  experiments  on  the  reac- 
tion of  steam  issuing  from  an  orifice." 
i  "  The  explanation  lies  in  the  fact  that  the  pressure  in  the 

*  When  the  difference  between  internal  and  external  pressures  is  great,-' 
should  be  added. 


782  MECHANICS   OF   ENGINEERING. 

centre  of  the  contracted  jet  is  not  the  same  as  that  of  the  sur- 
rounding medium.  The  jet  after  passing  the  contracted  sec- 
tion suddenly  expands,  and  the  change  of  direction  of  the  fluid 
particles  gives  rise  to  centrifugal  forces"  which  cause  the  pres- 
sures to  be  greater  in  the  centre  of  the  contracted  section  than 
at  the  circumference  ;  see  Fig.  620. 

Prof.  Cotterill  then  advises  the  assumption  that  j9m=.527j9n 
(for  air  and  perfect  gases)  as  the  mean  tension  in  the  jet  at  m 
(Fig.  619),  whenever  the  outside  medium  is  at  a  tension  less 
than  .527j9n  .  He  also  says,  "  Contraction  and  friction  must 
be  allowed  for  by  the  use  of  a  coefficient  of  discharge  the 
value  of  which,  however,  is  more  variable  than  that  of  the 
corresponding  coefficient  for  an  incompressible  fluid.  Little  is 
certainly  known  on  this  point."  See  §§  549  and  554. 

For  air  the  velocity  of  this  maximum  flow  of  weight  is 


Vel.ofmax.  G  =     w?\       ?    ft.  per  sec.,    .    (3) 


where  Tn  =  abs.  temp,  in  reservoir,  and  T0  =  that  of  freezing 
point.  Rankine's  Applied  Mechanics  (  p.  584)  mentions  ex- 
periments of  Drs.  Joule  and  Thomson,  in  which  the  circular 
orifices  were  in  a  thin  plate  of  copper  and  of  diameters  0.029 
in.,  0.053  in.,  and  0.084  in.,  while  the  outside  tension  was 
about  one  half  of  that  inside.  The  results  were  84  per  cent 
of  those  demanded  by  theory,  a  discrepancy  due  mainly,  as 
Rankine  says,  to  the  fact  that  the  actual  area  of  the  orifice  was 
used  in  computation  instead  of  the  contracted  section;  i.e.,  con- 
traction was  neglected. 

554.  Coefficients  of  Efflux  by  Experiment.  For  Orifices  and 
Short  Pipes.  Small  Difference  of  Tensions.  —  Since  the  discharge 
through  an  orifice  or  short  pipe  from  a  reservoir  is  affected 
not  only  by  contraction,  but  by  slight  friction  at  the  edges, 
even  with  a  rounded  entrance,  the  theoretical  results  for  the 
volume  and  weight  of  flow  per  unit  of  time  in  preceding  para- 
graphs should  be  multiplied  both  by  a  coefficient  of  velocity  0 
and  one  for  contraction  (7,  as  in  the  case  of  water  ;  i.e.,  by  a 
coefficient  of  efflux  /*,  =  0(7.  (Of  course,  when  there  is  no 


COEFFICIENTS   OF  EFFLUX.      GAS.  783 

contraction,  C  =  1.00,  and  then  //  =  0  as  with  a  well-rounded 
mouth-piece,  for  instance,  Fig.  541,  and  with  short  pipes.) 

Hence  for  practical  results,  with  orifices  and  short  pipes,  we 
should  write  for  the  weight  of  flow  per  unit  of  time 


(1) 


(from  the  equations  of  §  552  for  adiabatic  flow,  as  most  accu- 
rate; pm  -7-pn  may  range  from  -J-  to  1.00).  F=  area  of  orifice, 
or  of  discharging  end  of  mouth-piece  or  short  pipe.  yn  = 
heaviness  of  air  in  reservoir  and  =  T0pnyQ  ~  Tn  p0  ,  eq.  (13)  of 
§  437  ;  and  /*  =  the  experimental  coefficient  of  efflux. 

From  his  own  experiments  and  those  of  Koch,  D'Aubuis- 
eon,  and  others,  Weisbach  recommends  the  following  mean 
values  of  JJL  for  various  mouthpieces,  when  pn  is  not  more  than 
\  larger  than  pm  (i.e.,  about  Ml  %  larger),  for  use  in  eq.  (1)  : 

1.  For  an  orifice  in  a  thin  plate,     .......     /*=0.56 

2.  For  a  short  cylindrical  pipe  (inner  corners  not  rounded),/*  =0.75 

3.  For  a  well-rounded  mouth-piece  (like  that  in  Fig.  541),  /*=0.98 

4.  For  a  short  conical  convergent  pipe  (angle  about  6°),  /*=0.92 

EXAMPLE.  —  (Data  from  Weisbach's  Mechanics.)  "If  the 
sum  of  the  areas  of  two  conical  tuyeres  of  a  blowing-machine 
is  F  =  3  sq.  inches,  the  temperature  in  the  re&ervoir  15°  Cent., 
the  height  of  the  attached  (open)  mercury  manometer  (see 
Fig.  464)  3  inches,  and  the  height  of  the  barometer  la  the  ex- 
ternal air  29  inches,5'  we  have  (ft,  lb.,  sec.) 


pn  ^  (ft)  l±.r  X  144  Ibs.  per  sq.  ft  ; 
n  =  Ht  -ft  X  0.0807  =  0.0816  Ibs.  per  cob.  ft, 
While      ^'=  yfj-  sq.  ft.  and  (see  above)  /*  =  0.92  ;  hence 

Q  =  0.92  X  rl-r  (fl*1  V2X  322  X  3X  «  X  14.7  X  144  X  .0816  [1  -  j/fjjj 


784  MEOHAiaOS  OF  ENGINEERING. 

i.c.,  G  —  .6076  Ibs.  per  second ;  which  will  o  3upy  &  volume 

F  =  a  -f-  v.  =  G  -f-  .0807  =  7.59  cub.  ft. 


at  one  atmosphere  tension  and  freezing-point  teir  po 
while  at  a  temperature  of  Tn  =  288°  Abs.  Cent,  and  tcBoion  of 
pm  —  ||.  of  one  atmosphere  (i.e.,  in  the  state  in  which  it  was 
on  entering  the  blowing-engine)  it  occupied  a  volume 


F  = 


X  7.59  =  8.24  cub.  ft. 


(This  last  is  Weisbach's  result,  obtained  by  an  approximate 
formula*) 

555.  Coefficients  of  Efflux  for  Orifices  and  Short  Pipes  for  a 
Large  Difference  of  Tension.  —  For  values  >  f  and  <  2,  of  the 
ratio  pn  :  pm,  of  internal  to  external  tension,  Weisbach's  ex- 
periments with  circular  orifices  in  thin  plate,  of  diameters  (=d) 
from  0.4  inches  to  0.8  inches,  gave  the  following  results  : 


Pn'.pm           = 

1.05 

1.09 

1.40 

1.65 

1.90    1    2.00 

.  or  d  =  .4^;  V  = 

.55 

.59 

.69 

.72 

.76    1      .78 

'•    d  =.&*-;/*  = 

.56 

.57 

.64 

.68 

.72 

Whence  it  appears  that  ju  increases  somewhat  with  the  ratio  of 
Pn  to  pm ,  and  decreases  slightly  for  increasing  size  of  orifice. 

With  short  cylindrical  pipes,  internal  edges  not  rounded, 
and  three  times  as  long  as  wide,  Weisbach  obtained  p  as 
follows : 


diam.  =  .4in-  ;  ft  = 

1.05 
.73 

1.10 

.77- 

1.30 

.83 

1.40 

.81 

1.70    1    1.74 

.82    i 
.83 

When  the  inner  edges  of  the  0.4  in.  pipe  were  slightly 
founded,  /*  was  found  =  0.93 ;  while  a  well-rounded  mouth- 
piece of  the  form  shown  in  Fig.  541  gave  a  value  /*  ~  from 
.965  to  .968,  for  pn  :  pm  ranging  from  1.25  to  2.00.  These 
values  of  yu  are  for  use  in  eq.  (1),  above. 

556.  To  find  the  Discharge  when  the  Internal  Pressure  if 
measured  in  a  Small  Reservoir  or  Pipe,  not  much  larger  than  the 


VELOCITY  OF   APPROACH.     GASES.  785 

Orifice.-  --Fig.  621.    If  the  internal  pressure  pn,  and  tempera. 

fcure  Tn,  must  be  measured  in  a 

small  reservoir  or  pipe,  n,  whose 

sectional  area  Fn  is  not  very  large 

compared  with  that  of  the  oridce, 

F,  (or  of  the  jet  Fn  ,)  the  velocity 

vn  at  n  (velocity  of  approach)  can-  FIG.  esi. 

not  be  put  =  zero.     Hence,  in  applying  eq.  (/?),  §  550,  to  the 

successive  laminae  between  n  and  m,  and  integrating,  we  shall 

have,  ior  adiabatic  steady  flow, 

V*  Vn     _   tyn  I""-,          (Pm\ 

~'"- 


instead  of  eq.  (1)  of  §  552.    But  from  the  equation  of  continuity 
for  steady  flow  of  gases  [eq.  (a)  of  §  548],  Fnvnyn=  Fmvmym\ 

F  *v  a 
hence  vn*  —    ™     ,  ^OTa,  while  for  an  adiabatic  change  from  n 


to  m,  —  =  f— ) ;  whence  by  substitution  in  (1),  solving  for 
vm,  we  have 


before,  from  §§472  and  478, 


*»d  y- -(£)>.•  ...-.-..  (4) 


Having  observed  pn,  pm,  and  yn,  tiion,  and  knowing  the 
area  F  of  the  orifice,  we  may  compute  yn,  ^M9  and  vmy  and 
finally  the 

Weight  of  flow  per  tvine-unit  =  @  =  j&Fk)mym9  .   .    (5) 


786  MECHANICS   OF  ENGINEERING^ 

taking  //  from  §  554  or  555.  In  eq.  (2)  it  must  be  remembered 
that  for  an  orifice  in  "  thin  plate,"  Fm  is  the  sectional  area  of 

the  contracted  vein,  and  =  CF\  where  C  may  be  put  =  -^L  . 

."7 

.  EXAMPLE.—  If  the  diameter  of  AB,  Fig.  621,  is  3£  inches. 
and  that  of  the  orifice,  well  rounded,  =  2  in.  ;  if  pn  =  1^  at- 
mospheres  =  -ff  X  14.7  X  144  Ibs.  per  sq.  ft.,  while  pm  =  |J  of 

an  atmos.,  so  that  £2  =  ft,  and   Tn  =  283°  Abs.  Cent.,—  re- 

Pn 

qnired  the  discharge  per  second,  using  the  ft.,  lb.,  and  sea 
From  eq.  (3), 

Yn  =  if  -Hf  X  0.0807  «  .08433  Ibs.  per  cub.  ft; 
•whence  (eq.  (4)) 

ym  =  (U)Vn  =  -07544  Ibs.  per  cub.  ft. 
Then,  from  eq.  (2), 


=  558.1  ft.  per  sec.  ; 

.-.   a  =  0.98  |(|)S558.1  x  .07544  =  .9003  Ibs.  per  sec. 

557.  Transmission  of  Compressed  Air;  through  very  Long 
level  Pipes.  Steady  Flow. 

CASE  I.  When  the  difference  between  the  tensions  in  the 
reservoirs  at  the  ends  of  the  pipe  is  small.  —  Fig.  622.  Under 


FIG.  623. 


these  circumstances  it  is  simpler  to  employ  the  form  of  formula 
that  would  be  obtained  for  a  liquid  by  applying  Bernoulli's 
Theorem,  taking  into  account  the  "  loss  of  head  "  occasioned 


TRANSMISSION  OF  COMPRESSED  AIR.  787 

by  the  friction  on  the  sides  of  the  pipe.  Since  the  pipe  is 
very  long,  and  the  change  of  pressure  small,  the  mean  velocity 
in  the  pipe,  v',  assumed  to  be  nearly  the  same  at  all  points 
along  the  pipe,  will  not  be  large  ;  hence  the  difference  be- 
tween the  velocity-heads  at  n  and  m  will  be  neglected  ;  a  cer- 
tain mean  heaviness  y'  will  be  assigned  to  all  the  gas  in  the 
pipe,  as  if  a  liquid. 

Applying  Bernoulli's  Theorem,  with  friction,  §  516,  to  the 
ends  of  the  pipe,  n  and  m,  we  have  (as  for  a  liquid) 


Putting  (as  above  mentioned)  vm*  —  vj  =  0,  we  have,  more 
simply, 


The  value  of  f  as  coefficient  of  friction  for  air  in  long 
pipes  is  found  to  be  somewhat  smaller  than  for  water  ;  see  next 
paragraph. 

558.  Transmission  of  Compressed  Air.  Experiments  in  the  St. 
Gothard  Tunnel,  1878.—  [See  p.  96  of  Vol.  24  (Feb.  '81),  Tan 
Nostrand's  Engineering  Magazine.]  In  these  experiments, 
the  temperature  and  pressure  of  the  flowing  gas  (air)  were  ob- 
served at  each  end  of  a  long  portion  of  the  pipe  which  delivered 
the  compressed  air  to  the  boring-machines  three  miles  distant 
from  the  tunnel's  mouth.  The  portion  considered  was  selected 
at  a  distance  from  the  entrance  of  the  tunnel,  to  eliminate  the 
fluctuating  influence  of  the  weather  on  the  temperature  of  the 
flowing  air.  A  steady  flow  being  secured  by  proper  regulation 
of  the  compressors  and  distributing  tubes,  observations  were 
made  of  the  internal  pressure  (p\  internal  temperature  (T\  as 
well  as  the  external,  at  each  end  of  the  portion  of  pipe  con- 
sidered, and  also  at  intermediate  points  ;  also  of  the  weight 
of  flow  per  second  G  =  Q0y0,  measured  at  the  compressors 
under  standard  conditions  (0°  Cent,  and  one  atmos.  tension). 
Then  knowing  the  p  and  T  at  any  section  of  the  pipe,  the 


788 


MECHANICS   OF   ENGINEERING. 


heaviness  y  of  the  air  passing  that  section  can  be  computed 

rri  — | 

I  from  ~  =  —  •  -7^     and  the  velocity  v =  G  -f-  Fy,  F  being 

L       r0    PO   J-  -1 

the  sectional  area  at  that  point.  Hence  the  mean  velocity  v'9 
and  the  mean  heaviness  y' ',  can  be  computed  for  this  portion 
of  the  pipe  whose  diameter  =  d  and  length  =  Z.  In  the  ex- 
peri  men  ts  cited  it  was  found  that  at  points  not  too  near  the 
tunnel-mouth  the  temperature  inside  the  pipe  was  always 
about  3°  Cent,  lower  than  that  of  the  tunnel.  The  values  of 
/in  the  different  experiments  were  then  computed  from  eq. 
(2)  of  the  last  paragraph ;  i.e., 


(2) 


all  the  other  quantities  having  been  either  directly  observed, 
or  computed  from  observed  quantities. 

THE  ST.  GOTHARD  EXPERIMENTS. 

[Concrete  quantities  reduced  to  English  units.] 


No. 

I 

(feet.) 

d 

(ft.) 

Y 

(Ibs.  cub. 
ft.) 

Atmospheres. 

Pn-Pm 
Ibs.  sq.  in. 

^ 
ft.  per  sec. 

mean 
temp. 
Cent. 

f 

Pn 

Pm 

1 

15092 

I 

0.4058 

5.60 

5.24 

5.29 

19.32 

21° 

.0035 

2 

15092 

0.3209 

4.35 

4.13 

3.23 

16.30 

21° 

.0038 

8 

15092 

.2. 

.2803 

3.84 

3.65 

2.79 

15.55 

21° 

.0041 

4 

1712 

.3765 

5.24 

5.00 

3.52 

37.13 

26.5 

.0045 

5 

1712 

.3009 

4.13 

4.06 

1.03 

30.82 

26.5 

.0024(?) 

6 

1712 

.2641 

3.65 

3.54 

1.54 

29.34 

26.5 

.0045 

In  the  article  referred  to  (Van  Nostrand's  Mag.)  f  is  not 
computed.  The  writer  contents  himself  with  showing  that 
Weisbach's  values  (based  on  experiments  with  small  pipes  and 
high  velocities)  are  much  too  great  for  the  pipes  in  use  in  the 
tunnel.* 

With  small  tubes  an  inch  or  less  in  diameter  Weisbach 
found,  for  a  velocity  of  about  80  ft.  per  second,/ =.0060; 
for  still  higher  velocities/ was  smaller,  approximately,  in  ac- 
cordance with  the  relation 

/=             .0542 
iV  (in  ft.  per  sec.) 

*  See  also  experiments  described  in  Engineering  News,  Nov.  3,  1904,  p. 
387.  In  that  article  the  quantity  called  /  is  the/  of  this  chapter  divided 
by  64.4. 


TRANSMISSION   OF   COMPRESSED   AIR.  789 

On  p.  370,  vol.  xxiv,  Yan  Nostrand's  Mag.,  Prof.  Robinson 
of  Ohio  mentions  other  experiments  with  large  long  pipes. 

From  the  St.  Gothard  experiments  a  value  off=  .004  may 
be  inferred  for  approximate  results  with  pipes  from  3  to  8  in. 
in  diameter. 

EXAMPLE.  —  It  is  required  to  transmit,  in  steady  flow,  a  supply 
of  G  —  6.456  Ibs.  of  atmospheric  air  per  second  through  a  pipe 
30000  ft.  in  length  (nearly  six  miles)  from  a  reservoir  where 
the  tension  is  6.0  atmos.  to  another  where  it  is  5.8  atmos.,  the 
mean  temperature  in  the  pipe  being  80°  Fahr.,  =  24°  Cent. 
{i.e.  =  297°  Abs.  Cent.).  Required  the  proper  diameter  of 
pipe  ;.<#  =  ?  The  value  f—  .00425  will  be  used,  and  the  ft.- 
Ib.-sec.  system  of  units.  The  mean  volume  passing  per  second 
in  the  pipe  is 

Q'=G  +  y>  ........    (3) 

Ql  Q' 

The  mean  velocity  may  thus  be  written  :  v'  =  -~  =  -~-  .   (4) 

JT        \nd 

The  mean  heaviness  of  the  flowing  air,  computed  for  a  mean 
tension  of  5.9  atmospheres,  is,  by  §  472, 


x  -0807  =  °-431  Ibs-  ^roab-  ft-  ; 


-and  hence,  see  eq.  (3), 

xv       G-       6.456 


, 

cub" 


at  tension  of  5.9  atmos.,  and  temperature  297°  Abs.  Cent. 
Now,  from  eq.  (2), 


whence 

3*        4/  y'l          Q'* 


790  MECHANICS   OF   ENGINEERING. 

and  hence,  numerically, 

5  /  4  X  .004:25  X  0.431  X  30000  X  (14.T4)8" 

~  y  (.7854)2[U.7  X  M4(6.00  —  5.80)]2  X  32.2  ~ 

559.  (Case  II  of  §  557)  Long  Pipe,  with  Considerable  Differ- 
ence of  Pressure  at  Extremities  of  the  Pipe.  Flow  Steady. — Fig. 
623.  If  the  difference  between  the  end-tensions  is  compara- 
tively great,  we  can  no  longer  deal  with  the  whole  of  the  air 


in  the  pipe  at  once,  as  regards  ascribing  to  it  a  mean  velocity 
and  mean  tension,  but  must  consider  the  separate  laminae, 
such  as  AB(&  short  length  of  the  air-stream)  to  which  we  may 
apply  eq.  (2)  of  §  557  ;  A  and  B  corresponding  to  the  n  and 
m  of  Fig.  622.  Since  the  ^>n  —  pm,  Z,  y',  and  v'  of  §  557 
correspond  to  the  —  dp,  ds,  y,  and  v  of  the  present  case  (short 
section  or  lamina),  we  may  write 

dp       *  ,?  V*     7 

- 


But  if  G  =  weight  of  flow  per  unit  of  time,  we  have  at  any 
section,  Fvy  —  G  (equation  of  continuity)  ;  i.e.,  v  =  G  H-  Fy, 
whence  by  substitution  in  eq.  (1)  we  have 

dp        4/ 


Eq.  (2)  contains  three  variables,  y,  p,  and  s  (=  distance  of 
lamina  from  ri).  As  to  the  dependence  of  the  heaviness  y  on 
the  tension  p  in  different  laminae,  experiment  shows  that  in  most 
cases  a  uniform  temperature  is  found  to  exist  all  along  the 
pipe,  if  properly  buried,  or  shaded  from  the  sun  ;  the  loss  of 
heat  by  adiabatic  expansion  being  in  great  part  made  up  by 
the  heat  generated  by  the  friction  against  the  walls  of  the 


GAS   IN   LONG  PIPES.      LARGE  FALL   OF  TENSION.      791 

pipe.  This  is  due  to  the  small  loss  of  tension  per  unit  of 
length  of  pipe  as  compared  with  that  occurring  in  a  short  dis- 
charge pipe  or  nozzle.  Hence  we.  may  treat  the  flow  as  iso- 
thermal, and  write  p  -f-  y  —pn>-±-  yn>  (§  475,  Mariotte's  Law). 

Hence  y  =  —  p,  which   substituted  in  eq.  (2)  enables  us  to 

Pn' 


Performing  the  integration,  noting  that  at  n'  p—pn>,  5  =  0, 
and  at  m'  p  =pm>  and  s  =  I,  we  have 


ir*>  2  -  <n  '21  -  ^fl    —  £*L        I  isothermal  floiv 
~  Zgd'  F*'  yn,'      \    in  long  pipes 

It  is  here  assumed  that  the  tension  at  the  entrance  of  the  pipe 
is  practically  equal  to  that  in  the  head  reservoir,  and  that  at 
the  end  (m'}  to  that  of  the  receiving  reservoir;  which  is  not 
strictly  true,  especially  when  the  corners  are  not  rounded.  It 
will  be  remembered  also  that  in  establishing  eq.  (2)  of  §  557 
(the  basis  of  the  present  paragraph),  the  "inertia"  of  the  gas 
was  neglected  ;  i.e.,  the  change  of  velocity  in  passing  along 
the  pipe.  Hence  eq.  (4)  should  not  be  applied  to  cases  where 
the  pipe  is  so  short,  or  the  difference  of  end-tensions  so  great, 
as  to  create  a  considerable  difference  between  the  velocities  at 
the  two  ends  of  the  pipe.  (See  Addendum  on  p.  797.) 

EXAMPLE.  —  A  well  or  reservoir  supplies  natural  gas  at  a  ten- 
sion of  pn>  —  30  Ibs.  per  sq.  inch.  Its  heaviness  at  0°  Cent. 
and  one  atmosphere  tension  is  .0484  Ibs.  per  cub.  foot.  In 
piping  this  gas  along  a  level  to  a  town  two  miles  distant,  a 
single  four-inch  pipe  is  to  be  employed,  and  the  tension  in  the 
receiving  reservoir  (by  proper  regulation  of  the  gas  distributed 
from  it)  is  to  "be  kept  equal  to  16  Ibs.  per  sq.  in.  (which  would 
sustain  a  column  of  water  about  2  ft.  in  height  in  an  open 
water  manometer,  Fig.  465). 


792  MECHANICS   OF   ENGINEERING. 

The  mean  temperature  in  the  pipe  being  17°  Cent.,  required 
the  amount  (weight)  of  gas  delivered  per  second,  supposing 
leakage  to  be  prevented  (formerly  a  difficult  matter  in  practice). 
Solve  (4)  for  6?,  and  we  have 


(5) 


First,  from  §  472,  with  Tn,  =  Tm,  =  290°  Abs.  Cent.,  we 
compute 


Hence  with/  =.005, 


(16  X  144)] 


4X.005  X  10560  X  46154: 
=  0.337  Ibs.  per  sec. 

(For  compressed  atmospheric  air,  under  like  conditions,  we 
would  have  G  —  0.430  Ibs.  per  second.) 

Of  course  the  proper  choice  of  the  coefficient/  has  an  im- 
portant influence  on  the  result. 

From  the  above  result  (G-  =  0.337  Ibs.  per  second)  we  can 
compute  the  volume  occupied  by  this  quantity  of  gas  in  the 

/3 

receiving  reservoir,  using  the  relation  Qm>  =  —  . 

/  m' 

The  heaviness  ym>  of  the  gas  in  the  receiving  reservoir  is 
most  easily  found  from  the  relation  -SzL  =  S^  ,  which  holds 

7m'  Yn> 

good    since    the    flow    is    isothermal.     I.e.,   £™'=  46454ft.; 

Ymf 

whence  ym,  =  0.049  Ibs.  per  cubic  foot,  pm>  being  16  X  144 
Ibs.  per  sq.  ft. 
Hence 

cub-  ft-  per  sec- 


FLOW   OF   GAS   IN   PIPES.  793 

It  should  be  said  that  the  pressure  at  the  up-stream  end  of 
the  pipe  depends  upon  the  rate  of  flow  allowed  to  take  place. 

With  no  flow  permitted,  the  pressure  in  the  tube  of  a  gas- 
well  has  in  some  cases  reached  the  high  figure  of  500  or  600 
]bs.  per  sq.  in. 

560.  Rate  of  Decrease  of  Pressure  along  a  Long  Pipe. — Con- 
sidering further  the  case  of  the  last  paragraph,  that  of  a 
straight,  long,  level  pipe  of  uniform  diameter,  delivering  gas 
from  a  storage  reservoir  into  a  receiving  reservoir,  we  note 
that  if  in  eq.  (4)  we  retain  pm>  to  indicate  the  tension  in  the 
receiving  reservoir,  but  let  pn>  denote  in  turn  the  tension  at 
points  in  the  pipe  successively  further  and  further  (a  distance 
x)  from  the  receiving  reservoir  w',  we  may  write  x  for  I  and 
obtain  the  equation  (between  two  variables,  pn>  and  x) 

Pn*  —Pm*  =  Const.  X  os (6) 

This  can  be  used  to  bring  out  an  interesting  relation  men- 
tioned by  a  writer  in  the  Engineering  News  of  July  1887 
(p.  71),  viz.,  the  fact  that  in  the  parts  of  the  pipe  more  distant 
from  the  receiving  end,  m',  the  distance  along  the  pipe  in 
which  a  given  loss  of  pressure  occurs  is  much  greater  than 
near  the  receiving  end. 

To  make  a  numerical  illustration,  let  us  suppose  that  the 
pipe  is  of  such  size,  in  connection  with  other  circumstances, 
that  the  tension  pn,  at  A,  a  distance  x  =  six  miles  from  m' ',  is 
two  atmospheres,  the  tension  in  the  receiving  reservoir  being 
one  atmosphere ;  that  is,  that  the  loss  of  tension  between  A. 
and  m'  is  one  atmosphere.  If  we  express  tensions  in  atmos- 
pheres and  distances  in  miles,  we  have  for  the  value  of  the 
constant  in  eq.  (6),  for  this  case, 

Const.  =  (4  —  1)  —  6  =  f  ;  (for  assumed  units.)  .   .  (7) 

Now  \Qtpn'  =  the  tension  at  B,  a  point  18  miles  from  ra', 
and  we  have,  from  eqs.  (6)  and  (7),  the  tension  at  B  —  3.16 
atmospheres.  Proceeding  in  this  manner,  the  following  set  of 
values  is  obtained : 


794 


MECHANICS   OF   ENGINEERING. 


Point. 

Total  distance 
from  m'. 

Distance  be- 
tween consecu- 
tive points. 

Tension  at 
point. 

Loss  of  ten- 
sion in  each  in- 
terval. 

F 
E 
D 

C 
13 

A 

m' 

126m 
90 
60 
36 
18 
6 
0 

les. 

36  mi 
80 
24 
18 
12 
6 

ies. 

8.  00  at] 
6.78 
5.56 
4.35 
3.16 
2.00 
1.00 

en. 

1.22  at 
1.22 
1.21 
1.19 
1.16 
1.00 

tn. 

<  If  the  distances  and  tensions  in  the  second  and  fourth 
columns  be  plotted  as  abscissae  and  ordinates  of  a  curve,  the 
latter  is  a  parabola  with  its  axis  following  the  axis  of  the  pipe ; 
its  vertex  is  not  at  m',  however. 

561.  Long  Pipe  of  Variable  Diameter. — Another  way  of  stat- 
ing the  fact  mentioned  in  the  last  paragraph  is  as  follows  :  At 
the  up-strearn  end  of  the  pipe  of  uniform  diameter  the  gas  is 
of  much  greater  density  than  at  the  other  extremity  (the 
heaviness  is  directly  as  the  tension,  the  temperature  being  as- 
sumed the  same  throughout  the  pipe),  and  the  velocity  of  its 
motion  is  smaller  than  at  the  discharging  end  (in  the  same 
ratio).  It  is  true  that  the  frictional  resistance  per  unit  of 
iength  of  pipe  varies  directly  as  the  heaviness  [eq.  (1),  §  510], 
but  also  true  that  it  varies  as  the  square  of  the  velocity ;  so 
that,  for  instance,  if  the  pressure  at  a  point  A  is  double  that 
at  B  in  the  pipe  of  constant  diameter,  it  implies  that  the 
heaviness  and  velocity  at  A  are  double  and  half,  respectively, 
those  at  B,  and  thus  the  gas  at  A  is  subjected  to  only  half  the 
frictional  resisting  force  per  foot  of  length  as  compared  with 
that  at  B.  Hence  the  relatively  small  diminution,  per  unit  of 
length,  in  the  tension  at  the  up-stream  end  in  the  example  of 
the  last  paragraph. 

In  the  pipe  of  uniform  diameter,  as  we  have  seen,  the  greater 
part  of  the  length  is  subjected  to  a  comparatively  high  ten- 
sion, and  is  thus  under  a  greater  liability  to  loss  by  leakage 
than  if  the  decrease  of  tension  were  more  uniform.  The 
tetal  "hoop-tension"  (§  426)  in  a  unit  length  of  pipe,  also,  is 
proportional  to  the  gas  tension,*  and  thinner  walls  might  be 
employed  for  the  down-stream  portions  of  the  pipe  if  the  gas 

*  Or,  rather,  to  its  excess  over  that  of  external  air. 


FLOW  OF  GAS   IN  PIPES.  795 

tension  in  those  portions  could  be  made  smaller  than  as  shown 
in  the  preceding  example. 

To  secure  a  more  rapid  fall  of  pressure  at  the  up-stream  end 
of  the  pipe,  and  at  the  same  time  provide  for  the  same  delivery 
of  gas  as  with  a  pipe  of  uniform  diameter  throughout,  a  pipe 
of  variable  diameter  may  be  employed,  that  diameter  being 
considerably  smaller  at  the  inlet  than  that  of  the  uniform  pipe 
but  progressively  enlarging  down-stream.  This  will  require 
the  diameters  of  portions  near  the  discharging  end  to  be  larger 
than  in  the  uniform  pipe,  and  if  the  same  thickness  of  metal 
were  necessary  throughout,  there  would  be  no  saving  of  metal, 
but  rather  the  reverse,  as  will  be  seen  ;  but  the  diminished 
thickness  made  practicable  in  those  parts  from  a  less  total  hoop 
tension  than  in  the  corresponding  parts  of  the  uniform  pipe 
more  than  compensates  for  the  extra  metal  due  to  increased 
circumference,  aside  from  the  diminished  liability  to  leakage, 
which  is  of  equal  importance. 

A  simple  numerical  example  will  illustrate  the  foregoing. 
The  pipe  being  circular,  we  may  replace  F  by  \ndl  in  equation 
(4),  and  finally  derive,  G  being  given, 

d  =  Const.  X  f-  -1—-1*=  C.  \—  ^—  ,1!    .     (8) 

a  ' 


Let  A  be  the  head  reservoir,  and  m'  the  receiving  reservoir, 
and  B  a  point  half-way  between.  At  A  the  tension  is  10  at- 
nrospheres  ;  at  m',  2  atmospheres.  For  transmitting  a  given 
weight  of  gas  per  unit-time,  through  a  pipe  of  constant  diam- 
ter  throughout,  that  diameter  must  be  (tensions  in  atmospheres  ; 
2/0  being  the  length),  by  eq.  (8), 

d  =  Cl* 


If  we  substitute  for  the  pipe  mentioned,  another  having  a  con- 
stant diameter  d,  from  A  to  B,  where  we  wish  the  tension  to 
be  5  atmospheres,  and  a  different  constant  diameter  d^  from  B 
to  m',  we  derive  similarly 


796  MECHANICS   OF   ENGINEERING. 

and 


It  is  now  to  be  noted  that  the  sum  of  dl  and  dy  is  slightly 
greater  than  the  double  of  d  ;  so  that  if  the  same  thickness  of 
metal  were  used  in  both  designs  the  compound  pipe  would 
require  a  little  more  material  than  the  uniform  pipe;  but, 
from  the  reasoning  given  at  the  beginning  of  this  paragraph, 
that  thickness  may  be  made  considerably  less  in  the  down- 
stream part  of  the  compound  pipe,  and  thus  economy  secured. 

[In  case  of  a  cessation  of  the  flow,  the  gas  tension  in  the 
whole  pipe  might  rise  to  an  equality  with  that  of  the  head- 
reservoir  were  it  not  for  the  insertion,  at  intervals,  of  auto- 
matic regulators,  each  of  which  prevents  the  increase  of  ten- 
sion on  its  down-stream  side  above  a  fixed  value.  To  provide 
for  changes  of  length  due  to  rise  and  1'all  of  temperature,  the 
pipe  is  laid  with  slight  undulations.] 

It  is  a  noteworthy  theoretical  deduction  that  a  given  pipe  of 
variable  diameter  connecting  two  reservoirs  of  gas  at  specified 
pressures  will  deliver  the  same  weight  of  gas  as  before,  if 
turned  end  for  end.  This  follows  from  equation  (3)',  §  559. 
With  d  variable,  (3)'  becomes  (with  F=  \ptf) 

f™'-pdp)=G*C"fm'^',    i.e.,    &  =  -Pn'  ~^m/a  .    .    (9) 

WjJ—* 

(€"  is  a  constant.) 

r'ds 
—  is  evidently  the  same  in  value  if  the  pipe  be 

turned  end  for  end.  In  commenting  on  this  circumstance,  we 
should  remember  (see  §  559)  that  the  loss  of  pressure  along  the 
pipe  is  ascribed  entirely  tofrictional  resistance,  and  in  no  de- 
gree to  changes  of  velocity  (inertia). 

On  p.  73  of  the  Engineering  News  of  July  1887  are  given 
the  following  dimensions  of  a  compound  pipe  in  actual  use, 
and  delivering  natural  gas.  The  pressure  in  the  head-reservoir 
is  319  Ibs.  per  sq.  in.;  that  in  the  receiving  reservoir,  65.  For 
2.84  miles  from  the  head-reservoir  the  diameter  of  the  pipe  is 


NATURAL   GAS.      COEFFICIENT   OF  FLUID   FRICTION.     797 


8  in.  ;  throughout  the  next  2.75  miles,  10  in.;  while  in  the 
remaining  3.84  miles  the  diameter  is  12  in.  At  the  two 
points  of  junction  the  pressures  are  stated  to  be  185  and  132 
Ibs.  per  sq.  in.,  respectively,  during  the  flow  of  gas  under  the 
conditions  mentioned. 

561a.  Values  of  the  Coefficient  of  Fluid  Friction  for  Natural 
Gas. — In  the  Ohio  Keport  on  Economic  Geology  for  1888  may 
be  found  an  article  by  Prof.  S.  W.  Kobinson  of  the  University 
of  that  State  describing  a  series  of  interesting  experiments 
made  by  him  on  the  flow  of  natural  gas  from  orifices  and 
through  pipes.  By  the  insertion  of  Pitot  tubes  approximate 
measurements  were  made  of  the  velocity  of  the  stream  of  gas 
in  a  pipe.  The  following  are  some  of  the  results  of  these  ex- 
periments,^ —  p^  representing  the  loss  of  pressure  (in  Ibs.  per 
sq.  inch)  per  mile  of  pipe-length,  and /"the  coefficient  of  fluid 
friction,  in  experiments  with  a  six-inch  pipe : 


Pi  -p* 

1.00 

1.50 

2.25 

2.50 

5.75 

6.25 

f 

0.0025 

0.0037 

0.0052 

0.0059 

0.0070 

0.0060 

In  the  flow  under  observation  Prof.  Kobinson  concluded 
that  f  could  be  taken  as  approximately  proportional  to  the 
fourth  root  of  the  cube  of  the  velocity  of  flow  ;  though  calling 
attention  to  the  fact  that  very  reliable  results  could  hardly  be 
expected  under  the  circumstances. 

561b.  [Addendum  to  §  559.]  Isothermal  Flow  in  a  level  pipe, 
with  consideration  of  Inertia.  —  In  eq.  (1)  of  p.  697  neglect  dz, 
put  w  -T-  F=  4  -r-  d,  and  divide  through  by  v*.  In  the  second 
term  put  6P  -=-  F*y*  for  v*  and  then^?  (ym  -S-pm)  for  y.  We 
now  find  the  variables  separated,  and  on  integration  for  steady 
flow  obtain  (after  putting  vn-^-vm  =  Fym  -7-  Fyn  —  pm  -r-J?n)> 


* 


[Notation  as  in  §  557  with  G  = 


CHAPTEK  IX. 


IMPULSE  AND  RESISTANCE  OF  FLUIDS. 

562.  The  so-called  "  Reaction"  of  a  Jet  of  Water  flowing  from 
a  Vessel — In  Fig.  624,  if  a  frictionless  but  water-tight  plug  E 

be  inserted  in  an  orifice  in  the 
vertical  side  of  a  vessel  mounted 
on  wheels,  the  resultant  action  of 
the  water  on  the  rigid  vessel  (as  a 
whole)  consists  of  its  weight  G, 
and  a  force  P'  =  FJiy  (in  which 
F=  the  area  of  orifice)  which  is 
the  excess  of  the  horizontal  hydro- 
static pressures  on  the  vessel  wall 
toward  the  right  ( ||  to  paper)  over 
those  toward  the  left,  since  the 

pressure  P,  =  Fhy,  exerted  on  the  plug  is  felt  by  the  post  (7, 
and  not  by  the  vessel.     Hence  the  post  D  receives  a  pressure 


P'  =  Fhy. 


(1) 


Let  the  plug  B  be  removed.  A  steady  flow  is  then  set  up 
through  the  orifice,  and  now  the  pressure  against  the  post  D  is 
%Fhy  (as  will  be  proved  in  the  next  paragraph) ;  for  not  only 
IB  the  pressure  Fhy  lacking  on  the  left,  because  of  the  orifice, 
but  the  sum  of  all  the  horizontal  components  (  ||  to  paper)  of 
the  pressures  of  the  liquid  filaments  against  the  vessel  wall 
around  the  orifice  is  less  than  its  valup  before  the  flow  began, 
by  an  amount  Fhy.  A  resistance  E  —  ZFhy  being  provided, 
and  the  post  removed,  a  slow  uniform  motion  may  be  main- 
lined toward  the  right,  the  working  force  being  'zFliy  =  P" 

•as 


" REACTION"  OF  A  JET. 


799 


Fio.  625. 


(see  Yig.  625 ;  R  is  not  shown).     If  an  insufficient  resistance 

be  furnished  before  removing  the  post  Z>,  .    .  .  ^  . 

the  vessel  will  begin  to  move  toward  the 

right  with  an    acceleration,  which  will 

disturb  the  surface  of    the  water  and 

change  the  value  of  the  horizontal  force. 

This  force 

P"=:^Fhy     ...     (2) 

is  called  the  "  reaction"  of  the  water-jet ; 
y  is  the  heaviness  of  the  liquid  (§  7). 

Of  course,  as  the  flow  goes  on,  the 
water  level  sinks  and  the  '•  reaction"  diminishes  accordingly. 
Looked  upon  as  a  motor,  the  vessel  may  be  considered  to  be  a 
piston-less  and  valve-less  water-pressure  engine,  carrying  its 
own  reservoir  with  it. 

In  Case  II  of  §  500  we  have  already  had  a  treatment  of  the 
"  Reaction-wheel "  or  u  Barker's  mill,"  which  is  a  practical 
machine  operating  on  this  principle,  and  will  be  again  con- 
sidered in  "  Hydraulic  Motors."  * 

563.  "  Reaction"  of  a  Liquid  Jet  on  the  Vessel  from  which  it 
Issues. — Instead  of  showing  that  the  pressures  on  the  vessel 
close  to  the  orifice  are  less  than  they  were  when  there  was  no 
flow  by  an  amount  Fhy  (a  rather  lengthy  demonstration), 
another  method  will  be  given,  of  greater  simplicity  but  some- 
what fanciful. 

If  a  man  standing  on  the  rear  platform  of  a  car  is  to  take  up 
in  succession,  from  a  basket  on  the  car,  a  number  of  balls  of 
equal  mass  =  J/,  and  project  each  one  in  turn  horizontally 
backward  with  an  acceleration  =p,  he  can  accomplish  this 
only  by  exerting  against  each  ball  a  pressure  =  Mp,  and  in  the 
opposite  direction  against  the  car  an  equal  pressure  —  Mp.  If 
this  action  is  kept  up  continuously  the  car  is  subjected  to  a 
constant  and  continuous  forward  force  of  P"  =  Mp. 

Similarly,  the  backward  projection  of  the  jet  of  water  in  the 
case  of  the  vessel  at  rest  must  occasion  a  forward  force  against 
the  vessel  of  a  value  dependent  on  the  fact  that  in  each  small 
interval  of  time  At  a  small  mass  AM  of  liquid  has  its  velocity 
changed  from  zero  to  a  backward  velocity  of  v  —  Vtyh ;  that 

*  Hydraulic  Motors ;  with  related  subjects.  New  York,  1905,  John 
Wiley  &  Sons. 


800  MECHANICS   OF 

/y    _    Q- 

is,  has  been  projected  with  a  mean  acceleration  of  p  =  - 

At 

so  that  the  forward  force  against  the  vessel  is 


P"  =  mass  X  ace.  =  .  (3) 

At 

If  Q  =  the  volume  of  water  discharged  per  unit  time,  then 
AM  —  QZ.  At,  and  since  also  Q  =  Fv  =  F\/^gh,  eq.  (3)  be- 

comes "Reaction"  of  jet  =  P"  =  ZFhy.     ...     (4) 

(A  similar  proof,  resulting  in  the  same  value  for  P"  ,  is 
easily  made  if  the  vessel  has  a  uniform  motion  with  water  sur- 
face horizontal.) 

If  the  orifice  is  in  "  thin  plate,"  we  understand  by  F  the 
area  of  the  contracted  section.  Practically,  we  have  v  =  </>  V%gh 
(§  495),  and  hence  (4)  reduces  to 

P"  =  %<pFhy  ..........     (5) 

Weisbach  mentions  the  experiments  of  Mr.  Peter  Ewart  of 
Manchester,  England,  as  giving  the  result  P"  =  \.*V&Fliy 
with  a  well-rounded  orifice  as  in  Fig.  625.  He  also  found 
0  =  .94  for  the  same  orifice,  so  that  by  eq.  (4)  we  should  have 

P"  =  Z(MJFhy  =  l.WFhy. 

With  an  orifice  in  thin  plate  Mr.  Ewart  found  P"  — 
\.\kFliy.  As  for  a  result  from  eq.  (4),  we  must  put,  for  F^ 
the  area  of  the  contracted  section  .647^  (§  495),  which,  with 
0  =  .96,  gives 

P"  =  2(M)t.64Jfrhy  =  l.lSIthy.     ...     (6) 

Evidently  both  results  agree  well  with  experiment. 

Experiments  made  by  Prof.  J.  B.  Webb  at  the  Stevens 
Institute  (see  Journal  of  the  Franklin  Inst.,  Jan.  '88,  p.  35) 
also  confirm  the  foregoing  results.  In  these  experiments  the 
vessel  was  suspended  on  springs  and  the  jet  directly  down- 
ward, so  that  the  "reaction"  consisted  of  a  diminution  of  the 
tension  of  the  springs  during  the  flow. 

564.  Impulse  of  a  Jet  of  Water  on  a  Fixed  Curved  Vane  (with 
Borders).  —  The  jet  passes  tangentially  upon  the  vane.  Fig. 


IMPULSE   OF  JET. 


801 


626.     B  is  the  stationary  nozzle  from  which  a  jet  of  water  of 

cross-section  F  (area)  and  velocity  =  c  impinges  tangentially 

upon  the  vane,  which  has 

plane  borders,  parallel  to 

paper,  to  prevent  the  lat- 

eral   eicape    of    the    jet. 

The  curve  of  the  vane  is 

not    circular     necessarily. 

The  vane   being   smooth, 

the  velocity  of  the  water 

in  its  curved  path  remains 

=  c  at    ail   points    a*ong 

the  curve.     Conceive  the 

curve  divided  into  a  great 

number  of    small  lengths    each  =  ds,  and   subtending  some 

angle  =  rf'0  from  its  own  centre  of  curvature,  ite  radius  of 

curvature  being  —  r  (different  for  different  <fo's),  which  makes 

some  angle  =  0  with  the  axis  Y  (1  to  original  straight  jet 

BA).     At  any  instant  of  time  there  is  an  arc  of  water  AD  in 

contact  witli  the  vane,  exerting  pressure  upon  it.     The  pres- 

sure dP  of  any  ds  of  the  vane  against  the  small  ma^s  of  water 

Fds.  .  y  -T-  g  then  in  contact  with  it  is  the  "  deviating"  or  "  cen- 

tripetal "  force  accountable  for  its  motion  in  a  curve  of  radius 

=  r,  and  hence  must  have  a  value 


FIG.  626. 


dp  — 


(1) 


The  opposite  and  equal  of  this  force  is  the  dP  shown  in 
Fig.  626,  and  is  the  impulse  or  pressure  of  this  small  mass 
against  the  vane.  Its  ^-component  is  dX  =  dP  sin  0.  By 
making  0  vary  from  0  to  <*,  and  adding  up  the  corresponding 
values  of  dX  we  obtain  the  sum  of  the  JT-components  of  the 
small  pressures  exerted  simultaneously  against  the  vane  by  the 
src  of  water  then  in  contact  with  it  ;  i.e.,  noting  that  d#=r 

-  sin 


Lo 


802 


MECHANICS   OF   ENGINEERING. 


hence  the  X-  impulse  )  = 
against  fixed  vane    ) 


g 


^  a]= 


—  cos  a],  (2) 


in  which  Q  —  Fo—  volume  of  water  which  passes  through  the 
nozzle  (and  also  =  that  passing  over  the  vane,  in  this  case)  per 
unit  of  time,  and  a  =  angle  between  the  direction  of  the 
stream  leaving  the  vane  (i.e.,  at  D)  and  its  original  direction 
(BA  of  the  jet)  ;  i.e.,  a  =  total  angle  of  deviation.  Similarly, 
the  sura  of  the  ^-components  of  the  dP's  of  Fig.  626  may  be 
shown  to  be 

Y  -impulse  on  fixed  vane  =  JQ  dP  .  cos  0  =  52-2  sin  a...  (2)' 
Hence  the  resultant  impulse  on  tha  vane  is  a  force 


F"  = 


-  cos  a),      .    .     (3) 


and  makes  such  an  angle  <*',  Fig. 
that 

t      T          sin  a 

tan  a  =  -=?  = 


X      1  —  cos  a 


>  with  the  direction 


«?n 


FIG.  627. 


FIG.  628. 


For  example,  if  a  =  90°,  then  «'  =  45°;  while  ?f  at  = 
Fig.  628,  we  have  a'  =  0° ;  i.e.,  f*  is  parallel  to  the  jet  BAt 
and  its  value  is 


IMPULSE   OF   JET 


VANE. 


803 


565.  Impulse  of  a  Jet  on  a  Fixed  Solid  of  Revolution  whoso 
Axis  is  Parallel  to  the  Jet.— If  the 
curved  vane,  with  borders,  of  the  pre- 
ceding paragraph  be  replaced  by  a 
solid  of  revolution,  Fig.  629,  with  its 
axis  in  line  of  the  jet,  the  resultant 
pressure  of  the  jet  upon  it  will  simply 
be  the  sum  of  the  ^-components  (i.e., 
=  to  BA)  of  the  pressures  on  all  ele- 
ments of  the  surface  at  a  given  instant ;  Le., 


FIG.  629. 


(5) 


while  the  components  1  to  X,  all  directed  toward  the  axis  of 
the  solid,  neutralize  each  other.  For  a  fixed  plate,  then,  Fig. 
630,  at  right  angles  to  the  jet,  we  have  for  the  force,  or  "  im- 
pulse" (with  a  =  90°), 


,a=x7t 


FIG.  630. 


(6) 


The  experiments  of   Bidone,  made  in 
1838,  confirm   the  truth  of  eq.  (6)  quite 
closely,  as  do  also  those  of  two  students  of 
the  University  of  Pennsylvania  at  Phila- 
delphia (see  Jour,  of  the  Frank.  Inst.  for  Oct.  '87,  p.  258). 

Eq.(6)  is  applicable  to  the  theo- 
ry of  Pitot's  Tube  (see  §  539), 
Fig.  631,  if  we  consider  the  edge 
}f  the  tube  plane  and  quite  wide. 
The  water  in  the  tube  is  at  rest, 
and  its  section  at  A  (of  area = 
may  be  treated  as  a  flat  vertical 
plate  receiving  not  tmly  the 
hydrostatic  pressure  Fxy,  due 
to  the  depth  x  below  the  sur-  Fro.  en. 

face,  but  a  continuous  impulse  P"  —  F£y  -5-  g  [see  eq.  (6)].* 


*  This  implies  that  the  sectional  area  F ot  the  "equivalent  isolated  jet 
is  equal  to  that  of  the  extremity  of  tube  and  that  a  is  90°,  an  assumptioi 
which,  though  simple,  is  largely  conjectural. 


804 


MECHANICS    OF   ENGINEERING. 


For  the  equilibrium  of  the  end  A,  of  the  stationary  column 
AD,  we  must  have,  therefore, 


=  Fxy  +  Fh'Y;    i.e.    A'  =  (8.0)        .  (7) 


y 


«y 


The  relation  in  equation  (7)  corresponds  reasonably  well 
with  the  results  of  Weisbach's  experiments  with  the  instru- 
ment mentioned  in  §  539.  Pitot  himself,  on  trial  of  an  in- 
strument in  which  the  edges  of  the  tube  at  A  were  made  flar- 
ing or  conically  divergent,  like  a  funnel,  found 


(7)' 


while  Darcy,  desirous  that  the  end  of  the  tube  should  occasion 
as  little  disturbance  as  possible  in  the  surrounding  stream, 
made  the  extremity  small  and  conically 
convergent.*  The  latter  obtained  the 
relation 


h'  =  almost  exactly  (1.0) 


(See  §  539.) 

If  the  solid  of  revolution  is  made  cup- 
shaped^  as  in  Fig.  632,  we  have  (as  in 
Fig.  628)  a  =  180°,  and  therefore,  from 
eq.  (5),  ' 


FIG.  632. 


(8) 


EXAMPLE.—  Fig.  632.  If  c  —  30  ft.  per  sec.  and  the  jet 
(cylindrical)  has  a  diameter  of  1  inch,  the  liquid  being  water, 
so  that  y  =  62.5  Ibs.  per  cub.  ft.,  we  have  [ft.,  lb.,  sec.] 


the  impulse  (force)  =  P"  = 


ai(ia)-8 

32.2 


=  19.05  Ibs. 


Experiment  would  probably  show  a  smaller  result. 


See  p.  833  for  Mr.  Freeman's  Experiments. 


IMPULSE   OF   JET   ON   MOVING  VANE. 


805 


o- 


566.  Impulse  of  a  Liquid  Jet  upon  a  Moving  Vane  having 
Lateral  Borders  and  Moving  in  the  Direction  of  the  Jet. — Fig. 
633.  The  vane  has  a  motion  of  translation  (§  108)  in  the 
same  direction  as  the  jet.  Call  this  the  axis  X.  It  is  moving 
with  a  velocity  v  away  from  the  jet  (or,  if  toward  the  jet,  v 
is  negative).  We  con- 
sider v  constant,  its  ac- 
celeration being  prevented 
by  a  proper  resistance 
(such  as  a  weight  =  G) 
to  balance  the  JT-com- 
ponents  of  the  arc-pres- 
sures. Before  coming  in 
contact  with  the  vane, 
which  it  does  tangentially 
(to  avoid  sudden  devia- 
tion), the  absolute  velocity 
(§  83)  of  the  water  in  the 
jet  =  <?,  while  its  velocity 
relatively  to  the  vane  at  A  is  =  c  —  v9  and  it  will  now  be 
'proved  that  the  relative  velocity  along  the  vane  is  constant. 
See  Fig.  634.  Let  v  =  the  velocity  of  the  vane  (of  each 
point  of  it,  since  its  motion  is  one  of  translation),  and  u  =  the 
velocity  of  a  water  particle  (or  small  mass  of  water  of  length 
=  ds]  relatively  to  the  point  of  the  vane  which  it  is  passing. 
Then  10,  the  absolute  velocity  of  the  small  mass,  is  the  diago- 
nal formed  on  u  and  v.  Neglecting  friction,  the  only  actual 
force  acting  on  the  mass  is  P,  the  pressure  of  the  vane  against 
it,  and  this  is  normal  to  the  curve.  Now  an  imaginary  system 
of  forces,  equivalent  to  this  actual  system  of  one  force  jP,  i.e., 
capable  of  producing  the  same  motion  in  the  mass,  may  be 
conceived  of,  consisting  of  the  individual  forces  which  would 
produce,  separately,  the  separate  motions  of  which  the  actual 
motion  of  this  small  mass  M  is  compounded.  These  com- 
ponent motions  are  as  follows : 

1.  A  horizontal  uniform  motion  of  constant  velocity  =  v ; 
and 

2.  A  motion  in  the  arc  of  a  circle  of  radius  =  r  and  with  a 


806  MECHANICS    OF   ENGINEEKING. 

velocity  =  -w,  which  we  shall  consider  variable  until  proved 
otherwise. 

Motion  1  is  of  such  a  nature  as  to  call  for  no  force  (by  New- 
ton's first  law  of  motion),  while  motion  2  could  be  maintained 

bj  a  system  of  two  forces,  one  normal,  Pn ,  =  -  — ,  and  the 

it  ?/ 

other  tangential,  Pt—M-j-  [see  eq.  (5),  p.  76].     This  imagi- 

(Lt 

nary  system  of  forces  is  shown  at  (II.),  Fig.  634,  and  is  equiv- 


(I.) 


FIG.  634. 

alent  to  the  -actual  system  at  (I.).  Therefore  2  (tang,  com- 
pons.)  in  (I.)  should  be  equal  to  2  (tang,  compons.)  in  (II.) ; 
whence  we  have 

-  p<  =  Q->  «-•  ^J  =  °;  or  sH*  •  •  w 

i.e.,  u  is  constant  along  the  vane  and  is  equal  to  c  —  v  at  every 
point.  (The  weight  of  the  mass  has  been  neglected  since  the 
height  of  the  vane  is  small.)  In  Fig.  634  the  symbol  w0  has 
been  used  instead  of  <?,  and  the  point  0  corresponds  to  A  in 
Fig.  633. 

[N.B.  If  the  motion  of  the  vane  were  rotary,  about  an  axis 
1  to  AB  (or  to  c\  this  relative  velocity  would  be  different  at 
different  points.  See  Notes  on  Hydraulic  Motors.  If  the 
radius  of  motion  of  the  point  A,  however,  is  quite  large  com- 
pared with  the  projection  of  AD  upon  this  radius,  the  relative 
velocity  is  approximately  =  c  —  v  at  all  parts  of  the  vane, 
and  will  be  taken  =  c  —  v  in  treating  the  "  Hurdy-gurdy"  in 
§  567.] 


WORK   OF  JET   ON   VANE.  807 

By  putting  2  (normal  compons.)  of  (I.)  =  2  (normal  com- 
pons.)  in  (II.)  We  have 

P  =  Pn;    i.e.,    P  =  MU-=M(P-^;    .    .         (2) 


so  that  to  find  the  sum  of  the  ^-components  of  the  pressures 
exerted  against  the  vane  simultaneously  by  all  the  small  masses 
of  water  in  contact  with  it  at  any  instant,  the  analysis  differs 
from  that  in  §  564  only  in  replacing  the  c  of  that  article  by 
the  (c  —  v)  of  this.  Therefore 

2(Z-pressures)  =  Px  =  ~^(c  —  *>)8[1  —  cos  a],    .    (3) 

y 

(where  a  is  the  angle  of  total  deviation,  relatively  to  vane,  of 
the  stream  leaving  the  vane,  from  its  original  direction),  and 
is  seen  to  be  proportional  to  the  square  of  the  relative  velocity. 
F  is  the  sectional  area  of  jet,  and  y  the  heaviness  (§  7)  of  the 
liquid.  The  1^-cornponent  (or  PT)  of  the  resultant  impulse 
is  counteracted  by  the  support  EF^  Fig.  633.  Hence,  for  a, 
uniform  motion  to  le  maintained,  with  a  given  velocity  =  v, 
the  weight  G  must  be  made  =  Px  of  eq.  (3).  (We  here 
neglect  friction  and  suppose  the  jet  to  preserve  a  practically 
horizontal  direction  for  an  indefinite  distance  before  meeting 
the  vane.  If  this  uniform  motion  is  to  be  toward  the  jet,  v 
will  be  negative  in  eq.  (3),  making  Px(sa\d  .'.  G)  larger  than 
for  a  positive  v  of  same  numerical  value. 

As  to  the  doing  of  work  [§§  128,  etc.],  or  exchange  of 
energy,  between  the  two  bodies,  jet  and  vane,  during  a  uni- 
form motion  away  from  the  jet,  Px  exerts  a  power  of 


-  v)*v[l  -  cos  «],  .     .     .     (4) 

/ 

in  which  Z  denotes  the  number  of  units  of  work  done  per  unit 
of  time  by  Px\  i.e.,  i\\Q  power  (§  130)  exerted  by  P9. 
If  v  is  negative,  call  it  —  v',  and  we  have  the 

Power  expended  )         r>    ,       Fy  ,          /VI  /M  _        /tt 

b,,  vane  upon  jet  \  ==  P*V  =    -X*+«0  «  t1  ~  cos  «]•  •    (<* 


808 


MECHANICS   OF   ENGINEERING. 


Of  course,  practically,  we  are  more  concerned  with  eq.  (4) 
than  with  (5).  The  power  L  in  (4)  is  a  maximum  for  v  =  -Jc; 
but  in  practice,  since  a  single  moving  vane  or  float  cannot 
utilize  the  water  of  the  jet  as  fast  as  it  flows  from  the  nozzle, 
let  us  conceive  of  a  succession  of  vanes  coming  into  position 
consecutively  in  front  of  the  jet,  all  having  the  same  velocity 
v ;  then  the  portion  of  jet  intercepted  between  two  vanes  is  at 
liberty  to  finish  its  work  on  the  front  vane,  while  additional 
work  is  being  done  on  the  hinder  one ;  i.e.,  the  water  will  be 
utilized  as  fast  as  it  issues  from  the  nozzle. 

With  such  a  series  of  vanes,  then,  we  may  put  Q',  =  2<Jc9 
the  volume  of  flow  per  unit  of  time  from  the  nozzle,  in  place 
of  F(c  —  v)  =  the  volume  of  flow  per  unit  of  time  over  the 
vane,  in  eq.  (4) ;  whence 


Power  exerted  on  \  __ 
series  of  vanes     ) 


_  cos 


-?M   .    (6) 


Making  v  variable,  and  putting  dLf-s-dv=Q,  whence  c— 2^=0, 
we  find  that  for  v  =  -J-0,  Z',  the  power,  is  a  maximum.  As- 
suming different  values  for  a,  we  find  that  for  a  —  180°,  i.e., 
by  the  use  of  a  semicircular  vane,  or  of  a  hemispherical  cup, 
Fig.  635,  with  a  point  in  middle,  1  —  cos  or  is  a  max.,  =  2 ; 

whence,  with  v  =  £<?,  we  have,  as  the 

maximum  power, 


L' 


_Q'y  c*__ 

'~ 


in  which  M'  denotes  the  mass  of  the  flow 
per  unit   of  time   from   the    stationary 


nozzle.     Now 


is  the  entire  kinetic 


FIG. 


energy  furnished  per  unit  of  time  oy  the 
jet ;  hence  the  motor  of  Fig.  635  (series 
of  cups)  has  a  theoretical  efficiency  of 
unity,  utilizing  all  the  kinetic»energy  of  the  water.  If  this  is 
true,  the  absolute  velocity  of  the  particles  of  liquid  where  they 
leave  the  cup,  or  vane,  should  be  zero^  which  is  seen  to  be  true, 


POWER   OF   JETS. 


809 


as  follows :  At  ff,  or  Hr,  the  velocity  of  the  particles  rela- 
tively to  the  vane  is  =  G  —  v  =  what  it  was  at  A,  and  hence 

is  =  G =  — ;    hence  at  H  the  absolute  velocity  is  w  = 

2       2 

G  G 

(rel.  veloc.  —  toward  left)— (veloc.  —  of  vane  toward  right)  =  0  °9 

2  2 

Q.E.D.     For  v  >  or  <  \c  this  efficiency  will  not  be  attained. 

567    The  California  "Hurdy-gurdy;"  or  Pelton  Wheel. — The 

efficiency  of  unity  in  the  series  of  cups  just  mentioned  is  in 
practice  reduced  to  80  or  8,5  per  cent  from  friction  and  lateral 
escape  of  water.  The 
Pelton  wheel  or  Cali- 
fornia "  Hurdy-gur- 
dy," shown  (in  prin- 
ciple only)  in  Fig.  636, 
is  designed  to  utilize 
the  mechanical  rela- 
tion just  presented, 
and  yields  results  con- 
firming the  above  the- 
ory, viz.,  that  with  the 
linear  velocity  of  the 

cup-centres  regulated  to  equal  ^,  and  with  a  =  180°,  the  effi- 
ciency approaches  unity  or  100  per  cent.  Each  cup  has  a  pro- 
jecting sharp  edge  or  rib  along  the  middle,  to  split  the  jet ;  see 
Fig.  635. 

This  wheel  was  invented  to  utilize  small  jets  of  very  great 
velocities  (c)  in  regions  just  deserted  by  "  hydraulic  mining" 
operators.  Although  c  is  great,  still,  by  giving  a  large  value 

y> 

to  r,  the  radius  of  the  wheel,  the  making,  of  v  =  -  does  not 

necessitate  an  inconveniently  great  speed  of  rotation  (i.e., 
revolutions  per  unit  of  time).  The  plane  of  the  wheel  may 
be  in  any  convenient  position. 

In  the  London  Engineer  of  May  '84,  p.  397,  is  given  an  ac- 
count of  a  test*  made  of  a  "  Hurdy-gurdy,"  in  which  the  motor 

*  See  p.  834  for  further  details  of  this  test  and  a  perspective  view  of  wheel 


810 


MECHANICS    OF   ENGINEERING. 


showed  an  efficiency  of  87  per  cent.  The  diameter  of  the 
wheel  was  only  6  ft.,  that  of  the  jet  1.89  in.,  and  the  head  of 
the  supply  reservoir  386  ft.,  the  water  being  transmitted 
through  a  pipe  of  22  inches  diameter  and  6900  ft.  in  length, 
107  H.  P.  was  developed  by  the  wheel. 

EXAMPLE. — If  the  jet  in  Fig.  636  has  a  velocity  c  =  60  ft. 
per  second,  and  is  delivered  through  a  2-inch  nozzle,  the  total 
power  due  to  the  kinetic  energy  of  the  water  is  (ft.,  lb.,  sec.) 

~  -1  =  34- 

and  if,  by  making  the  velocity  of  the  cups  =  -  —  30  ft.  per 

2 

sec.,  85  per  cent  of  this  power  can  be  utilized,  the  t>ower  of 
the  wheel  at  this  most  advantageous  velocity  is 

L  —  .85  X  4566.9  =  3881  ft  Ibs.  per  sec.  =  7.05  horse-power 

[since  3881  -r-  550  ==  7.05]  (§  132).  For  a  cup-velocity  of  30 
ft.  per  sec.,  if  we  make  the  radius,  /*,  =  10  feet,  the  angular 
velocity  of  the  wheel  will  be  GO  =  v  -f-  r  =  3.0  radians  per 
sec.  (for  radian  see  Example  in  §  428 ;  for  angula**  velocity, 
§  110),  which  nearly  =  TT,  thus  implying  nearly  a  half-revolu- 
tion per  sec. 

568.  Oblique  Impact  of  a  Jet  on  a  Moving  Plate  having 

no  Border. — The  plate 
has  a  motion  of  trans- 
lation with  a  uniform 
veloc.  ==  v  in  a  direc- 
tion parallel  to  jet, 
whose  velocity  is  —  c. 
At  0  the  filaments  of 
liquid  are  deviated,  so 
that  in  leaving  the  plate 
their  particles  are  an 
found  in  the  moving 

FIG.  637.  plane  BB'  of  the  plate 

surface,  but  the  respective  absolute  velocities  of  these  particles 


OBLIQUE  JET   AND   PLATE.  811 

depend  on  the  location  of  the  point  of  the  plate  where  they 
leave  it,  being  found  by  forming  a  diagonal  on  the  relative 
veloc.  c  —  v  and  the  velocity  v  of  the  p]ate.  For  example,  at 
B  the  absolute  velocity  of  a  liquid  particle  is 


w  =  BE  =  4V  +  (c  —  vf  +  2v(c  —  v)  cos  or, 
while  at  B1  it  is 


w'  =  B'E'  —  Vv*  +  (c  —  v)*  —  2v(c  —  v)  cos  a  ; 


but  evidently  the  component  1  to  plate  (the  other  component 
being  parallel)  of  the  absolute  velocities  of  all  particles  leaving 
the  plate,  is  the  same  and  —  v  sin  a.  The  skin-friction  of  the 
liquid  on  the  plate  being  neglected,  the  resultant  impulse  of 
the  let  against  the  plate  must  be  normal  to  its  surface,  and  its 

JO  L 

amount,  jP,  is  most  readily  found  as  follows  : 

Denoting  by  AM  the  mass  of  the  liquid  passing  over  the 
plate  in  a  short  time  At,  resolve  the  absolute  velocities  of  all 
the  liquid  particles,  before  and  after  deviation,  into  com- 
ponents "I  to  the  plate  (call  this  direction  T)  and  ||  to  the 
plate.  Before  meeting  the  plate  the  particles  composing  AM 
have  a  velocity  in  the  direction  of  T  of  cy  =  c  sin  a  ;  on  leav- 
ing the  plate  a  velocity  in  direction  of  Y  of  v  sin  a  :  they  have 
therefore  lost  an  amount  of  velocity  in  direction  of  Y  = 
(c  —  v)  sin  a  in  time  At\  i.e.,  they  have  suffered  an  average 
retardation  (or  negative  acceleration)  in  a  Indirection  of 

.  accelera-  )        (c  —  v)sm  a  /<iv 

to  Y     -—£-  -•  •    •    W 


Hence  the  resistance  in  direction  of  I^(i.e.,  the  equal  and  op- 
posite of  P  in  figure)  must  be 

PY  =  mass  X  '1^-accel.  =  -—  (c—  v)  sin  a  ;  .     .     (2) 

At 


and  therefore,  since  —  —  -  =  M  =  ~-  —  mass  of  liquid  passing 


812  MECHANICS   OF   ENGINEERING. 

over  the  plate  per  unit  of  time  (not  that  issuing  from  nozzle), 
we  have 


in  which  F  =  sectional  area  of  jet  before  meeting  plate. 

[N.B.  Since  eq.  (3)  can  also  be  written  P  —  Me  sin  a  — 
Mv  sin  a,  and  Me  sin  ex  may  be  called  the  I^-momentum  before 
contact,  while  Mv  sin  a  is  the  Z-momentum  after  contact  (of 
the  mass  passing  over  plate  per  unit  of  time),  this  method  may 
be  said  to  be  founded  on  \hv  principle  of  momentum  which  is 
nothing  more  than  the  relation  that  the  accelerating  force  in 
any  direction  —  mass  X  acceleration  in  that  direction  ;  e.g., 
P.  =  Mpx  ;  Pv  =  Mpv  ;  see  §  74.] 

If  we  resolve  P,  Fig.  637,  into  two  components,  one,  P',  \\ 
to  the  direction  of  motion  (v  and  c\  and  the  other,  P"9  ~|  to 
the  same,  we  have 


=(c-v)sm*<x,  .    ...    (4) 

and 

P"=  P  cos  a  =  -^-  (c  —  v)  sin  a  cos  a.     .     .     (5) 

(Q  =  F(c  —  v)=  volume  passing  over  the  plate  per  unit  of 
time.)  The  force  P"  does  no  work,  while  the  former,  P'9 
does  an  amount  of  work  P'v  per  unit  of  time  ;  i.e.,  exerts  a 
power  (one  plate) 

=  Z  =  P/tf  =  -^(<?--y)vsinV.    .    .    .    (6) 

If,  instead  of  a  single  plate,  a  series  of  plates,  forming  a 
regular  succession,  is  employed,  then,  as  in  a  previous  paragraph, 
we  may  replace  Q,  =  F(c  —  v),  by  Q'  —  Fc,  obtaining  as  the 

Power  exerted  ~by  jet  )        T,      Fey  ,         . 
on  series  of  plates     \  =  L  =  "^(c  ^  ^  8m  «•  •  W 


IMPULSE   OF  JET   ON  OBLIQUE  PLATE.  813 

Q 

For  v  —  —  and  a  =  90°  we  have 

2i 

\M'<?  ,» 


T,        _ 

m"~ 


max. 


=  only  half  the  kinetic  energy  (per  time-unit)  of  the  jet. 

569.  Rigid  Plates  Moving  in  a  Fluid,  Totally  Submerged. 
Fluid  Moving  against  a  Fixed  Plate.  Impulse  and  Resistance. — 
If  a  thin  flat  rigid  plate  have  a  motion  of  uniform  translation 
with  velocity  =  v  through  a  fluid 

which   completely  surrounds  it,  Fig.  "" ' 

638,  a  resistance  is  encountered  (which  ~^> 
must  be  overcome  by  an-  equal  and  op-  - 
posite  force,  not  shown  in  figure,  to 
preserve  the  uniform  motion)  consist- 
ing of  a  normal  component  N,  1  to 
plate,  and  a  (small)  tangential  com- 
ponent, or  skin-friction,  T,  \\  to  plate.  FIG.  638. 
Unless  the  angle  #,  between  the  surface  of  plate  and  the  direc- 
tion of  motion  0  ...  v,  is  very  small,  i.e.  unless  the  plate  is 
moving  nearly  edgewise  through  the  fluid,  N  is  usually  much 
greater  than  T.  The  skin-resistance  between  a  solid  and  a  fluid 
has  already  been  spoken  of  in  §  510. 

When  the  plate  and  fluid  are  at  rest  the  pressures  on  both 
sides  are  normal  and  balance  each  other,  being  ordinary  static 
fluid  pressures.  When  motion  is  in  progress,  however,  the 
normal  pressures  on  the  front  surface  are  increased  by  the 
components,  normal  to  plate,  of  the  centrifugal  forces  of  the 
curved  filaments  (such  as  AB)  or  "  stream-lines,"  while  on 
the  back  surface,  D,  the  fluid  does  not  close  in  fast  enough  to 
produce  a  pressure  equal  to  that  (even)  of  rest.  In  fact,  if  the 
moticr  is  sufficiently  rapid,  and  the  fluid  is  inelastic  (a  liquid), 
a  vacuum  may  be  maintained  behind  the  plate,  in  which  case 
there  is  evidently  no  pressure  on  that  side  of  the  plate. 

Whatever  pressure  exists  on  the  back  acts,  of  course,  to 
diminish  the  resultant  resistance.  The  water  on  turning  the 
sharp  corners  of  the  plate  is  broken  up  into  eddies  forming  a 


814  MECHANICS   OF   ENGINEERING. 

"  wake"  behind.  From  the  accompaniment  of  these  eddies, 
the  resistance  in  this  case  (at  least  the  component  N  normal  to 
plate)  is  said  to  be  due  to  "  eddy-making ;"  though  logically 
we  should  say,  rather,  that  the  body  does  not  derive  the  assist- 
ance (or  negative  resistance)  from  behind  which  it  would  ob- 
tain if  eddies  were  not  formed  ;  i.e.,  if  the  fluid  could  close  in 
behind  in  smooth  curved  stream-lines  symmetrical  with  those 
in  front. 

The  heat  corresponding  to  the  change  of  temperature  pro- 
duced in  the  portion  of  fluid  acted  on,  by  the  skin-friction 
and  by  the  mutual  friction  of  the  particles  in  the  eddies,  is  the 
equivalent  of  the  work  done  (or  energy  spent)  by  the  motive 
force  in  maintaining  the  uniform  motion  (§  149).  (Joule's 
experiments  to  determine  the  Mechanical  Equivalent  of  Heat 
were  made  with  paddles  moving  in  water.) 

If  the  fluid  is  sea-water,  the  results  of  Col.  Beaufoy's  ex- 
periments are  applicable,  viz.: 

The  resistance,  per  square  foot  of  area,  sustained  by  a  sub- 
merged plate  moving  normally  to  itself  [i.e.,  a  =  90°]  in  sea- 
water  with  a  velocity  qfv  =  10  ft.  per  second  is  112  Ibs.  He 
also  asserts  that/0/1  other  velocities  the  resistance  varies  as  the 
square  of  the  velocity.  This  latter  fact  we  would  be  led  to 
suspect  from  the  results  obtained  in  §  568  for  the  impulse  of 
jets;  also  in  §565  [see  eq.  (6)].  Also,  that  when  the  plate 
moved  obliquely  to  its  normal  (as  in  Fig.  638)  the  resistance 
was  nearly  equal  to  (the  resistance,  at  same  velocity,  when 
a  =  90°)  X  (the  sine  of  the  angle  a)  ;  also,  that  the  depth  of 
submersion  had  no  influence  on  the  resistance. 

Confining  our  attention  to  a  plate  moving  nor- 
mally to  itself,  Fig.  639,  let  F  =  area  of  plate, 
y  —  heaviness  (§  409)  of  the  fluid,  v  =  the  uni- 
form velocity  of  plate,  and  g  —  the  acceleration 
of  gravity  (=32.2  for  the  foot  and  second ; 
=  9.81  for  the  metre  and  second).  Then  from 
the  analogy  of  eq.  (6),  §  565,  where  velocity  c  of 
the  jet  against  a  stationary  plate  corresponds  to 
the  velocity  v  of  the  plate  in  the  present  case 
moving  through  a  fluid  at  rest,  we  may  write. 


PLATES  MOVING  IN  FLUIDS.  815 

Resistance  of  fluid  \  __  r>  _  „  ™    v*          (v  normal )          ,^ 
to  moving  plate     j  =         ~       ^  2^~ '  '  '  (  to  plate  j  *  *  *  W 

And  similarly  for  the  impulse  of  cm  indefinite  stream  of  fluid 
against  a  fixed  plate  ( ~|  to  velocity  of  stream),  v  being  the 
velocity  of  the  current, 

Impulse  of  current  \  _  p  _  -,-&    1?          (v  normal  j        /QX 
upon  fixed  plate    )"          ~  ^     ^2^'*'(to  plate  j  '  "  °  ^  f 

The  2y  is  introduced  simply  for  convenience;  since,  having 
<y  given,  we  may  easily  find  v*  -r-  2g  from  a  table  of  velocity- 
heads  ;  and  also  (a  ground  of  greater  importance)  since  the  co- 
efficients C  and  C,f  which  depend  on  experiment  are  evidently 
abstract  numbers  in  the  present  form  of  these  equations  (for 
22  and  P  are  forces,  and  Fytf  -i-  2^  is  the  weight  (force)  of 
an  ideal  prism  of  fluid ;  hence  C  and  C'  must  be  abstract 
numbers.) 

From  Col.  Beaufoy's  experiments  (see  above),  we  have  for 
sea-water  [ft.,  lb.,  sec.],  putting  R  =  112  Ibs.,  F=  1  sq.  ft., 
y  =  64  Ibs.  per  cub.  ft.,  and  v  =  10  ft.  per  second, 

2X32.2X112      ,13 
"  1.0  X  64  X  102 

Hence  in  eq.  (1)  for  sea-water,  we  may  put  C  =  1.13  (with 
y  =  64  Ibs.  per  cub.  ft.). 

From  the  experiments  of  Dubuat  and  Thibault,  Weisbach 
computes  that  for  the  plate  of  Fig.  639,  moving  through  either 
water  or  air,  C  =  1-25  for  eq.  (1),  in  which  the  y  for  air  must 
be  computed  from  §  473 ;  while  for  the  impulse  of  water  or 
air  on  fixed  plates  he  obtains  C'  =  1-86  for  use  in  eq.  (2).  It 
is  hardly  reasonable  to  suppose  that  C  and  C'  should  not  be 
identical  in  value,  and  Prof.  Unwin  thinks  that  the  difference 
in  the  numbers  just  given  must  be  due  to  errors  of  experi- 
ment.* The  latter  value,  C'  =  1.86,  agrees  well  with  equation 
(6)  below.  For  great  velocities  C  and  Cr  are  greater  for  air 
than  for  water,  since  air,  being  compressible,  is  of  greater 
heaviness  in  front  of  the  pkte  than  would  be  computed  for 

.  *  Flamant  thinks  that  this  difference  is  due  to  the  fact  that  the  relative 
conditions  are  not  identical  in  the  two  cases;  since  when  a  current  of  liquid 
impinges  against  a  stationary  plate  there  is  much  intricacy  of  internal  motion 
among  the  particles  of  fluid,  to  which  there  is  nothing  to  correspond  when 
a  plate  is  moved  through  stationary  liquid. 


816  MECHANICS   OF  ENGINEERING. 

the  given  temperature  and  barometric  height  for  use  in  eqs. 
(1)  and  (2) 

The  experiments  of  Borda  in  1763  led  to  the  formula 

P  =  [0.0031  +  0.0003.5^'      ....     (3) 

for  the  total  pressure  upon  a  plate  moving  through  the  air 
in  a  direction  1  to  its  own  surface.  P  is  the  pressure  in 
pounds,  c  the  length  of  the  contour  of  the  plate  in  feet,  and  S 
its  surface  in  square  feet,  while  v  is  the  velocity  in  miles  per 
hour.  Adopting  the  same  form  of  formula,  Hagen  founds 
from  experiments  in  1873,  the  relation 

P=  [0.002894  +  0.00014c]$y"       .     .     .     (4) 

for  the  same  case  of  fluid  resistance. 

Hagen's  experiments  were  conducted  with  great  care,  but 
like  Borda's  were  made  with  a  "whirling  machine,"  in  which 
the  plate  was  caused  to  revolve  in  a  horizontal  circle  of  only 
7  or  8  feet  radius  at  the  end  of  a  horizontal  bar  rotating  about 
a  vertical  axis.  Hagen's  plates  ranged  from  4  to  40  sq.  in.  in 
area,  and  the  velocities  from  1  to  4  miles  per  hour. 

The  last  result  was  quite  closely  confirmed  by  Mr.  H.  Allen 
Hazen  at  "Washington  in  November  1886,  the  experiments 
being  made  with  a  whirling  machine  and  plates  of  from  16  to 
576  sq.  in.  area.  (See  the  American  Journal  of  Science,  Oct. 
1887,  p.  245.) 

In  Thibault's  experiments  plates  of  areas  1.16  and  1.531  sq. 
ft.  were  exposed  to  direct  wind-pressure,  giving  the  formula 

.T  =  0.00475$y3 (5) 

Recent  experiments  in  France  (see  R.  R.  and  Eng.  Journal, 
FeK  '87),  where  flat  boards  were  hung  from  the  side  of  a  rail- 
way train  run  at  different  velocities,  gave  the  formula 

P  =  0.00535$y2 (6) 

The  highest  velocity  was  44  miles  per  hourc  The  magnitude 
of  the  area  did  not  seemingly  affect  the  relation  given.*  More 

*  Langley  found  P  =  0.000327/SV.  See  also  Irminger's  experiments 
^Engineering  News,  Feb.  1895,  p.  109). 


PLATES   IN   FLUIDS.  817 

extended  and  elaborate  experiments  are  needed  in  this  field, 
those  involving  a  motion  of  translation  being  considered  the 
better,  rather  than  with  whirling  machines,  in  which  "centrif- 
ugal action"  must  have  a  disturbing  influence.* 

The  notation  and  units  for  eqs.  (4),  (5),  and  (6)  are  the  same 
as  ihose  givsn  for  (3). 

It  may  be  of  interest  to  note  that  if  equation  (3)  of  §  568  be 
considered  applicable  to  this  case  of  the  pressure  of  an  un- 
limited stream  of  fluid  against  a  plate  placed  at  right-angles  to 
the  current,  with  F  put  equal  to  the  area  of  the  plate,  we  ob- 
tain, after  reduction  to  the  units  prescribed  above  for  the  pre- 
ceding equations  and  putting  a  =  90°, 


(7) 


The  value  y  =  0.0807  Ibs.  per  cub.  ft.  has  been  used  in  the 
substitution,  corresponding  to  a  temperature  of  freezing  and 
a  barometric  height  of  30  inches.  At  higher  temperatures, 
of  course,  y  would  be  less,  unless  with  very  high  barometer. 

569a.  Example.  —  Supposing  each  blade  of  the  paddle-wheel 
of  a  steamer  to  have  an  area  of  6  sq.  ft.,  and  that  when  in  the 
lowest  position  its  velocity  [relatively  to  the  water,  not  to  the 
vessel]  is  5  ft.  per  second  ;  what  resistance  is  it  overcoming  in 
salt  water  ? 

From  eq.  (1)  of  §  569,  with  C  =  1.13  and  y  =  64  Ibs.  per 
cubic  foot,  we  have  (ft.,  lb.,  sec.) 


If  on  the  average  there  may  be  considered  to  be  three  pad- 
dles always  overcoming  this  resistance  on  each  side  of  the 
boat,  then  ,the  work  lost  (work  of  "  slip")  in  overcoming  these 
resistances  per  second  (i.e.,  power  lost)  is 

LI  =  [6  X  169.4]  Ibs.  X  5  ft.  per  sec.  =  5082  ft.-lbs.  per  sea. 
or  9.24  Horse  Power  (since  5082  H-  550  =  9.24).  _ 

*  See  Capt.  Bixby's  article  on  p.  175  et  seq.  of  the  Engineering  New* 
MarcLi  1895. 


818 


MECHANICS   OF   ENGINEERING. 


If.  further,  the  velocity  of  the  boat  is  uniform  and  =  20  ft. 
per  sec.,  the  resistance  of  the  water  to  the  progress  of  the  boat 
at  this  speed  being  6  X  169.4,  i.e.  1016.4  Ibs.,  the  power  ex- 
pended in  actual  propulsion  is 

Z2  =  1016.4  X  20  =  20328  ft.-lbs.  per  sec. 

Hence  the  power  expended  in  both  ways  (usefully  in  propul- 
sion, uselessly  in  "  slip")  is 

Z3  +  L,  =  25410  ft.-lbs.  per  sec.  =  46.2  H..P. 
Of  this,  9.24  H.  P.,  or  about  20  per  cent,  is  lost  in  "  slip." 


570.  Wind-pressure 
on  the  surface  of  a 
roof  inclined  at  an 
angle  =  a  with  the 
horizontal,  i.e.,  with 
the  direction  of  the 
B  wind,  is  usually  esti- 
mated according  to 
the  empirical  formula 


FIG.  640. 


(Button's) 


p  —pf  [sin  a]  [i-84cosa- 1]^ 


(1) 


in  which  p'  =  pressure  of  wind  per  unit  area  against  a  vertical 
surface  (  ~|  to  wind),  and  p  =  that  against  the  inclined  plane 
(and  normal  to  it)  at  the  same  velocity.  For  a  value  of 
jp'='40  Ibs.  per  square  foot  (as  a  maximum),  we  have  the 
following  values  for  j?,  computed  from  (1) : 


For  a—        5° 

10° 

15° 

20° 

25° 

30° 

35° 

40° 

45° 

50° 

55° 

60° 

p=(lbs.  sq.  ft.)  5.  2 

9.6 

14 

18.3 

22.5 

26.5 

30.1 

33.4 

36.1 

38.1 

39.6 

40. 

Duchemin's  formula  for  the  normal  pressure  per  unit-area  is 

/      2  sin  of 


'  I  +  sin2  a 


(2) 


WIND   AND   SAIL. 


with  the  same  notation  as  above.  Some  experimenters  in 
London  tested  this  latter  formula  by  measuring  the  pressure 
on  a  metal  plate  supported  in  front  of  the  blast-pipe  of  a  blow- 
ing engine;  the  results  were  as  follows: 


a  = 

15° 

20° 

60° 

)0° 

p  by  experiment  =  (in  Ibs.  per  sq.  ft.) 

1.65 

2.05 

3.01 

3.31 

By  Duchemin's  formula  p  = 

1.60 

2.02 

3.27 

3.31 

The  scale  of  the  Smithsonian  Institution  at  Washington  for 
the  estimation  and  description  of  the  velocity  and  pressure  of 
the  wind  is  as  follows  : 


Grade. 

Velocity  iu 
miles  per  hour. 

Pressure  per 
sq.  foot  in  Ibs. 

.• 
Name. 

0 

0 

0.00 

Calm. 

1 

2 

0.02 

Very  light  breeze. 

2 

4 

0.08 

Gentle  breeze. 

3 

12 

0.75 

Fresh  wind. 

4 
5 

25 

35 

3.00 
6 

Strong  wind. 
High  wind. 

6 

45 

10 

Gale. 

7 

60 

18 

Strong  gale.    • 

8 
9 

75 

90 

Violent  gale. 
Hurricane. 

10 

100 

Most  violent  hurricane. 

571.  Mechanics  of  the  Sail-boat. — The  action  of  the  wind  on  a 
sail  will  be  understood  from  the  following.  Let  Fig.  641 
represent  the  boat  in  horizontal  projection  and  OS  the  sail,  O 
being  the  mast.  For  /  / 

simplicity  we  consider 
the  sail  to  be  a  plane 
and  to  remain  vertical. 
At  this  instant  the  boat 
is  moving  in  the  direc- 
tion MB  of  its  fore-and- 
aft  line  with  a  velocity  F™- 
=  0,  the  wind  having  a  velocity  of  the  direction  and  magni- 
tude represented  by  w  (purposely  taken  at  an  angle  <  90°  with 
the  direction  of  motion  of  the  boat).  "We  are  now  to  inquire 
the  nature  of  the  action  of  the  wind  on  the  boat,  and  whether 


820  MECHANICS   OF   ENGINEERING. 

in  the  present  position  its  tendency  is  to  accelerate,  or  retard, 
the  motion  of  the  boat.  If  we  form  a  parallelogram  of  which 
w  is  the  diagonal  and  c  one  side,  then  the  other  side  OK,  mak- 
iag  some  angle  a  with  BM,  will  be  the  velocity  v  of  the  wind 
relatively  to  the  boat  (and  sail),  and  upon  this  (and  not  upon  w) 
depends  the  action  on  the  sail.  The  sail,  being  so  placed  that 
the  angle  0  is  smaller  than  a,  will  experience  pressure  from 
"he  wind ;  that  is,  from  the  impact  of  the  particles  of  air  which 
strike  the  surface  and  glance  along  it.  This  pressure,  P,  is 
normal  to  the  sail  (considered  smooth),  and  evidently,  for  the 
position  of  the  parts  in  the  figure,  the  component  of  P  along 
MB  points  in  the  same  direction  as  c,  and  hence  if  that  com- 
ponent is  greater  than  the  water-resistance  to  the  boat  at  this 
velocity,  c  will  be  accelerated;  if  less,  G  will  be  retarded. 
Any  change  in  c,  of  course,  gives  a  different  form  to  the 
parallelogram  of  velocities,  and  thus  the  relative  velocity  v 
and  the  pressure  P,  for  a  given  position  of  the  sail,  will  both 
change.  [The  component  of  P  ~]  to  MB  tends,  of  course,  to 
cause  the  boat  to  move  laterally,  but  the  great  resistance  to 
such  movement  at  even  a  very  slight  lateral  velocity  will  make 
the  resulting  motion  insignificant.] 

As  c  increases,  a  diminishes,  for  a  given  amount  and  position 
of  w ;  and  the  sail  must  be  drawn  nearer  to  the  line  MB,  i.e. 
6  must  be  made  to  decrease,  to  derive  a  wind-pressure  having 
a  forward  fore-and-aft  component ;  and  that  component  be- 
comes smaller  and  smaller.  But  if  the  craft  is  an  ice-boat,  this 
small  component  may  still  be  of  sufficient  magnitude  to  exceed 
the  resistance  and  continue  the  acceleration  of  c  until  c  is 
larger  than  w ;  i.e.,  the  boat  may  be  caused  to  go  as  fast  as,  or 
faster  than,  the  wind,  and  still  be  receiving  from  the  latter  a 
forward  pressure  which  exceeds  the  resistance.  And  it  is 
plain  that  there  is  nothing  in  the  geometry  of  the  figure  to 
preclude  such  a  relation  (i.e.,  c  >  w,  with  6  <  a  and  >  0). 

672.  Resistance  of  Still  Water  to  Moving  Bodies,  Completely 
Immersed. — This  resistance  depends  on  the  shape,  position,  and 
velocity  of  the  moving  body,  and  also  upon  the  roughness  of 
its  surface.  If  it  is  pointed  at  both  ends  (Fig.  642)  with  its 


MOVING  BODIES,    IMMERSED. 


821 


FIG  642 

so  that  the  resistance 


axis  parallel  to  the  velocity,  v,  of  its  uniform  motion,  the 
stream  lines     on     closing    to- 
gether smoothly  at  the  hinder 
extremity,  or  stern,  B,  exert 
normal   pressures  against  the 
surface  of  the  portion  CD...B 
whose     longitudinal     compo- 
nents   approximately   balance 
the  corresponding  components 
of  the  normal  pressures  on  CD 
R,  which  must  be  overcome  to  maintain  the  uniform  velocity 
-y,  is  mainly  due  to  the  "  skin-friction"  alone,  distributed  along 
the  external  surface  of  the  body ;  the  resultant  of  these  resist- 
ances is  a  force  R  acting  in  the  line  AB  of  symmetry  (sup- 
posing the  body  symmetrical  about  the  direction  of  motion). 
If,  however,  Fig.  643,  the  stern,    E..B..F\*  too   bluff, 

. eddies  are  formed  round  the  corners 

^and  jF'j  and  the  pressure  on  the 
surface  E .  .  .  F  is  much  less  than 
in  Fig.  642;  i.e.,  the  water  pres- 
sure from  behind  is  less  than  the 
backward  (longitudinal)  pressures 
from  in  front,  and  thus  the  resultant 
resistance  R  is  due  partly  to  skin- 
friction  and  partly  to  "  eddy-making"  (§  569). 

[NOTE. — The  diminished  pressure  on  EF\$>  analogous  to  the 
loss  of  pressure  of  water  (flowing  in  a  pipe)  after  passing  a  nar- 
row section  the  enlargement 
from  which  to  the  original 
section  is  sudden.  E.g.,  Fig. 
644,  supposing  the  velocity  v 
and  pressure  p  (per  unit-area) 
to  be  the  same  respectively 
at  A  and  A',  in  the  two 
pipes  shown,  with  diameter 
AL  =  WK=A'L'  =  WE'  • 
then  the  pressure  at  M 
equal  to  that  at  A  (disregarding  skin-friction),  whereas  that  at 


FIG.  643. 


K' 


FIG.  644. 


822 


MECHANICS   OF   ENGINEERING. 


that  at  M  is  considerably  less  than  that  at  A'  on  account 
of  the  head  lost  in  the  sudden  enlargement.    (See  also  Fig. 

575.)] 

It  is  therefore  evident  that  bluff  ness  of  stern  may  be  a  large 
factor  in  the  production  of  resistance. 

In  any  case  experiment  shows  that  for  a  given  body  sym- 
metrical  about  an  axis  and  moving  through  a  fluid  (not  only 
water,  but  any  fluid)  in  the  direction  of  its  axis  with  a  uni- 
form velocity  =  -y,we  may  write  approximately  the  resistance 


R  =  (resistance  at  vel.  v)  = 


2? 


(i) 


As  in  preceding  paragraphs,  F  —  area  of  the  greatest  section, 
"I  to  axis,  of  the  external  surface  of  body  (not  of  the  sub- 
stance;  i.e.,  the  sectional  area  of  the  circumscribing  cylinder 
(cylinder  in  the  most  general  sense)  with  elements  parallel 
to  the  axis  of  the  body,  y  =  the  heaviness  (§  409)  of  the 
fluid,  and  v  =  velocity  of  motion ;  while  C  is  an  abstract 
number  dependent  on  experiment. 

According  to  Weisbach,  who  cites  different  experimenters, 
we  can  put  for  spheres,  moving  in  water,  C  =  about  0.55 ; 
for  cannon-balls  moving  in  water,  C  =  .467. 

According  to  Robins  and  Hutton,  for  spheres  in  air,  we 
have 


For  v  in  mets.  [ 
per  sec.        f  * 

5 

25 

100 

200 

300 

400 

KAA  j  meters 
5UU1  per  sec. 

C  =  .59 

.63 

.67 

.71 

.77 

88 

.99 

1.04 

For  musket-balls  in  the  air,  Piobert  found 

C  =  0.451  (1  +  0.0023  X  veloc.  in  metres  per  sec.). 

From  Dubuat's  experiments,  for  the  resistance  of  water  to 
a  right  prism  moving  erudwise  and  of  length  =  Z, 


For(J:  VF)  =    0 

1 

2 

3 

C  =  125 

1.26 

1.31 

1.33 

For  a  circular  cylinder  moving  perpendicularly  to  its  axis 
Borda  claimed  that  C  is  one-half  as  much  as  for  the  circun> 


MOVING  BODIES,    IMMERSED.  823 

scribing  right  parallelopiped  moving  with  four  faces  parallel 
to  direction  of  motion. 

EXAMPLE. — The  resistance  of  the  air  at  a  temperature  of 
freezing  and  tension  of  one  atmosphere  to  a  musket-ball  J  inch 
in  diameter  when  moving  with  r  velocity  of  328  ft.  per  sec. 
is  thus  determined  by  Piobert's  formula,  above : 

C  =  0.451(1  +  .0023  X  100)  =  0.554 ; 
hence,  from  eq.  (1), 


R  =  0.554  X  x  -0807  X          -=  0.1018  Ibs. 


572a.  Deviation  of  a  Spinning  Ball  from  a  Vertical  Plane  in 
Still  Air.  —  It  is  a  well-known  fact  in  base-  ball  playing  that  if  a 
rapid  spinning  motion  is  given  to  the  ball  about  a  vertical  axis 
as  well  as  a  forward  motion  of  translation,  its  path  will  not 
remain  in  its  initial  vertical  plane,  but  curve  out  of  that  plane 
toward  the  side  on  which  the  absolute  velocity  of  an  external 
point  of  the  ball's  surface  is  least.  Thus,  if  the  ball  is  thrown 
from  North  to  South,  with  a  spin  of  such  character  as  to  ap- 
pear "  dock-wise"  seen  from  above,  the  ball  will  curve  towa/rd 
the  West,  out  of  the  vertical  plane  in  which  it  started. 

This  could  not  occur  if  the  surface  of  the  ball  were  perfectly 
smooth  (there  being  also  no  adhesion  between  that  surface  and 
the  air  particles),  and  is  due  to  the  fact  that  the  cushion  of  com- 
pressed air  which  the  ball  piles  up  in  front  during  its  progress, 
and  which  would  occupy  a  symmetrical  position  with  respect 
to  the  direction  of  motion  of  the  centre  of  the  ball  if  there 
were  no  motion  of  rotation  of  the  kind  indicated,  is  now  piled 
up  somewhat  on  the  East  of  the  centre  (in  example  above), 
crerting  constantly  more  obstruction  on  that  side  than  on  the 
right  ;  the  cause  of  this  is  that  the  absolute  velocity  of  the  sur- 
face-points, at  the  same  level  as  the  centre  of  ball,  is  greatest, 
and  the  friction  greatest,  at  the  instant  when  they  are  passing- 
through  their  extreme  Easterly  positions;  since  then  that 
velocity  is  the  sum  of  the  linear  velocity  of  translation  and 
that  of  rotation  ;  whereas,  in  the  position  diametrically  oppo- 


824  MECHANICS   OF  ENGINEERING. 

Bite,  on  the  West  side,  the  absolute  velocity  is  the  difference  ; 
hence  the  greater  accumulation  of  compressed  air  on  the  left 
(in  the  case  above  imagined,  ball  thrown  from  North  to  South, 
etc.). 

573.  Robinson's  Cup-anemometer. — This  instrument,  named 
after  Dr.  T.  K.  Kobinson  of  Armagh,  Ireland,  consists  of  four 
hemispherical  cups  set  at  equal  intervals  in  a  circle,  all  facing 
in  the  same  direction  round  the  circle,  and  so  mounted  on  a 
light  but  rigid  framework  as  to  be  capable  of  rotating  with 
but  little  friction  about  a  vertical  axis.  When  in  a  current  of 
air  (or  other  fluid)  the  apparatus  begins  to  rotate  with  an  ac- 
celerated velocity  on  account  of  the  pressure  against  the  open 
mouth  of  a  cup  on  one  side  being  greater  than  the  resistance 
met  by  the  back  of  the  cup  diametrically  opposite.  Very  soon, 
however,  the  motion  becomes  practically  uniform,  the  cnp- 
centre  having  a  constant  linear  velocity  v"  the  ratio  of  which 
to  the  velocity,  v',  of  the  wind  at  the  same  instant  must  be 
found  in  some  way,  in  order  to  deduce  the  value  of  the  latter 
from  the  observed  amount  of  the  former  in  the  practical  use 
of  the  instrument.  After  sixteen  experiments  made  by  Dr. 
Kobinson  on  stationary  cups  exposed  to  winds  of  varying  in- 
tensities, from  a  gentle  breeze  to  a  hard  gale,  the  conclusion 
was  reached  by  him  that  with  a  given  wind- velocity  the  total 
pressure  on  a  cup  with  concave  surface  presented  to  the  wind 
was  very  nearly  four  times  as  great  as  that  exerted  when  the 
convex  side  was  presented,  whatever  the  velocity  (see  vol. 
xxu  of  Transac.  Irish  Royal  Acad.,  Part  /,  p.  163). 

Assuming  this  ratio  to  be  exactly  4.0  and  neglecting  axle- 
friction,  we  have  the  data  for  obtaining  an  approximate  value 
of  m,  the  ratio  of  v'  to.the  observed  v",  when  the  instrument  is 
in  use.  The  influence  of  the  wind  on  those  cups  the  planes  of 
whose  mouths  are  for  the  instant  ||  to  its  direction  will  also  be 
neglected. 

If,  then,  Fig.  645,  we  write  the  impulse  on  a  cup  when  the 
hollow  is  presented  to  the  wind  [§  572,  eq.  (1)] 

^»  =  C*Fy, (i) 


ROBINSON'S  CUP-ANEMOMETER.  825 

and  the  resistance  when  the  convex  side  is  presented 

P,  =  ^Fr^,    .......   (2) 

we  may  also  put 

In  (1)  and  (2)  v  and  va  are  relative  velocities. 

Eegarding  only  the  two  cups  A  and  B, 
whose  centres  at  a  definite  instant  are  mov-  - 

ing  in  lines  parallel  to  the  direction  of  the  

wind,  it  is  evident  that  the  motion  of  the  ~_ 
cups  does  not  become  uniform  until  the  rel- 
ative velocity  v'  —  v"  of  the  wind  and  cup 
A  (retreating  before  the  wind)  has  become 
so  small,  and  the  relative  velocity  v'  4-  v" 
with  which  B  advances  to  meet  the  air- 
particles  has  become  so  great,  that  the  im- 
pulse of  the  wind  on  A  equals  the  resist- 
ance encountered  by  B\  i.e.,  these  forces, 
Ph  and  PC)  must  be  equal,  having  equal 
lever-arms  about  the  axis.  Hence,  for  uniform  rotary  motion, 


i.e.  [see  eq.  (3)], 

+  1J;    °r'   ^-l)'  =  (^  +  l)'.  •  (5) 
Solving  the  quadratic  for  m,  we  obtain 

m  =  3.00 (6) 

That  is,  the  velocity  of  the  wind  is  about  three  times  that  of 
the  cup-centre. 

574.  Experiments  with  Robinson's  Cup-anemometei.  -The 
ratio  3.00  just  obtained  is  the  one  in  common  use  in  connec- 
tion with  this  instrument  in  America.  Experiments  by  Mr. 


826  MECHANICS   OF   ENGINEERING. 

Hazen  at  Washington  in  1886  (Am.  Jour.  Science,  Oct.  '87, 
p.  248)  were  made  on  a  special  type  devised  by  Lieut.  Gibbon. 
The  anemometer  was  mounted  on  a  whirling  machine  at  the 
end  of  a  16-ft.  horizontal  arm,  and  values  for  ra  obtained,  with 
velocities  up  to  12  miles  per  hour,  from  2.84  to  3.06 ;  average 
2.94.  The  cups  were  4  in.  in  diameter  and  the  distance  of  their 
centres  from  the  axis  6.72  in.,  these  dimensions  being  those 
usually  adopted  in  America.  This  instrument  was  nearly  new 
and  was  well  lubricated. 

Dr.  Robinson  himself  made  an  extensive  series  of  experi- 
ments, with  instruments  of  various  sizes,  of  which  an  account 
may  be  found  in  the  Philos.  Transact,  for  1878,  p.  797  (see 
also  the  volume  for  1880,  p.  1055).  Cups  of  4  in.  and  also  of 
9  in.  were  employed,  placed  first  at  24  and  then  at  12  in.  from 
the  axis.  The  cup-centres  revolved  in  a  (moving)  vertical 
plane  perpendicular  to  the  horizontal  arm  of  a  whirling- 
machine  ;  this  arm,  however,  was  only  9  ft.  long.  A  friction- 
brake  was  attached  to  the  axis  of  the  instrument  for  testing  the 
effect  of  increased  friction  on  the  value  of  m.  At  high  speeds 
of  30  to  40  miles  per  hour  (i.e.,  the  speed  of  the  centre  of  the 
instrument  in  its  horizontal  circle,  representing  an  equal  speed 
of  wind  for  an  instrument  in  actual  use  with  axis  stationary) 
the  effect  of  friction  was  relatively  less  than  at  low  velocities. 
That  is,  at  high  speeds  with  considerable  friction  the  value  of 
m  was  nearly  the  same  as  with  little  friction  at  low  speeds. 
With  the  large  9  in.  cups  at  a  distance  of  either  24  or  12  in. 
from  the  axis  the  value  of  in  at  30  miles  per  hour  ranged 
generally  from  2.3  to  2.6,  with  little  or  much  friction  ;  while 
with  the  minimum  friction  m  rose  slowly  to  about  2.9  as  the 
velocity  diminished  to  10  miles  per  hour.  At  5  miles  per 
hour  with  minimum  friction  m  was  3.5  for  the  24  in.  instru- 
ment and  about  5.0  for  the  12  in.  The  effect  of  considerable 
friction  at  low  speeds  was  to  increase  m,  making  it  as  high  as 
8  or  10  in  some  cases.  With  the  4  in.-cups  no  value  was  ob- 
tained f or  m  less  than  3.3.  On  the  whole,  Dr.  Robinson  con- 
cluded that  m  is  more  likely  to  have  a  constant  value  at  all 
velocities  the  larger  the  cups,  the  longer  the  arms,  and  the  less 
the  friction,  of  the  anemometer.  But  few  straight-line  experi- 


VAEIOUS   ANEMOMKTEKS.  827 

ments  have  been  made  with  the  cup-anemometer,  the  most 
noteworthy  being  mentioned  on  p.  308  of  the  Engineering 
News  for  October  1887.  The  instrument  was  placed  on  the 
front  of  the  locomotive  of  a  train  running  between  Baltimore 
and  Washington  on  a  calm  day.  The  actual  distance  is  40 
miles  between  the  two  cities,  while  from  the  indications  of  the 
anemometer,  assuming  m  =  3.00,  it  would  have  been  in  one  trip 
46  miles  and  in  another  47.  The  velocity  of  the  train  was  20 
miles  per  hour  in  one  case  and  40  in  the  other. 

575  Other  Anemometers. — Both  Biram's  and  Castello's  ane- 
mometers consist  of  a  wheel  furnished  with  radiating  vanes 
set  obliquely  to  the  axis  of  the  wheel,  forming  a  small  "wind- 
mill," somewhat  resembling  the  current-meter  for  water  shown 
in  Fig.  604 ;  having  six  or  eight  blades,  however.  The  wheel 
revolves  with  but  little  friction,  and  is  held  in  the  current  of 
air  with  its  axis  parallel  to  the  direction  of  the  latter,  and  very 
quickly  assumes  a  steady  motion  of  rotation.  The  number  of 
revolutions  in  an  observed  time  is  read  from  a  dial.  The  in- 
struments must  be  rated  by  experiment,  and  are  used  chiefly 
in  measuring  the  velocity  of  the  currents  of  air  in  the  galleries 
of  mines,  of  draughts  of  air  in  flues  and  ventilating  shafts,  etc. 

To  quote  from  vol.  v  of  the  Eeport  of  the  Geological  Sur- 
vey of  Ohio,  p.  370  :  fc<  Approximate  measurements  (of  the 
velocity  of  air)  are  made  by  miners  by  flashing  gunpowder, 
and  noting  with  a  watch  the  speed  with  which  the  smoke 
moves  along  the  air-way  of  the  mine.  A  lighted  lamp  is 
sometimes  used,  the  miner  moving  along  the  air-gallery,  and 
keeping  the  light  in  a  perfectly  perpendicular  position,  noting 
the  time  required  to  pass  to  a  given  point." 

Another  kind  makes  use  of  the  principle  of  Pitot's  Tube 
(p.  751),  and  consists  of  a  U-tube  partially  filled  with  water, 
one  end  of  the  tube  being  vertical  and  open,  while  the  other 
turns  horizontally,  and  is  enlarged  into  a  wide  funnel,  whose 
mouth  receives  the  impulse  of  the  current  of  air;  the  differ- 
ence of  level  of  the  water  in  the  two  parts  of  the  U  is  a  meas- 
ure of  the  velocity. 


828  MECHANICS   OF   ENGINEERING. 

576.  Resistance  of  Ships.* — We  shall  first  suppose  the  ship  to 
l)e  towed  at  a  uniform  speed ;  i.e.,  to  be  without  means  of  self- 
propulsion  (under  water).  This  being  the  case,  it  is  found  that 
at  moderate  velocities  (under  six  miles  per  hour),  the  ship 
being  of  "fair"  form  (i.e.,  the  hull  tapering  both  at  bow  and 
stern,  under  water)  the  resistance  in  still  water  is  almost  wholly 
due  to  skin-friction,  " eddy-making"  (see  §  569)  being  done 
away  with  largely  by  avoiding  a  bluff  stern. 

When  the  velocity  is  greater  than  about  six  miles  an  hour 
the  resistance  is  much  larger  than  would  be  accounted  for  by 
skin-friction  alone,  and  is  found  to  be  connected  with  the  sur- 
face-disturbance or  waves  produced  by  the  motion  of  the  hull 
in  (originally)  still  water.  The  recent  experiments  of  Mr. 
Froude  and  his  son  at  Torquay,  England,  with  models,  in  a  tank 
300  feet  long,  have  led  to  important  rules  (see  Mr.  White's 
Naval  Architecture  and  "  Hydromechanics"  in  the  Ency. 
Britann.}  of  so  proportioning  not  only  the  total  length  of  a 
ship  of  given  displacement,  but  the  length  of  the  entrance  (for- 
ward tapering  part  of  hull)  and  length  of  run  (hinder  tapering 
part  of  hull),  as  to  secure  a  minimum  "wave-making  resist- 
ance" as  this  source  of  resistance  is  called. 

To  quote  from  Mr.  White  (p.  460  of  his  Naval  Architecture, 
London,  1882):  "Summing  up  the  foregoing  remarks,  it 
appears : 

"  (1)  lL\\&t  frictional  resistance,  depending  upon  the  area  of 
the  immersed  surface  of  a  ship,  its  degree  of  roughness,  its 
length,  and  (about)  the  square  of  its  *  speed,  is  not  sensibly 
affected  by  the  forms  and  proportions  of  ships;  unless  there 
be  some  unwonted  singularity  of  form,  or  want  of  fairness. 
For  moderate  speeds  this  element  of  resistance  is  by  far  the 
most  important ;  for  high  speeds  it  also  occupies  an  important 
position — from  50  to  60  per  cent  of  the  whole  resistance, 
probably,  in  a  very  large  number  of  classes,  when  the  bottoms 
are  clean  •  and  a  larger  percentage  when  the  bottoms  become 
foul. 

"  (2)  That  eddy-making  resistance  is  usually  small,  except  in 
special  cases,  and  amounts  to  8  or  10  per  cent  of  the  frictiona) 

*  Not  in  canals,  but  in  water  of  indefinite  width  and  deptli 


RESISTANCE   OF   SHIPS.  829 

resistance.  A  defective  form  *of  stern  causes  largely  increased 
eddy-making. 

"  (3)  That  wave-making  resistance  is  the  element  of  the 
total  resistance  which  is  most  influenced  by  the  forms  and  pro- 
portions of  ships.  Its  ratio  to  the  frictional  resistance,  as  well 
as  ite  absolute  magnitude,  depend  on  many  circumstances  ;  the 
Tnost  important  being  the  forms  and  lengths  of  the  entrance 
and  run,  in  relation  to  the  intended  full  speed  of  the  ship. 
For  every  ship  there  is  a  limit  of  speed  beyond  which  each 
small  increase  in  speed  is  attended  by  a  disproportionate  in- 
crease in  resistance  ;  and  this  limit  is  fixed  by  the  lengths  of 
the  entrance  and  run  —  the  '  wave-making  features  '  of  a  ship. 

"The  sum  of  these  three  elements  constitutes  the  total  re- 
sistance offered  by  the  water  to  the  motion  of  a  ship  towed 
through  it,  or  propelled  by  sails  ;  in  a  steamship  there  is  an 
1  augment  '  of  resistance  due  to  the  action  of  the  propel- 
lers." 

In  the  case  of  a  steamship  driven  by  a  screw  propeller,  this 
augment  to  the  resistance  varies  from  20  to  45  per  cent  of  the 
"tow-rope  resistance,"  on  account  of  the  presence  and  action 
of  the  propeller  itself  ;  since  its  action  relieves  the  stern  of 
some  of  the  forward  hydrostatic  pressure  of  the  water  closing 
in  around  it.  Still,  if  the  screw  is  placed  far  back  of  the  stern, 
the  augment  is  very  much  diminished  ;  but  such  a  position  in- 
volves risks  of  various  kinds  and  is  rarely  adopted. 

We  may  compute  approximately  the  resistance  of  the  water 
to  a  ship  propelled  by  steam  at  a  uniform  velocity  v,  in  the 
following  manner  :  Let  L  denote  the  power  developed  in  the 
engine  cylinder  ;  whence,  allowing  10  per  cent  of  L  for  engine 
friction,  and  15  per  cent  for  "  work  of  slip"  of  the  propeller- 
blade,  we  have  remaining  0.75Z,  as  expended  in  overcoming 
the  resistance  R  through  a  distance  =  v  each  unit  of  time  ;  i.e., 


(ypprox.)  0.75Z  =  Jfo  ........     (1) 

EXAMPLE.—  If  3000  indicated  H.  P.  (§  132)  is  exerted  by  the 
engines  of  a  steamer  at  a  uniform  speed  of  15  miles  per  hour 


830  MECHANICS    OF    ENGINEERING. 

(=  22  ft.  per  sec.),  we  have  (with  above  allowances  for  slip  and 
engine  friction)  [foot-lb.-sec.] 

i  X  3000  X  550  =  E  X  22 ;     /.    R  =  56250  Ibs. 

Further,  since  R  varies  (roughly)  as  the  square  of  the  veloc- 
ity, and  can  therefore  be  written  R  =  (Const.)  X  ^a,  we  have 

from  (1) 

L  =  a  constant  X  v*      .     .     .     .     .     .     (2) 

as  a  roughly  approximate  relation  between  the  speed  and  the 
power  necessary  to  maintain  it  uniformly.  In  view  of  eq.  (3) 
involving  the  cube  of  the  velocity  as  it  does,  we  can  understand 
why  a  large  increase  of  power  is  necessary  to  secure  a  propor, 
tionally  small  increase  of  speed. 

577,  "  Transporting  Power,"  or  Scouring  Action,  of  a  Current, 
—The  capacity  or  power  .of  a  current  of  water  in  an  open 
channel  to  carry  along  with  it  loose  particles,  sand,  gravel, 
pebbles,  etc.,  lying  upon  its  bed  was  investigated  experiment 
tally  by  Dubuat  about  a  century  ago,  though  on  a  rather  small 
scale.  His  results  are  as  follows : 

The  velocity  of  current  must  be  at  least 

0.25  ft.  per  sec.,  to  transport  silt ; 

0.50        "          ."          "        loam; 

1.00        "  "          "        sand; 

2.00        "  «          "        gravel; 

3.5         "  "          "        pebbles  1  in.  in  diam.; 

4.0  "          "        broken  stone ; 

5.0          "  "          «        chalk,  soft  shale. 

However,  more  modern  writers  call  attention  to  the  fact 
that  in  some  instances  beds  of  sand  are  left  undisturbed  by 
currents  of  greater  velocity  than  that  above  indicated  for  sand, 
and  explain  this  fact  on  the  theory  that  the  water-particles 
may  not  move  parallel  to  the  bed,  but  in  cycloids,  approxi- 
mately, like  the  points  in  the  rim  of  a  rolling  wheel,  so  as  to 
have  little  or  no  scouring  action  on  the  bed  in  those  cases. 

In  case  the  particles  move  in  filaments  or  stream-lines 
parallel  to  the  axis  of  the  stream  the  statement  is  sometimes 
made  that  the  "  transporting  power"  varies  as  the  sixth  powef 


"TRANSPORTING  POWER"  OF  CURRENTS. 


831 


of  the  velocity  of  the  current,  by  which  is  meant,  more  defi- 
nitely, the  following:  Fig.  646.  Conceive  a  row  of  cubes  (or 
other  solids  geometri- 
cally similar  to  each 
other)  of  many  sizes, 
all  of  the  same  heavi- 
ness (§  7),  and  simi- 
larly situated,  to  be 
placed  on  the  horizon- 

sSt^S^^ 


FIG.  646. 


tal  bottom  of  a  trough 
and  there  exposed  to 
a  current  of  water, 
being  completely  im- 
mersed. Suppose  the  coefficient  of  friction  between  the  cubes 
and  the  trough-bottom  to  be  the  same  for  all.  Then,  as  the 
current  is  given  greater  and  greater  velocity  v,  the  impulse 
Pm  (corresponding  to  a  particular  velocity  vm)  against  some 
one,  m,  of  the  cubes,  will  be  just  sufficient  to  move  it,  and  at 
some  higher  velocity  vn  the  impulse  Pn  against  some  larger 
cube,  ft,  will  be  just  sufficient  to  move  it,  in  turn.  "We  are  to 
prove  that  Pm  :  Pn  ::  vm6  :  vn\ 

Since,  when  a  cube  barely  begins  to  move,  the  impulse  is 
equal  to  the  friction  on  its  base>  and  the  frictions  under  the 
cubes  (when  motion  is  impending)  are  proportional  to  their 
volumes  (see  above),  we  have  therefore 


3  * 


(i) 


Also,  the  impulses  on  the  cubes,  whatever  the  velocity,  are  pro- 
portional to  the  face  areas  and  to  the  squares  of  the  velocities 
(nearly ;  see  §  572) ;  hence 


(2) 


From  (1)  and  (2)  we  have 


11     "»    •        Tj  />  '*     •"• 

" ""  ~~ ~~ 


(8) 


832 


MECHANICS   OF  ENGINEERING. 


while  from  (3)  and  (2)  we  have,  finally, 


Thus  we  see  in  a  general  way  why  it  is  that  if  the  velocity 
of  a  stream  is  doubled  its  transporting  power  is  increased 
about  sixty -four-fold  ;  i.e.,  it  can  now  impel  along  the  bottom 
pebbles  that  are  sixty-four  times  as  heavy  as  the  heaviest  which 
it  could  move  before  (of  same  shape  and  heaviness). 

Though  rocks  are  generally  from  two  to  three  times  as 
heavy  as  water,  their  loss  of  weight  under  water  causes  them  to 
encounter  less  friction  on  the  bottom  than  it  not  immersed. 

578.  Recent  Experiments  with  Fire-hose,  Nozzles,  etc.  (Ad- 
dendum to  §  520.) — The  very  full  and  careful  investigations  of 
Mr.  J.  R.  Freeman,  hydraulic  engineer  of  Boston,  Mass.,  in  this 
line  (see  Transac.  A.  S.  0.  E.,  Nov.  1889)  furnish  the  following 
results:  By  taking  piezometer  readings  at  the  ends  of  a  portion 
of  fire-hose  conducting  a  steady  flow  of  water,  the  values  of  loss 
of  pressure  due  to  fluid  friction  per  100  feet  of  length  could  be 
computed ;  a  careful  measurement  being  also,  made  of  the 
diameter  of  the  hose  and  of  the  volume  of  water  transmitted  in 
an  observed  time.  The  table  here  given  presents  results  applica- 
ble to  hose  of  exactly  2.5  inches  diameter,  for  a  delivery  of  water 
at  the  rate  of  240  gallons  per  minute  (that  is,  for  a  velocity  in 
the  hose  of  15.68  ft.  per  sec.).  (The  value  of  /  has  been  com- 
puted by  the  writer.) 


Sample. 

Description. 

Loss  of  pressure, 
per  100  ft.  of 
length,  in  Ibs.  per 
sq.  in. 

Coefficient/. 
(See  §  520.) 

Velocity  of 
water  in 
hose. 

L 
K 

I 

E 

Unlined  linen  hose  
Woven    cotton,     rubber-lined, 
"Mill  Hose"  
K;  "-,  cotton,  rubber-lined,  hose. 
Ditto,  but  interstices  between 
threads  well  filled  

33.2 

25.5 
19.4 

16.0 

0.01045 

0.00802 
0.00610 

0.00508 

15.68 

15.68 
>  15.68 

15.68 

C 

Wo  /en     cotton,    rubber-lined, 
hose.    So  well  filled  with  the 
rubber  that  the  inner  surface 
remained  smooth  under  pres- 
sure 

14  1 

0  00443 

15  68 

It  was  found  that  with  other  rates  of  flow  the  friction-head 
varied  nearly  as  the  square  of  the  velocity.  The  great  importance 
of  a  smooth  interior  of  hose  is  well  shown  by  this  table. 

A  short  section  of  each  kind  of  hose  was  filled  with  liquid 
plaster  underpressure.  After  the  setting  of  the  plaster  the  hose 
was  removed  and  photographs  taken  of  the  cast,  thus  conveying 
a  definite  idea  of  the  degree  of  roughness  of  interior  of  hose. 


EXPERIMENTS    WITH   NOZZLES. 


833 


As  to  nozzles,  it  was  found  that  the  plain  conical  nozzle  gave 

the  best  results,  jets  from  the  ring-nozzles  being  slightly  inferior 
in  range. 

By  means  of  a  very  delicate  form  of  Pitot's  tube  measurement! 
were  made  of  the  velocity  in  different  parts  of  the  section  of  jets, 
near  the  nozzle,  with  the  interesting  result  that  in  "about  two- 
thirds  the  whole  distance  from  centre  to  circumference  the  ve- 
locity remains  the  same  as  at  centre,"  and  that  at  ^  inch  from 
tb:  wall  of  most  of  the  orifices  the  velocity  was  only  5$  less  than 
at  centre  of  jet.  With  a  jet  from  a  5-foot  length  of  brass  tubing 
1£  inch  in  diameter  and  used  as  a  nozzle  the  velocity  fell  off 
rapidly  for  filaments  further  from  the  centre;  e.g.,  at  half  the 
distance  from  centre  to  circumference  the  velocity  was  90$  of 
that  at  the  centre,  and  at  the  outside  edge  60$.  Most  of  the 
nozzles  ranged  from  1  in.  to  1£  in-  in  diameter  of  orifice. 

By  using  these  velocity  measurements  to  "  gauge  "  the  flow  it 

02 

was  found  that  the  relation  h'  =  — -  was  quite  closely  borne  out 

*9 

(within  1$)  (see  eq.  (7)",  p.  804).  The  point  of  the  Pitot  tube 
was  conically  convergent,  its  extremity  being  0.017  in.  in  external 
diameter  and  containing  an  orifice  of  0.006  in.  diameter.  A 
minute  passage-way  led  from  the  orifice  to  a  Bourdon  gauge. 

Based  on  his  experiments,  Mr.  Freeman  gives  tables  for  the 
maximum  vertical  height,  V,  and  also  the  maximum  horizontal 
range,  H,  of  "good  effective  fire-streams"  delivered  from  smooth 
conical  nozzles  of  various  sizes  and  with  different  piezometer 
pressures  p  (in  Ibs.  p.  sq.  in.  above  atmosphere)  at  the  base  of: 
play-pipe,  the  gauge  being  at  same  level  as  nozzle.  (The  dis- 
tances reached  by  the  extreme  drops  are  very  inucn  greater  with, 
the  high  pressures.  V  and  H  are  in  feet.) 

The  following  is  a  brief  synopsis  of  this  table,  d  is  the  internal 
diameter  of  the  extremity  of  nozzle.  The  maximum  horizontal 
range  was  obtained  at  an  angle  of  elevation  of  about  32°. 


For  p  = 

10 

20 

40 

60 

80 

100 

.g:  s  s  t  s 

wUB8? 

II  II  II  II  II  II 

•e-e-e'c'e'a  1 

V   H 

V   H 

V   H 

V   H 

V   H 

V   H 

~83   «T 
90   7ft 
96   83 
99   8fr 
101   98 
103   98 

17  ft.  19 
18   21 
18   21 
18   22 
19   22 
20   23 

33   29 
34   33 
35   37 
36   38 
37   40 
38   42 

60   44 
62   49 
64   55 
65   59 
67   63 
69   66 

72   54 
77   61 
79   67 
83   72 
85   76 
87   79 

79   62 
85   70 
89   76 
92   81 
95   85 
97   88 

834 


MECHANICS   OF   ENGINEERING. 


579  Addendum  on  the  Pelton  Water-wheel. — The  annexed  cut  and 
additional  details  of  the  test  alluded  to  on  p.  810  are  taken  from  circulars 
of  the  manufacturers,  of  San  Franciso,  Cal. 


The  water  was  measured  over  an  iron  weir  I"  thick  and  3.042  feet  long 
Without  end  contraction. 

The  depth  was  measured  by  a  Boyden  hook  gauge  reading  to  .001",  and 
was  .4146  foot.  The  quantity  of  water  discharged  was  found  to  be  2.819 
cubic  feet  per  second — Fteley's  formula.  The  head  lost  by  friction  in  pipe 
was  1.8  feet,  reducing  the  effective  head  to  384.7  feet. 

The  work  done  was  measured  by  a  Prony  brake  bearing  vertically  down 
upon  a  platform  scale  and  which  showed  a  weight  of  200  pounds  upon  the 
scale-beam  when  the  brake  gear  was  suspended  by  a  cord  from  a  point  im« 
mediately  above  the  wheel-shaft.  This  made  a  constant  minus  correction 
of  200  pounds.  The  friction  pulley  had  a  face  of  12",  and  being  kept  wet 
by  a  jet  of  clear  cold  water,  it  developed  very  little  heat  and  ran  without 
much  jumping.  Thirteen  tests  were  made  showing  very  uniform  results, 
the  first  four  of  which  were  as  follows : 


Tests. 

Weight  shown  by 
scale 

Net  weight 
(—200  Ibs.) 

Rev.  wheel-  shaft 
per  min. 

i 

665 
665 
660 
660 

0  =  465 
465 
460 
460 

u  =  254* 
255 
256 
256J 

Gu  s?  118342 
118575 
117760 
117990 
472667 
118167 

Totals    1022 
Means     255£ 

The  arm  of  the  Prony  brake  was  4.775  feet  from  centre  of  wheel-shaft  to 
point  of  contact  on  scale  and  hence  described  a  circle  with  a  circumference 
of  30  feet.  The  work  done  per  minute  was  therefore  Gu(2*tr)  =r=  (118167 
x  30)  or  3,545,000  foot-pounds,  equal  to  107.4  horse-power.  The  theoret- 
ical power  of  the  water  was  (2.819  x  60  x  384.7  x  62.4)  or  4S  0^0,2??  foot- 
pounds. The  useful  effect  was  therefore  87.3  per  cent. 


INDEX  TO  MECHANICS  OF  SOLIDS. 

(For  index  to  "  Hydraulics"  see  p.  xxii.) 


1>AGE. 

Aberration  of  Light 90 

Absolute  Velocity 89 

Abutment-Line 414 

Abutments  of  Arches 430,  435 

Acceleration,  Angular 107 

Acceleration,  Linear 49 

Acceleration,  Normal 75,  77 

Acceleration  of  Gravity  51, 160, 179 

Acceleration,  Tangential 74,  80 

Action  and  Reaction 1,  53 

Angular,  or  Rotary,  Motion. . .  107 

Anomalies  in  Friction 192 

Anti-Derivative,    see    Preface 

and  p.  253 

Anti-Resultant 402 

Anti-Stress-Resultant 411 

Apparent  Weight 78,  79 

Arches  Linear 386,  396 

Arches  of  Masomy 421,  437 

Arch-Ribs 438,483 

Arch-Ribs,  Classification  of . ...  458 
Arch-Rib  of  Three  Hinges .  458,460 
Arch-Rib  of  Hinged  Ends  440, 

458,  461 

Arch-Rib  of  Fixed  Ends  439,459,465 
Arch  Truss,  or  Braced  Arch.. .  478 

At  wood's  Machine 159 

Autographic  Testing  Machine.  240 
Beams,  Rectangular,  Compar- 
ative Strength  and  Stiffness 

272,273,277 

Belting,  Pressure  of 181 

Belting,  slip  of 182 

Bent  Lever  with  Friction 173 

Boat-Rowing 160 

Bow's  Notation 467 

Box-Girder 275,292 

Braced  Arch 438,478 

Brake,  Prony 158 

Brakes,  Railroad 190 

Bridge,     Arch 430 

Bridge-Pier 141 

Bridge,  Suspension 46 

Bridge  Truss,  Warren 35 

Buckling  of  Web-Plates 383 

Built  Beams,   designing    Sec- 
tions of 295 

Built  Columns 878 

Burr,  Prof. ,  Citations  from.224,  229 
Butt- Joint 226 


FAOK. 

Cantilevers 260,  276,  341 

Cantilever,  Oblique 353,  356 

Catenary,  Common 46 

Catenary  Inverted 387 

Catenary,  Transformed 395 

Cast  Iron 220,  279 

Cast  Iron,  Malleable 224 

Centre  of  Gravity,  18, 19,  etc.. .  .336 

Centre  of  Oscillation 119 

Centre  of  Percussion  of  Rod. ..  Ill 

Centrifugal  Action 125 

Centrifugal  Force 77,  78 

Centripetal  Force 77,  78,  79 

Centrobaric  Method 24 

Cheval-  Vapaur 136 

Chrome  Steel 224 

Circle  as  Elastic  Curve 262,  343 

Circular  Arc  as  Linear  Arch. .  391 

Closing  Line 414 

Coblenz,  Bridge  at 459,  478 

Columns,  Long 363 

Composition  of  Forces... 4,  8,  31,  38 
Compression  of  Short  Blocks. .  218 

Concurrent  Forces 6,  8,  397 

Cone  of  Friction 168 

Conical  Pendulum 78 

Connecting-Rod 59,  69,  70 

Conservation  of  Energy. .....  156 

Conservation  of  Momentum 66 

Continuous  Girders,  by  Analysis 

320-333 

Continuous  Girders,  by 

Graphics 484—514 

Copying-Press 71 

Cords,  Flexible 42 

Cords,  Rigidity  of 192 

Couples 27,  30 

Cover-Plates 226 

Crane 362 

Crank-Shaft,  Strength  of 314 

Creeping  of  Belts 186 

Crushing,  Modulus  of. ...  219,  424 

Curvilinear  Motion 72 

Dangerous  Section 262,  332 

Dash-Pot 158 

Deck-Beam 275 

Deflections,  (Flexure)  252—262,  342 
Derivatives,  (Elastic  curve)....  310 

Derived  Quantities 2 

Deviating  Force 77 


XVU1  INDEX  TO   MECHANICS    OF   SOLIDS. 

(For  index  to  "  Hydraulics"  see  p.  xxii.) 


Diagrams,   Strain 209,  241 

Displacement  of  Point  of  Arch 

Rib 447 

Dove-Tail    Joint 269 

Duchayla's  Proof  of  the  Parallelo- 
gram of  Forces 4 

Dynamics,  Definition 4 

Dynamics  of  a  Rigid  Body 105 

Dynamics,  of  Material  point. ...  49 

Dynamometers 157,  158,  159 

Eddy,  Prof.,  Graphic  Methods, 

(See  Preface) 426 

Elastic  Curve  a  Circle  . . .  262,  343 
Elastic  Curve  an  Equilibrium 

Polygon 484 

Elastic  Curves. .  245,  252—262,  358 
Elastic  Curves,  the  Four  x-De- 

rivatives  of 310 

Elasticity-Line 241 

Elasticity,  Modulus  of 203,  22V 

Elastic  Limit 202 

Elevation  of  Outer  Rail  on  Rail- 
road Curves 78 

Ellipse  of  Inertia 94 

Ellipsoid  of  Inertia 104 

Elliptical  Beam 340 

Elongation     of    Wrought-Iron 

Rod 207 

Energy 137 

Energy,  Conservation  of 156 

Energy,  Kinetic...  137, 144,  147,  150 

Energy,   Potential 155 

Equations,  Homogeneous 2 

Equator,  Apparent  Weight  at  the  78 

Equilibrium.. ...   4,  33,  39 

Equilibrium  Polygon 401,450 

ICquilibrium  Polygon  Through 

Three  Points 418,419 

Equivalent  Systems 7,  105,  145 

Exaggeration  of   Vertical  Di- 
mensions   in  Arch-Ribs, . . .  470 

Examples  in  Flexure 280—284 

Examples  in  Shearing. . .  .231,  232 
Examples  in  Tension  and 

Compression 222,  223 

Examples  in  Torsion 241—243 

Experiments    of    an    English 

Railroad  Commission 314 

Experiments  of  Hodgkinsbn 

207,369 

Experiments  of  Prof.  Lanza...  280 
Experiments  on  Building  Stone  424 

Experiments  on  Columns 378 

Extrados 421 

Euler's Formula  for  Columns...  364 

Factor  of  Safety 223 

Falling    Bodies 51 

"False  Polygons" 497,  501 

Fatigue  of  Metals 224 


"  Fixed  Points  " 503 

Flexural  Stiffness 250 

Flexure 244—886 

Flexure  and  Torsion  Combined  314 
Flexure,    Beams    of    Uniform 

Strength 335 

Flexure,  Common  Theory 244 

Flexure,  Eccentric  Load. .  .256,  301 
Flexure,  Elastic  Curves  in..  245, 

251,252—262 

Flexure,  Examples  in 280—284 

Flexure,  Hydrostatic  Pressure  308 

Flexure,    Moving  Loads 298 

Flexure,     Non-Prismatic 

Beams 332,    335 

Flexure  of  Long  Columns 363 

Flexure    of    PrismaticBeams 

Under  Oblique  Forces. .  .347—362 

Flexure,  Safe  Loads  in 262 — 284 

Flexure,  Safe  Stress  in 279 

Flexure,  Shearing  Stress  in 

284—295 

Flexure,  Special  Problems 

in 295—319 

Flexure,  Strength  in 249 

Flexure,  the  Elastic  Forces 246 

Flexure,   the  "Moment" 249 

Flexure,  the  "Shear" 248 

Flexure,  Uniform  Load . . .  258,  267 

302,  305,  307,  324,  329,  340 

Flow  of  Solids 212 

Fly-Wheel 121,151 

Fly-Wheel,  Stresses  in. ..  .126,  127 

Force 1 

Force   Diagram 400 

Force    Polygons S97 

Forces,  Concurrent 6,  8 

Forces,  Distributed 197 

Forces,  Non-Concurrent 6,  31 

Forces,  Parallel 13 

Forces,  Parallelogram  of 4 

Forces,  Varieties  of 7 

Free  Axes 129 

Free-Body   Method,  the. . .  .11,  105 

Friction 164—194,  422,  423 

Frictional  Gearing 172 

Friction,  Anomalies  in 192 

Friction  Axle 175 

Friction  Brake 158 

Friction,  Cone  of 168 

Friction  in  Machinery 191 

Friction,    Sliding 164—168 

Friction  of  Pivots 179 

Friction,  Rolling 186 

Friction- Wheels 1 77 

Friction,  Work  Spent  in 149' 

General  Properties  of  Materials 

204 

Governor  Ball.  .  78 


INDEX   TO   MECHANICS   OF   SOLIDS. 

(For  index  to  "  Hydraulics"  see  p.  xxii.) 


xix 


Graphic  Representations  of  Uni- 

formally  Accelerated  Motion. .  57 
Graphic  Treatment  of  Arch  ....  431 
Gravity,  Acceleration  of  51,  79, 160 

Gravity,  Centre  of 18,  336,453 

Gravity- Vertical 453 

Graphical   Statics,  Elemants 

of 397—420 

Graphical  Statics  of  Vertical 

Forces 412—420 

Guide  Curve 83 

Gyroscope 132 

Harmonic  Motion 58,  81,  117 

Heat  Energy 156 

Heaviness,  Table,  Etc 3 

Height  Due  to  Velocity 52,  84 

Hodgkinson's     Formulae     for 

Columns 369 

Hoir  ogeneous  Equations 2 

Hooke's  Law. 201,  203,  207 

Hooks,  Strength    of 362 

Horizontal  Straight  Girders  by 

Graphics 479—483 

Horse-Power 136,  239,242 

I-Beam 275,  292,295,337 

Ice-Boat,  Speed  of 90 

Impact 63 

Impact,  Loss  of  Energy  in. ...  141 

Inclined  Beam 359 

Inclined  Plane  83,  135, 151,  166, 169 

Indicator 159 

Inertia 53 

Inertia  of  Piston-Rod 59 

Instantaneous    Rotation,  Axis 

of 112 

Internal  Stress,  General  Prob- 
lem of 205 

Intrados 421 

Isochronal  Axes 120 

Isotropes 204 

Kinetic  Energy..,  137,144,  147,  150 

Knot,    Fixed 43 

Knot,  Slip 43 

Lanza,  Experiments  of  Prof....  280 

Lateral  Contraction 211,  229 

Lateral  Security  of  Girders  280,  298 

Lever 18,  71 

Lever,  Bent,  With  Friction  173, 174 
Linear  Arches...  386—396,  417,  425 

Live  Loads 298,  430 

Load-Line 413 

Locomotive  on  Arch 430 

Locomotive  on  Girder 298 

Locomotive.  Parallel-Rod  of...  131 

Malleable  Cast  Iron 224 

Mass 2,  53 

Material  Point 3 

Mechanical  Equivalent  of  Heai;  156 
liechanics.  Definition  of 1 


Mechanics,  Divisions  of ,  4 

Middle  Third 423 

Modulus  of  Elasticity 203,  227 

Modulus  of  Resilience 213 

Modulus  of  Rupture  (Flexure)  278 

Modulus  of  Tenacity 212 

Moduli  of  Compression 219 

Mohr's  Theorem 486 

Moment- Area 485 

Moment  of  a  Force , 14 

Moment-Diagram 263 

Moment  of  Flexure. . .  248,  348,  351 

Moment  of  Inertia 91 

Moment  of  Inertia  by  Graphics  454 
Moment    of    Inertia    of   Box- 
Girder 276 

Moment    of    Inertia    of    Built 

Beam 296 

Moment    of  Inertia    of   Built 

Column 379 

Moment  of    Inertia  of    Plane 

Figures. 91—98,  249,  274 

Moment    of     Inertia  of  Rigid 

Bodies     98,  103 

Moment  of  Inertia  of  Truss 478 

Moment  of  Torsion 236 

Momentum 66 

Mortar • 422 

Motion,  Curvilinear 72 

Motion,  Rectilinear 50 

Motion,  Rotary 107 

Moving  Loads  (Flexure 298 

Naperian  Base 183,  357,  387 

Naperian  Logarithms 47 

Navier's  Principle 422,  436 

Neutral  Axis 245,  247,  347,  355 

Neutral  Line  (see  Elastic  Curve.) 

Newton's  Laws 1,  53 

Non-Concurrent  Forces  in  a 

Plane 31,  399 

Non-Concurrent  Forces  in  Space  37 

Normal  Acceleration 75,  76,  77 

Normal  Stress 200 

Oblique  Section  of  Rod  in  Ten- 
sion   200 

Parabola  as  Linear  Arch 391 

Parabolic  Cord 45 

Parabolic  Working-Beam.  336,  344 

Parallel  Forces . .  13 

Parallel- Rod  of  Locomotive. .  131 

Parallelogram  of  Forces 4 

Parallelogram  of  Motions 7? 

Parallelogram  of  Velocities 72 

Pendulum,  Compound 118 

Pendulum,  Conical .' 78 

Pendulunii  Cycloidal, 80 

Pendulum,  Simple  Circular 81 

Phoenix  Columns 378 

Pier  Reactions 404 


INDEX  TO   MECHANICS    OF   SOLIDS. 

(For  index  to  "  Hydraulics"  see  p.  xxii.) 


Piers  of  Arches  430,  435 

Pile-Driving " 140 

Pillars  (see  columns) 

"  Pin-and-Square  "  Columns...  364 

Planet,  Velocity  of 82 

Polar  Moment  of  Inertia 97,  238 

Pole  (in  Graphics) 401 

Pole-Distance 416,  417 

Practical  Notes 223 

Potential  Energy 155 

Power 134 

Power  of  Motors. . . .  153,  157,  158 
Power,  Transmission  of,  by 

Belting 184 

Power,  Transmission  of  by 

Shafts  238,  318 

Principal  Axes 104,  129 

Projectile  in  Vacuo 83,  84—87 

Prony  Friction  Brake 158 

Pulley 43,103 

Punching  Rivet  Holes 229 

Quantity,  Kinds  of 1 

Quantities,  Derived 2 

Radius    of    Curvature,  75,  76, 

250,353 

Radius  of  Gyration  91,  92,  115, 

,....313,376 

Rankine's  Formula  for  Col- 
umns   372,375 

Rays  of  Force  Diagram 401 

Reaction     . ,   1, 18,  36,  53,  404 

Reduced  Load-Contour 429 

Regulation  of  Machines 153 

Relative  and  Absolute  Veloci- 
ties  89 

Resilience.... 204, 213,  237,  251,  313 
Resultant  of  Parallel  Forces,  13,  15 
Resultant  of  Two  or  More 

Forces 4,  6 

Reversal  of  Stress 514 

Rigid  Body 4 

Rigidity  of  Ropes 192 

Riveting  for  Built  Beams. ...  292 
Rivets  and  Riveted  Plates.  225,  292 
Robinson,  Prof. ,  Integration  by  357 

Rod  in  Tension 198,  200 

Rolling  Friction 186 

Rolling  Motion 130 

Roof  Truss 37,  405 

Rotary  Motion 68,  107,  129 

Rotation  and  Translation  Com- 
bined     , .130,  150 

Rupture 202 

Safe  Limit  of  Stress 202 

Safe  Loads  in  Flexure  ...  262,  284 

St.  Louis  Bridge 459,  467,  478 

Schiele's"  Anti-Friction"  Pivots  181 
Set,  Permanent...  202,  209,  208  241 
Shafts 233—239 


Shafts,  Non-Circular 239 

Shear  Diagram  (Flexure) 265 

Shearing 225—232 

Shearing  Distortion 227 

Shear,  Distribution  of  in  Flex- 
ure  287 

Shearing  Forces 7,  225 

Shearing  Stress, 

7,200,201,  225,  234,284 

Shear,  the  First  ^-Derivative 

of  Moment  (Flexure) 264 

Skidding 190 

Slip  (of  Oar,  etc.) 161 

Slope  (in  Flexure) 253 

Soffit 421 

Spandrel. 421 

Special  Equilibrium   Polygon 

409,424,440 

Specific  Gravity 3 

Statics,  Definition  of 4 

Statics,  Graphical 397—420 

Statics  of  a  Material  Point 8 

Statics  of  a  Rigid  Body 27 

Statics  of  Flexible  Cords 42 

Steam  Engine   Problems...  5$, 

61,  69,70,121,131,  151 

Steam  Hammer 138 

Stiffening  of  Web-Plates 383 

Stone,  Strength  of 221,  424 

Stress  Diagrams  for  Arch-Ribs.  471 
Stresses  Due  to   Rib  Shorten- 
ing  476 

Strain  Diagrams 209,  241 

Strains,  two  kinds  only.   196 

Stress 197,  198 

Stress    and    Strain,    Relation 

Between 201 

Stress-Couple 253,  348 

Stress,  Normal  and  Shearing...  200 

Strength  of  Materials ;  195 

Stretching  of  a  Prism  Under 

its  Own  Weight 215 

"  Sudden  "  Application  of  a 

Load 214,  255 

Summation    of    Products    by 

Graphics 451 

Suspension  Bridge 46 

Table  for  Flexure 279 

Table  for  Shearing 228 

Tables  for  Tension  and  Com- 
pression  221 

Tackle 43 

Temperature  Stresses  206,  217, 

..    . 222,473 

Tenacity.  Moduli  of 212 

Testing'  Machine,  Autographic  240 
Theorem  of  Three  Moments....  332 

Thrust  (in  Flexure) 348,  350 

Torsion..  ....233—243 


INDEX  TO  MECHANICS  0^  SOLIDS. 
(For  Index  to  "  Hydraulics"  see  p.  rrii.) 


XXI 


Torsion,  Angle  of 233 

Torsion  Balance 116 

Torsion,  Helix  Angle  in 233 

Torsion,  Moment  of 236 

Torsional  Resilience 237 

Torsional  Stiffness 236 

Torsional  Strength 235 

Tractrix,  The 181 

Transformed  Catenary 395 

Transmission  of  power  by 

Belting 184 

Transmission  of  Power  by  Shafting 

238,  318 

Translation,  Motion  of 68, 

105,106,  133,  137 

Trussed  Girders 381 

Uniformly  Accelerated  Motion 

Uniform  Motion.'.'.' .'.'.' .' 48,'  107',  129 
Uniform  Strength,  Beams  of..  335 
Uniform  Strength,  Solid  of,  in 
Tension 216 


Units,  Proper  Use see  §  6,  p.  2 

Velocity,  Absolute 89 

Velocity,  Angular 107 

Velocity,  Linear 49 

Velocity,  Relative.. . 89 

Virtual  Moment 67 

Virtual  Velocities 67 

Voussoir 386, 421 

Warren  Bridge  Truss 35 

Water,  Jets  of ...  87 

Web  of  I-Beam 274 

Wsb  of  I-Beam,  Buckling  of..  383 
Web  of  I-Beam,  Shear  in.. ....  290 

Wedge,  with  Friction 171 

Weight 2,3,7,  79 

Weight,  Apparent. 78,  79 

Wind  and  Sail-Boat 89,  90 

Work 133,  134,  etc. 

Work  and  Energy  in  Machines 

146,147,  etc. 

Working-Beam 336,  844 

Working  Strength 202 


INDEX  TO  HYDBAULIC& 

(For  index  to  "Mechanics  of  Solids"  see  p.  xvii.) 


Absolute  Temperature 606 

Absolute  Zero 606 

Accumulator,  Hydraulic 700 

Adiabatic  Change,  621,  629,  631,  636 
Adiabatic  Expansion  in  Com- 
pressed-air Engine 631 

Adiabatic  Flow  of  Gases  from 

an  Orifice 778 

Air  Collecting  in  Water-pipes, 

731,  736 
Air,  Compressed,  Transmission 

of 786-790 

Air- compressor 636 

Air -profile 749 

Air-pump,  Sprengel's 656 

Air-thermometer 604 

Amplitude  of  Backwater 772 

Anemometer,  Robinson's. .  824,  826 

Anemometer,  Biram's 827 

Anemometer,  Castello's 827 

Angle  of  Repose 572 

Angular  Stability  of  Ships. ...  597 

Atkinson  Gas-engine 642,  643 

Atmosphere  as  a  Unit  Pressure,  519 
Augment  of  Resistance  of  Screw 

Propeller 829 

Backwater 771 

Ball,  Spinning,  Deviation  from 

Vertical  Plane 823 

Balloon '. 644 

Barker's  Mill 672 

Barometers 530 

Barometric  Levelling 619 

Bazin,  Experiments 688 

Beaufoy's  Experiments 814 

Bellinger,    Prof.,   Experiments 

with  Elbows 729 

Bends,  Loss  of  Head  due  to. ...  728 

Bends  in  Open  Channels 770 

Bent  Tube,  Liquids  in 629 

Bernoulli's  Theorem  and   the 

Conservation  of  Energy 717 

Bernoulli's  Theorem  for  Gases,  773 
Bernoulli',*    Theorem,    General 

Form 706 

Bernoulli's     Theorem,     Steady 

Flow  without  Friction. .  .652,  654 


Bernoulli's  Theorem  with  Fric- 

tion 696 

Bidone,  Experiments  on  Jets. . .  803 

Blowing-engine,  Test 776 

Borda's  Formula 722 

Bourdon  Steam-gauge 532 

Boyle's  Law , . ; . .  615 

Bramah  Press 52t 

Branching  Pipes . .  731 

Bray  ton's  Petroleum-engine  . . .  641 

Buoyant  Effort 586 

Buoyant  Effort  of  the  Atmos- 
phere   644 

Canal  Lock,  Time  of  Filling,  739,  740 

Centigrade  Scale 605 

Centre  of  Buoyancy 586 

Centre  of  Pressure 546 

Change  of  State  of  Gas 610 

Chezy's  Formula 714 

Chezy's  Formula  for  Open  Chan- 
nels   758 

Church  and  Fteley,  Report  on 
Quaker  Bridge  Dam,  558,  563,  564 

Clearance 627 

Closed  Air-manometer 614 

Coal  Consumption 643 

Coal,  Heat  of  Combustion 643 

Coefficient  of  Contraction 659 

Coefficieats  of  Efflux,  661,  712, 

676,  734,  738,  784 
Coefficient  of  Fluid  Friction,  707,  797 

Coefficient  of  Resistance 704 

Coefficient  of  Roughness. .  .759,  760 
Coefficient  of  Velocity,  661,  689, 

704,  712,  723,  734 

Collapse  of  Tubes. 538 

Communicating  Prismatic  Ves- 
sels   739 

Complete  and  Perfect  Contrac- 
tion   676 

Component  of  Fluid  Pressure.    525 

Compressed-air  Engine 631 

Compressed  Air,  Transmission 

of 78^ 

Compressibility  of  Water 516 

Conical  Short  Tubes 692 

Conservation    of    En°rgy    and 
xxii 


INDEX  TO   HYDRA JLICS. 

(For  index  to  "  Mechanics  of  Solids"  see  p.  xvii.) 
PAGE  J 


xxni 


Bernoulli's  Theorem .  717 

Contraction 659 

Contraction,  Perfect 676 

Cooling  in  Sudden  Expansion 

of  Gas 622 

Critical  Temperature  of  a  Vapor,  608 

Croton  Aqueduct,  Slope 749 

Cunningham,   Experiments  on 

the  River  Ganges 759 

Cup-anemometer 834,  826 

Cups,  Impulse  of  Jet  on. ..  804,  808 
Current, Transporting  Power  of,  830 

Current-meters „ 750 

Curved  Dams 562 

Cylinders,       Thin        Hollow, 

Strength 5rS 

Dams,  High  Masonry 562 

Darcy,  Experiments  with  Pitot's 

Tube 804 

Darcy  and  Bazin,  Experiments 

with  Open  Channels 758 

Decrease    of    Tension    of  Gas 

along  a  Pipe 793 

Depth  of  Flotation 592 

Deviation  of  Spinning  Ball  from 

Vertical  Plane 823 

Diaphragm  in  a  Pipe,  Loss  of 

Head 724 

Displacement  of  a  Ship 596 

Divergent  Tubes,  Flow  through,  692 
Dividing  Surf  ace  of  TwoFluids,  528 

Diving-bell 617 

Double  Floats 750 

Draught  of  Ships 596 

Dubuat's  Experiments,  707,  815, 

822,  830 

Duchemin's  Formula  for  Wind- 
pressure 818 

Duty  of  Pumping-enginee 644 

Dynamics  of  Gaseous  Fluids,  773-797 

Earth  Pressure 572 

Earthwork  Dam 566 

Eddy-making  Resistance. .  814,  828 

Efficiency 637,  642 

Efflux  of  Gases 773-797 

Efflux  from  Steam-boiler 664 

Efflux  from  Vessel  in  Motion. .  670 

Efflux  into  Condenser 665 

Efflux  under  Water 669 

Egg-shaped  Section  for  Sewers,  765 

Elastic  Fluids 516 

Elbows,  Loss  of  Head  due  to,  727,  729 
Ellis,  Book  on  Fire-streams  . . .  716 
Emptying  Vessels,  Time  of. .  737-746 

Engine,  Gas- 641 

Engine,  Hot-air 639 

Engine,  Petroleum- 641 

Engine,  Steam- 624 

Enlargement,  Sudden,  in  Pipe,  721 


Equtu    Transmission    of  Fluid 

Pressure 524 

Equation  of  Continuity,  648,  737,  756 
Equation    of    Continuity     for 

Gaset 773 

Equation    of    Continuity     for 

Open  Channels 756 

Equilibrium  of  Flotation 590 

Ericsson's  Hot-air  Engine 640 

Ewart's  Experiments  on  Jets. . .  800 

Expanding  Steam 625 

Fahrenheit  Scale 605 

Fairbairn's  Experiments  on  Col- 
lapse of  Tubes 538 

Fanning,  Table  of  Coefficients 

of  Fluid  Friction 709 

Feet  and  Meters,  Table 677 

Fire-engine  Hose,  Friction  in. . .  716 

Fire-streams,  by  Ellis 716 

Floating  Staff 750 

Flood-gate 553 

Flotation 590 

Flow  in  Plane  Layers 648,  652 

Flow  in  Open  Channels 749 

Flow  of  Gas  in  Pipes 786,  790 

Fluid  Friction 695,  797,  828 

Fluid  Friction,    Coefficient  for 

Natural  Gas 797 

Fluid  Pressure,   Equal  Trans- 
mission of 524 

Force-pump , 667 

Francis'  Formula  for  Overfalls,  687 
Free  Surface  of  Liquid  at  Rest,  528 

Free  Surface  a  Paraboloid 544 

Fresh  Water,  Heaviness,  Table,  518 
Friction-head  in  Open  Channels,  757 

Friction-head  in  Pipes 699 

Friction,  Fluid 695,  797,  828 

Fronde's  Experiments  on  Fluid 

Friction 696 

Fronde's  Experiments  on  Grad- 
ual Enlargement  in  Pipes  . . .  725 
Froude's     Experiments     with 

Piezometers 720 

Fteley    and   Stearns's    Experi- 
ments on  Overfalls 687 

Fteley    and    Stearns's    Experi- 
ments with  Open  Channels. .  758 

Gas  and  Vapor 607 

Gas-engines 641 

Gaseous  Fluids 604-645 

Gas,  Flow  through  Shoi)  Pipes,  784 
Gas,  Flow  through  Orifi.  us.. 773-784 

Gas,  Illuminating 517,  533 

Gas,  Natural,  Flow  in  Pj  »es,  786,  790 

Gas,  Steady  Flow 773 

Gas,  Velocity  of  Approach 784 

Gases,  Definition 515 

Gauging-  of  Streams 755 


XXIV 


INDEX   TO    HYDRAULICS. 
(For  index  to  "  Mechanics  of  Solids"  see  p.  xvfl 


PAGE 

Gay-Lussac's  Law 609 

Gradual  Enlargement  in  Pipe..  725 

Granular  Materials 572 

Graphic  Representation  of 

Change  of  State  of  Gas 628 

Head  of  Water 530 

Heat-engines,  Efficiency 642 

Heaviness  of  Fluids 517 

Heaviness  of  Fresh  Water  at 

Different  Temperatures 518 

Height  Due  to  Velocity 649 

Height  of  the  Homogeneous 

Atmosphere. 820 

Herschel'sVenturi Water-meter,  726 

High  Masonry  Dams 562 

Hoop-tension 537 

Hose,  Rubber  and  Leather,  Fric- 
tion in 716 

Hot-air  Engines 639 

Humphreys  and  Abbot's  Survey 

of  the  Mississippi  River.. .754,  759 

Hurdy-gurdy,  California 809 

Button's  Formula  for  Wind- 
pressure  818 

Hydraulic  Grade-line 715 

Hydraulic  Mean  Depth,  698,757,  764 

Hydraulic  Press 526 

Hydraulic  Radius 698 

Hydraulic  Radius  for  Minimum 

Friction 764 

Hydraulics,  Definition 518 

Hydrodynamics. 646-832 

Hydrodynamics  or  Hydrokinet- 

ics,  Definition 518 

Hydromechanics,  Definition...  519 

Hydrometers 591 

Hydrornetric  Pendulum 753 

Hydrostatic  Pressure 522 

Hydrostatics,  Definition 518 

Ice-making  Machine  624 

Illuminating  Gas 517,  533 

Immersion  of  Rigid  Bodies. . . .  586 

Imperfect  Contraction 680,  684 

Impulse  and  Resistance  of 

Fluids 797-832 

Impulse  of  Jet  on  Vanes 801,  805 

Inclined  Short  Tubes,  Efflux 

through 691 

Incomplete  Contraction 679,  684 

Inelastic  Fluids 516 

r  Shape,  Emptying  of 

Vessels  of 746 

Isothermal  Change 615,  629,  639 

Isothermal  Expansion 624,  635 

Isothermal  Flow  of  Gas  in 

Pipes 790 

Isothermal  Flow  of  Gases 

through  Orifices 777 

Jacket  of  Hot  Water 635 


Jackson's  Wbrks  on  Hydraulics,  761 

Jet  from  Force-pump 667 

Jets,  Impulse  of 800,  803,  810 

Jets  of  Water 660,  662,  833 

Joule,  Experiment  on  Flow  of 

Gas 782 

Kansas  City   Water-works;    Si- 
phon  731,  736 

Kinetic  Energy 672,  718 

Kinetic  Energy  of  Jet 672,  808 

Kinetic  Theory  of  Gases 516, 

>506,  622 

Kutter's  Diagram ...  761 

Kutter's  Formula  759 

Kutter's  Hydraulic  Tables 761 

Laminated  Flow 648,  652 

Land-ties 585 

Law  of  Charles 609 

Levelling,  Barometric 619 

Liquefaction  of  Oxygen 609 

Liquid,  Definition 515 

Long    Pipes,   Flow    of    Water 

through 710-716 

Loss  of  Head 698,  703,  721,  etc. 

Loss  of  Head  Due  to  Bends 728 

Loss  of  Head  Due  to  Elbows, 

727,  729 

Loss  of  Head  Due  to  Throttle- 
valves 730 

Loss  of  Head  Due  to  Valve- 
gates 730 

Manometers 530 

Mariotte's  Law 615,  777 

Mechanics  of  the  Sail-boat 819 

Mendelejeff 's  Device  for  Spec- 
ula    544 

Metacentre  of  a  Ship 599 

Metres  and  Feet,  Table 677 

Mill,  Barker's 672 

Minimum  Frictional  Resistance 

in  Open  Channel 764,  766 

Mississippi     River,     Hydraulic 

Survey 754,  759,  770 

Mixture  of  Gases 618 

Momentum,  Principle  of 812 

Moving  Pistons 524 

Moving  Vane,  Impulse  of  Jet  on  805 

Napier  on  Flow  of  Steam 781 

Natural  Gas,  Flow  in  Pipes,  791,  797 

Natural  Slope,  of  Earth 573 

Non-planar  Pistons 526 

Notch,Rectangular,Efflux  from, 

683,  741 
Obelisk-shaped  Vessel;  Time  of 

Emptying 744 

Oblique  Impact  of  Jet  on  Plate,    810 

Open  Channels,  Flow  in 749-797 

Orifices  in  Thin  Plate 658,  773 

Otto  Gas-engine 641,  643 


INDEX  TO   HYDRAULICS. 
(For  index  to  "  Mechanics  of  Solids"  see  p.  xvii.) 


XXV 


Overfall,    Emptying    Reservoir 

through 741 

Overfall  Weirs. . .  .677,  683,  688,  756 
Overfall    Weirs,     Actual    Dis- 
charge   683 

Paddle-wheel  of  a  Steamer 817 

Paraboloid  as  Free  Surface. . . .  544 

Paraboloidal  Vessel 742 

Parallelopipedical       Reservoir 

Walls 555 

Pelton  Wheel  or  Hurdy-gurdy,  809 

Pendulum,  Hydrometric 753 

Perfect  Fluid,  Definition 515 

Permanent  Gases 605,  608 

Petroleum-engine . , 641 

Petroleum  Pumping 708 

Piezometer 649,  ,657,  700 

Pipes,  Clean 708 

Pipes,  Foul 708 

Pipes,       Thickness      of,      for 

Strength 538 

Pipes,  Tuberculated 708 

Piston  Pressures 523 

Pistons,  Moving .- . .  524 

Pistons,  Non-planar 526 

Pitot's  Tube 751,  804,  827 

Plate  between  Two  Levels  of 

Water 568,569 

Plates,  Impulse  of  Jets  on.. 801, 

805,  810 

Plates  Moving  in  a  Fluid 813 

Plates,  Resistance  in  Sea-water.  814 

Pneumatics,  Definition 518 

Poisson's  Law 621 

Poncelet's    Experiments    with  830 

Overfalls 677 

Power     Required    to      Propel 

Ships 830 

Pressure,  Centre  of 546 

Pressure  on  Bottom  of  a  Vessel,  545 
Pressure  on  Curved  Surfaces. . ,  569 

Pressure  on  Sluice-gate 551 

Pressure  per  Unit-area 519 

Pressure-energy 717 

Pressure-head 650 

Principle  of  Momentum 812 

Pyramidal  Vessel,  Emptying  of,  743 
Quaker  Bridge  Dam,  Proposed,  564 

Radian,  Definition 544 

Reaction  of  a  Water- jet 798 

Rectangular  Orifices 672,  676 

Refrigerator  of  Hot-air  Engine,  640 
Regenerator  of  Hot-air  Engine,  640 
Relative  Equilibrium  of  a 

Liquid 540 

Reservoir  of  Irregular  Shape, 

Emptying  of 746 

Reservoir  Walls 554-567 

Resistance,  Eddy-making.  .814,  828 


PARK 

Resistance,  Wave-making 828 

Resistance  of  Fluid  to  Moving 

Bodies : 820 

Resistance  of  Fluid  to  Moving 

Plates 814 

Resistance  of  Ships 828 

Resistance    of    Still  Water   to 

Moving  Solids 820 

Retaining  Walls 572-580 

Righting  Couple,   of  Floating 

Body 597 

Ritchie-Haskell  Direction  Cur- 
rent-meter   750 

Rivers,  Flow  in 749-797 

Robinson,    Prof.,   Experiments 

on  Flow  of  Natural  Gas 797 

Robinson's  Anemometer 824,  826 

Rounded  Orifice 663 

Safety-valves 534 

Sail-boat,  Mechanics  of  the 819 

Sail-boatiMoving  Faster  than  the 

Wind 820 

Saturated  Steam,  Heaviness  . . .  628 

Saturated  Vapor 607 

Scouring  Action  of  a  Current. .  830 
Screw  Propeller,  Augment  of ..  829 

Sewers,  Flow  in 761,  765 

Ships,  Resistance  of 828 

Ships,  Stability  of 597,  599 

Short  Cylindrical  Pipes,  Efflux 

through 689,  704 

Short  Pipes 722,  723 

Short  Pipe,  Minimum  Head  for 

Full  Discharge 724 

Simpson's  Rule 603,  747,  748 

Siphons 735 

Skin-friction 695,  828 

"Slip" 829 

Slope,  in  Open  Channels 749 

Smith,  Mr.  Hamilton;  Hydrau- 
lics   689 

Smithsonian     Scale  of    Wind- 
pressures 819 

Solid  of  Revolution,  Impulse  of 

Jet  on ,...803 

Specific  Gravity 589 

SprengeFs  Air-pump 656 

St.   Gothard    Tunnel,    Experi- 
ments in 787 

Stability  of  Rectangular  Wall. .  554 

Stability  of  Ships 597 

State  of  Permanency  of  Flow. . .  647 

Steady  Flow,  Definition 647 

Steady     Flow,     Experimental 

Phenomena 646 

Steady  Flow  of  a  Gas 773 

Steam,  Expanding 624 

Steam.  Flow  of 781 

Steam,  Saturated,  Heaviness  of  .  628 


XXVI 


INDEX   TO    HYDRAULICS. 
(For  index  to  "  Mechanics  of  Solids  "  see  p.  xvii.) 


PAGE 

Steam-engine,  Examples 627 

Steam-gauge,  Bourdon 532 

Stirling's  Hot-air  Engine 639 

Stream-line 658 

Sudbury  Conduit,  Experiments, 

758,  762 
Sudden   Diminution  of  Section 

in  a  Pipe 727 

Sudden  Enlargement  of  Section 

in  a  Pipe 721 

Surface  Floats 750 

Survey,    Hydraulic,    of    Missis- 
sippi River 754,  759 

Tachometer. 750 

Temperature,  Absolute 606 

Temperature,       Influence       on 

Flow  of  Water 703 

Tension  of  Gas 519 

Tension  of  Illuminating  Gas 533 

Thermodynamics 606 

Thermometers 604 

Thickness  of  Pipes 538 

Thickness  of  Pipe  for  Natural 

Gus 796 

Thin  Hollow  Cylinders 535 

Thin  Plate,  Orifices  in 658 

Throttle-valves,    Loss*  of   Head 

Due  to 730 

Time  of  Emptying  Vessels   of 

Various  Forms 737-746 

Transmission      of     Compressed 

Air 786,  790 

Transporting  Power  of  a  Cur- 
rent  830 

Trapezoidal   Section    for    Open 

Channel 765 

Trapezoidal  Wall,  Stability  of. .  559 

Triangular  Orifices 675 

Triangular  Wall,  Stability 561 

Uniform  Motion  in  Open  Chan- 
nel    756 

Uniform  Rotation  of  Liquid 542 

Uniform  Translation  of  Liquid..  540 

Upsetting  Couple 599 

Vacuum-chamber,  in  Siphon . . .  736 
Valve-gaies,  Loss  of  Head  due 

to 730 

Vanes,  Impulse  of  Jets  on... 801,  805 


PAGE 

Vapors 516,  607 

Variab]e  Diameter,  Long  Pipe  of,   794 
Variable  Motion  in  Open  Chan- 
nels   768 

Velocities  in  Section  of  River. . .   754 
Velocity  of  Efflux  as  related  to 

Density 668 

Velocity -head 649 

Velocity  Measurements  in  Open 

Channel 750 

Vena  Contracta 659 

Veuturi's  Tube  693 

Venturi  Tube,  New  Forms 694 

Venturi  Water-meter 725 

Volume  of  Reservoir  Found  by 
Observing  Time  of  Emptying,  748 

Water,  Compressibility  of 516 

Water-formula      for    Flow    of 

Gases 774 

Water,  Heaviness,  at  Different 

Temperatures 518 

Water  in  Motion 646 

Water-meter,  Venturi 725 

Water-ram 530,  538 

Wave-making  Resistance 828 

Webb,   Prof.,   Experiments   on 

the  Reaction  of  Jets 800 

Wedge  of  Maximum  Thrust.  . . .   573 
Wedge-shaped  Vessel,  Time  of 

Emptying 742 

Weirs,  Overfall. .  .677,  683,  688, 

756,  772 

Weisbach's  Experiments 682, 

685,  686,  691.  707,  721,  804 
Weisbach's    Experiments    with 

Elbows  and  Bends 727,  728 

Weser,  River,  Backwater  in. ...  772 

Wetted  Perimeter 697,  749 

Wex,  von,  Hydrodynamik 688 

Whirling  Machine 816,  826 

Wind-pressure 818 

Wind-pressure,         Smithsonian 

Scale  of 819 

Woltmann's  Mill 750 

Work    of    Compressed-air    En- 
gine   631 

Work  of  Expanding  Steam 624 


Work  of  Jet  on  a  Vane 


807 


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*  Davis's  Elements  of  Law Svo,  2  50 

*  Treatise  on  the  Military  Law  of  United  States Svo,  7  oo 

Sheep,  7  50 

De  Brack's  Cavalry  Outposts  Duties.     (Carr.) 24mo,  morocco,  2  oo 

Dietz's  Soldier's  First  Aid  Handbook i6mo,  morocco,  i  25 

*  Dredge's  Modern  French  Artillery 4to,  half  morocco,  15  oo 

Durand's  Resistance  and  Propulsion  of  Ships Svo,  5  oo 

>  Dyer's  Handbook  of  Light  Artillery i2mo,  3  oo 

Eissler's  Modern  High  Explosives Svo,  4  oo 

*  Fiebeger's  Text-book  on  Field  Fortification Small  Svo,  2  oo 

Hamilton's  The  Gunner's  Catechism i8mo,  i  oo 

*  Eoff's  Elementary  Naval  Tactics Svo,  i  50 

Ingalls's  Handbook  of  Problems  in  Direct  Fire Svo,  4  oo 

*  Ballistic  Tables Svo,  i  50 

*  Lyons's  Treatise  on  Electromagnetic  Phenomena.   Vols.  I.  and  II.  .Svo,  each,  6  oo 

*  Mahan's  Permanent  Fortifications.    (Mercur.) Svo,  half  morocco,  7  50 

Manual  for  Courts-martiaL i6mo,  morocco,  i  50 

*  Mercur's  Attack  of  Fortified  Places i2mo,  2  oo 

*  Elements  of  the  Art  of  War Svo,  4  oo 

Metcalf's  Cost  of  Manufactures — And  the  Administration  of  Workshops.  .Svo,  5  oo 

*  Ordnance  and  Gunnery.     2  vols i2mo,  5  oo 

Murray's  Infantry  Drill  Regulations iSmo,  paper,  10 

Nixon's  Adjutants'  Manual 241110,  i  oo 

Peabody's  Naval  Architecture Svo,  7  50 

2 


*  I'helps's  Practical  Marine  Surveying 8vo,  2  50 

Powell's  Army  Officer's  Examiner i2mo,  4  oo 

Sharpe's  Art  of  Subsisting  Armies  in  War i8mo.  morocco,  i  50 

*  Walke's  Lectures  on  Explosives 8vo,  4  oo 

*  Wheeler's  Siege  Operations  and  Military  Mining 8vo,  2  oo 

Winthrop's  Abridgment  of  Military  Law i2mo,  2  50 

WoodhulPs  Notes  on  Military  Hygiene i6mo,  i  50 

Young's  Simple  Elements  of  Navigation i6mo,  morocco,  i  oo 

Second  Edition,  Enlarged  and  Revised i6mo,  morocco,  2  oa 

ASSAYING. 
Fletcher's  Practical  Instructions  i.^  Quantitative  Assaying  with  the  Blowpipe. 

1 2 mo,  morocco,  i  50 

Furman's  Manual  of  Practical  Assaying •. 8vo,  3  oo 

Lodge's  Notes  on  Assaying  and  Metallurgical  Laboratory  Experiments.  .  .  .8vo,  3  oo 

Miller's  Manual  of  Assaying I2mo,  i  oo 

O'Driscoli's  Notes  on  the  Treatment  of  Gold  Ores 8vo,  2  oo 

Ricketts  and  Miller's  Notes  on  Assaying 8vo,  3  oo 

Ulke's  Modern  Electrolytic  Copper  Refining 8vo,  3  oo 

Wilson's  Cyanide  Processes i2mo,  i  50 

Chlorination  Process 12 mo,  i  50 

ASTRONOMY. 

Comstock's  Field  Astronomy  for  Engineers 8vo,  2  50 

Craig's  Azimuth 410,  3  50 

Doolittle's  Treatise  on  Practical  Astronomy 8vo,  4  oo 

Gore's  Elements  of  Geodesy Cvo,  2  50 

Hayford's  Text-book  of  Geodetic  Astronomy 8vo,  3  oo 

Merriman's  Elements  of  Precise  Surveying  and  Geodesy 8vo,  2  50 

*  Michie  and  Harlow's  Practical  Astronomy 8vo,  3  oo 

*  White's  Elements  of  Theoretical  and  Descriptive  Astronomy i2mo,  2  oo 

BOTANY. 

Davenport's  Statistical  Methods,  with  Special  Reference  to  Biological  Variation. 

i6mo,  morocco,     i  25 

Thome'  and  Bennett's  Structural  and  Physiological  Botany. i6mo,    2  25 

Westermaier's  Compendium  of  General  Botany.     (Schneider.) 8vo,    2  oo 

CHEMISTRY. 

Adriance's  Laboratory  Calculations  and  Specific  Gravity  Tables i2mo,  i  25 

Allen's  Tables  for  Iron  Analysis 8vo,  3  oo 

Arnold's  Compendium  of  Chemistry.     (Mandel.) Small  8vo,  3  50 

Austen's  Notes  for  Chemical  Students i2mo,  i  50 

Bernadou's  Smokeless  Powder. — Nitro-cellulose,  and  Theory  of  the  Cellulose 

Molecule i2mo,  2  50 

Bolton's  Quantitative  Analysis 8vo,  i  50 

*  Browning's  Introduction  to  the  Rarer  Elements. 8vo,  i  50 

Brush  and  Penfield's  Manual  of  Determinative  Mineralogy. 8vo,  4  oo 

Classen's  Quantitative  Chemical  Analysis  by  Electrolysis.    (Bolrwood.).  .8vo,  3  oo 

Cohn's  Indicators  and  Test-papers i2mo,  2  oo 

Tests  and  Reagents 8vo,  3  oo 

Crafts's  Short  Course  in  Qualitative  Chemical  Analysis.   (Schaeffer.). .  .i2mo,  i  50* 
Dolezalek's  Theory  of  the  Lead  Accumulator  (Storage  Battery).        (Von 

Ende.) i2mo,  2  50 

Drechsel's  Chemical  Reactions.     (Merrill.) I2mo,  i  25 

Duhem's  Thermodynamics  and  Chemistry.     (Burgess.) 8vo,  4  oo 

Eissler's  Modern  High  Explosives 8vo,  4  oo 

Effront's  Enzymes  and  their  Applications.     (Prescott.) 8vo,  3  oo 

Erdmann's  Introduction  to  Chemical  Preparations.     (Dunlap.) i2mo,  i  25 

3 


Fletcher's  Practical  Instructions  in  Quantitative  Assaying  with  the  Blcvpipe. 

i2mo,  morocco,  i  50 

Fowler's  Sewage  Works  Analyses i2mo,  2  oo 

Fresenius's  Manual  of  Qualitative  Chemical  Analysis.     (Wells.) 8vo,  5  oo 

Manual  of  Qualitative  Chemical  Analysis.  Part  I.  Descriptive.  (Wells.)  8vo,  3  oo 
System   of   Instruction   in    Quantitative    Chemical   Analysis.      (Cohn.) 

2  vols .8vo,  12  50 

Fuertes's  Water  and  Public  Health „ i2mo>  i  50 

Furman's  Manual  of  Practical  Assaying 8vo,  3  oo 

*  Getman's  Exercises  in  Physical  Chemistry i2mo,  2  oo 

Gill's  Gas  and  Fuel  Analysis  for  Engineers 12010,  i  25 

Grotenfelt's  Principles  of  Modern  Dairy  Practice.     (Woll.) i2mo,  2  oo 

Hammarsten's  Text-book  of  Physiological  Chemistry.     (Mandel.) 8vo,  4  oo 

Helm's  Principles  of  Mathematical  Chemistry.     (Morgan.) i2mo,  i  50 

Bering's  Ready  Reference  Tables  (Conversion  Factors) i6mo  morocco,  2  50 

Hind's  Inorganic  Chemistry 8vo,  3  oo 

*  Laboratory  Manual  for  Students i2mo,  i  oo 

Holleman's  Text-book  of  Inorganic  Chemistry.     (Cooper.) 8vo,  2  50 

Text-book  of  Organic  Chemistry.     (Walker  and  Mott.) 8vo,  2  50 

*  Laboratory  Manual  of  Organic  Chemistry.     (Walker.) i2mo,  i  oo 

Hopkins's  Oil-chemists'  Handbook 8vo,  3  oo 

Jackson's  Directions  for  Laboratory  Work  in  Physiological  Chemistry.  .8vo,  i  25 

Keep's  Cast  Iron '. 8vo,  2  50 

Ladd's  Manual  of  Quantitative  Chemical  Analysis i2mo,  i  oo 

Landauer's  Spectrum  Analysis.     (Tingle.) 8vo,  3  oo 

*  Langworthy  and  Austen;        The   Occurrence   of  Aluminium  in  Vege  able 

Products,  Animal  Products,  and  Natural  Waters 8vo,  2  oo 

Lassar-Cohn's  Practical  Urinary  Analysis.  (Lorenz.) i2mo,  i  oo 

Application  of  Some  General  Reactions  to  Investigations  in  Organic 

Chemistry.  (Tingle.) i2rco,  i  oo 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control 8vo,  7  50 

Lob's  Electrolysis  and  Electrosynthesis  of  Organic  Compounds.  (Lorenz, ).i2mo,  i  oo 

Lodge's  Notes  on  Assaying  and  Metallurgical  Laboratory  Experiments.  ..  .8vo,  3  oo 

Lunge's  Techno-chemical  Analysis.  (Cohn.) i2mo,  i  oo 

Mandel's  Handbook  for  Bio-chemical  Laboratory i2mo,  i  50 

*  Martin's  Laboratory  Guide  to  Qualitative  Analysis  with  the  Blowpipe .  .  i2mo,  60 
Mason's  Water-supply.     (Considered  Principally  from  a  Sanitary  Standpoint.) 

3d  Edition,  Rewritten 8vo,  4  oo 

Examination  of  Water.     (Chemical  and  Bacteriological.). i2rco,  i  25 

Matthew's  The  Textile  Fibres 8vo,  3  50 

Meyer's  Determination  of  Radicles  in  Carbon  Compounds.     (Tingle.).  .  i2mo,  i  oo 

Miller's  Manual  of  Assaying i2mo,  i  oo 

Mixter's  Elementary  Text-book  of  Chemistry i2mo,  i  50 

Morgan's  Outline  of  Theory  of  Solution  and  its  Results i2mo,  i  oo 

Elements  of  Physical  Chemistry i2mo,  2  oo 

Morse's  Calculations  used  in  Cane-sugar  Factories i6mo,  morocco,  i  50 

Mulliken's  General  Method  for  the  Identification  of  Pure  Organic  Compounds. 

Vol.  I Large  8vo,  5  oo 

O'Brine's  Laboratory  Guide  in  Chemical  Analysis 8vo,  2  oo 

O'DriscolTs  Notes  on  the  Treatment  of  Gold  Ores 8vo,  2  oo 

Ostwald'?  Conversations  on  Chemistry.     Part  One.     (Ramsey.) i2mo,  150 

Ostwald's  Conversations  on  Chemistry.     Part  Two.     (Turnbull ).     (In  Press.) 

*  Penfield's  Notes  on  Determinative  Mineralogy  and  Record  of  Mineral  Tests. 

8vo,  paper,  50 

Pictet's  The  Alkaloids  and  their  Chemical  Constitution.     (Biddle.) 8vo,  5  oo 

Pinner's  Introduction  to  Organic  Chemistry.     (Austen.) i2mo,  i  50 

Poole's  Calorific  Power  of  Fuels 8vo,  3  oo 

Prescott  and  Winslow's  Elements  of  Water  Bacteriology,  with  Special  Edcr- 

ence  to  Sanitary  Water  Analysis i2mo,  i  25 


*  Reisig's  Guide  to  Piece-dyeing 8vo,  25  oo 

Richards  and  Woodman's  Air,  Water,  and  Food  from  a  Sanitary  Standpoint  8vo,  2  oo 

Richards's  Cost  of  Living  as  Modified  by  Sanitary  Science i2mo,  i  oo 

Cost  of  Food,  a  Study  in  Dietaries i2mo,  i  oo 

*  Richards  and  Williams's  The  Dietary  Computer 8vo,  i  50 

Ricketts  and  Russell's  Skeleton  Notes  upon  Inorganic   Chemistry.     (Part  I. 

Non-metallic  Elements.) 8vo,  morocco,  75 

Ricketts  and  Miller's  Notes  on  Assaying 8vo,  3  oo 

Rideal's  Sewage  and  the  Bacterial  Purification  of  Sewage 8vo,  3  50 

Disinfection  and  the  Preservation  of  Food 8vo,  4  oo 

Rigg's  Elementary  Manual  for  the  Chemical  Laboratory 8vo,  i  25 

Rostoski's  Serum  Diagnosis.  (Bolduan.) i2mo,  i  oo 

Ruddiman's  Incompatibilities  in  Prescriptions 8vo,  2  oo 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish 8vo,  3  oo 

Salkowski's  Physiological  and  Pathological  Chemistry.  (Orndorff.) 8vo,  2  50 

Schimpf's  Text-book  of  Volumetric  Analysis i2mo,  2  50 

Essentials  of  Volumetric  Analysis i2mo,  i  25 

Spencer's  Handbook  for  Chemists  of  Beet-sugar  Houses i6mo,  morocco,  3  oo 

Handbook  for  Sugar  Manufacturers  and  their  Chemists.  .  i6mo,  morocco,  2  oo 

Stockbridge's  Rocks  and  Soils 8vo,  2  50 

*  Tillman's  Elementary  Lessons  in  Heat 8vo,  i  50 

*  Descriptive  General  Chemistry 8vo,  3  oo 

Treadwell's  Qualitative  Analysis.     (Hall.) 8vo,  3  oo 

Quantitative  Analysis.     (Hall.) 8vo,  4  oo 

Turneaure  and  Russell's  Public  Water-supplies 8vo,  5  oo 

Van  Deventer's  Physical  Chemistry  for  Beginners.     (Boltwood.) i2mo,  i  50 

*  Walke's  Lectures  on  Explosives 8Tro,  4  oo 

Washington's  Manual  of  the  Chemical  Analysis  of  Rocks 8"o,  2  oo 

Wassermann's  Immune  Sera :  Haemolysins,  Cytotoxins,  and  Precipitins.    (Bol- 
duan.)'  i2mo,  i  oo 

Well's  Laboratory  Guide  in  Qualitative  Chemical  Analysis 8vo,  i  50 

Short  Course  in  Inorganic  Qualitative  Chemical  Analysis  for  Engineerirg 

Students i2mo,  i   50 

Text-book  of  Chemical  Arithmetic I2mo,  i  25 

Whipple's  Microscopy  of  Drinking-water 8vo,  3  50 

Wilson's  Cyanide  Processes i2mo,  i  50 

Chlorination  Process , i2mo,  i  50 

Wulling's    Elementary    Course    in  Inorganic,  Pharmaceutical,  and  Medical 

Chemistry i2mo,  2  oo 

CIVIL  ENGINEERING. 

BRIDGES    AND    ROOFS.       HYDRAULICS.       MATERIALS    OF    ENGINEERING. 
RAILWAY  ENGINEERING. 

Baker's  Engineers'  Surveying  Instruments i2mo,  3  oo 

Bixby's  Graphical  Computing  Table Paper  19^X24!  inches.  25 

**  Burr's  Ancient  and  Modern  Engineering  and  the  Isthmian  Canal.     (Postage, 

27  cents  additional.) 8vo,  3  50 

Comstock's  Field  Astronomy  for  Engineers 8vo,  2  50 

Davis's  Elevation  and  Stadia  Tables 8vo,  i  op 

Elliott's  Engineering  for  Land  Drainage i2mo,  i  50 

Practical  Farm  Drainage i2mo,  i  oo 

*Fiebeger's  Treatise  on  Civil  Engineering 8vo,  5  oo 

Folwell's  Sewerage.     (Designing  and  Maintenance.) 8vo,  3  oo 

Freitag's  Architectural  Engineering.     2d  Edition,  Rewritten 8vo,  3  50 

French  and  Ives's  Stereotomy 8vo,  2  50 

Goodhue's  Municipal  Improvements I2mo,  i   75 

Goodrich's  Economic  Disposal  of  Towns'  Refuse 8vo,  3  50 

Gore's  Elements  of  Geodesy 8vo,  2  50 

Hayford's  Text-book  of  Geodetic  Astronomy 8vo,  3  oo 

Bering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,  2  50 

5 


Howe's  Retaining  Walls  for  Earth I2mo,  i  25 

Johnson's  (J.  B.)  Theory  and  Practice  of  Surveying Small  8vo,  4  oo 

Johnson's  (L.  J.)  Statics  by  Algebraic  and  Graphic  Methods 8vo,  2  oo 

Laplace's  Philosophical  Essay  on  Probabilities.    (Truscott  and  Emory.) .  i2mo,  2  oo 

Mahan's  Treatise  on  Civil  Engineering.     (1873.)     (Wood.) 8vo,  5  oo 

*  Descriptive  Geometry 8vo,  i  50 

Merriman's  Elements  of  Precise  Surveying  and  Geodesy 8vo,  2  50 

Elements  of  Sanitary  Engineering 8vo,  2  oo 

Merriman  and  Brooks's  Handbook  for  Surveyors i6mo,  morocco,  2  oo 

KTugent's  Plane  Surveying 8vo,  3  50 

Ogden's  Sewer  Design / • i2mo,  2  oo 

Patton's  Treatise  on  Civil  Engineering 8vo  half  leather,  7  50 

Reed's  Topographical  Drawing  and  Sketching 4to,  5  oo 

Rideal's  Sewage  and  the  Bacterial  Purification  of  Sewa{,3 8vo,  3  50 

Siebert  and  Biggin's  Modern  Stone-cutting  and  Masonry 8vo,  i  50 

Smith's  Manual  of  Topographical  Drawing.     (McMillan.) 8vo,  2  50 

Sondericker's  Graphic  Statics,  with  Applications  to  Trusses,  Beams,  and  Arches. 

8vo,  2  oo 

Taylor  and  Thompson's  Treatise  on  Concrete,  Plain  and  Reinforced 8vo,  5  oo 

*  Trautwine's  Civil  Engineer's  Pocket-book i6mo,  morocco,  5  oo 

Wait's  Engineering  and  Architectural  Jurisprudence 8vo,  6  oo 

Sheep,  6  50 

Law  of  Operations  Preliminary  to  Construction  in  Engineering  and  Archi- 
tecture  8vo,  5  oo 

Sheep,  5  50 

Law  of  Contracts .8vo,  3  oo 

Warren's  Stereotomy — Problems  in  Stone-cutting 8vo,  2  50 

Webb's  Problems  in  the  Use  and  Adjustment  of  Engineering  Instruments. 

i6mo,  morocco,  i   25 

*  Wheeler  s  Elementary  Course  of  Civil  Engineering 8vo,  4  oo 

Wilson's  Topographic  Surveying 8vo,  3  50 

BRIDGES  AND  ROOFS. 

Boiler's  Practical  Treatise  on  the  Construction  of  Iron  Highway  Bridges.  .8-vo,  2  oo 

*  Thames  River  Bridge 4*0,  paper,  5  oo 

Burr's  Course  on  the  Stresses  in  Bridges  and  Roof  Trusses,  Arched  Ribs,  and 

Suspension  Bridges 8vo,  3  50 

Burr  and  Falk's  Influence  Lines  for  Bridge  and  Roof  Computations.  .  .  .8vo,  3  oo 

Du  Bois's  Mechanics  of  Engineering.     Vol.  II Small  4to,  10  oo 

Foster's  Treatise  on  Wooden  Trestle  Bridges 4to,  5  oo 

Fowler's  Ordinary  Foundations 8vo,  3  50 

Greene's  Roof  Trusses 8vo,  i  25 

Bridge  Trusses 8vo,  2  50 

Arches  in  Wood,  Iron,  and  Stone 8vo,  2  50 

Howe's  Treatise  on  Arches 8vo,  4  oo 

Design  of  Simple  Roof-trusses  in  Wood  and  Steel 8vo,  2  oo 

Johnson,  Bryan,  and  Turneaure's  Theory  and  Practice  in  the  Designing  of 

Modern  Framed  Structures Small  4to,  10  oo 

Merriman  and  Jacoby's  Text-book  on  Roofs  and  Bridges: 

Part  I.     Stresses  in  Simple  Trusses 8vo,  2  50 

Part  II.     Graphic  Statics .8vo,  2  50 

Part  IH.     Bridge  Design 8vo,  2  50 

Part  IV.     Higher  Structures 8vo,  2  50 

Morison's  Memphis  Bridge 4to,  10  oo 

WaddelTs  De  Pontibus,  a  Pocket-book  for  Bridge  Engineers.  .  i6mo,  morocco,  3  oo 

Specifications  for  Steel  Bridges i2mo,  i  25 

Wood's  Treatise  on  the  Theory  of  the  Construction  of  Bridges  and  Roofs .  .  8vo,  2  oo 
Wright's  Designing  of  Draw-spans : 

Part  I.     Plate-girder  Draws 8vos  2  50 

Part  II.     Riveted-truss  and  Pin-connected  Long-span  Draws 8vo,  2  50 

Two  parts  in  one  volume ~ .  .8vo,  3  50 

6 


Briggs's  Elements  of  Plane  Analytic'  Geometry i2mo, 

Compton's  Manual  of  Logarithmic  Computations i2mo, 

Davis's  Introduction  to  the  Logic  of  Algebra 8vo, 

*  Dickson's  College  Algebra Large  i2mo, 

*  Introduction  to  the  Theory  of  Algebraic  Equations Large  i2mo, 

Emch's  Introduction  to  Protective  Geometry  and  its  Applications 8vo, 

Halsted's  Elements  of  Geometry 8vo, 

Elementary  Synthetic  Geometry '. „ 8vo, 


Wolff's  Windmill  as  a  Prime  Mover 8vo,    3  oo 

Wood's  Rustless  Coatings:   Corrosion  and  Electrolysis  of  Iron  and  Steel.  .8vo,    4  oo 

MATHEMATICS. 

Baker's  Elliptic  Functions 8vo,    I  50 

Elements  of  Differential  Calculus i2mo,    4  oo 

oo 
50 
So 
50 
25 
50 
75 
50 
Rational  Geometry i2mo,  75 

*  Johnson's  (J.  B.)  Three-place  Logarithmic  Tables:   Vest-pocket  size. paper,         15 

too  copies  for    5  oo 

*  Mounted  on  heavy  cardboard,  8X  TO  inches,        25 

10  copies  for  2  oo 

Johnson's  (W.  W.)  Elementary  Treatise  on  Differential  Calculus.  .Smah  8vo,  3  oo 

Johnson's  (W.  W.)  Elementary  Treatise  on  the  Integral  Calculus. Small  8vo,  i  50 

Johnson's  (W.  W.)  Curve  Tracing  in  Cartesian  Co-ordinates i2mo,  i  oo 

Johnson's  (W.  W.)  Treatise  on  Ordinary  and  Partial  Differential  Equations. 

Small  8vo,  3  50 

Johnson's  (W.  W.)  Theory  of  Errors  and  the  Method  of  Least  Squares.  i2mo,  i  50 

*  Johnson's  (W.  W.)  Theoretical  Mechanics 1200,  3  oo 

Laplace's  Philosophical  Essay  on  Probabilities.     (Truscott  and  Emory.) .  i2mo,  2  oo 

*  Ludlow  and  Bass.     Elements  of  Trigonometry  and  Logarithmic  and  Other 

Tables 8vo,  3  oo 

Trigonometry  and  Tables  published  separately Each,  2  oo 

*  Ludlow's  Logarithmic  and  Trigonometric  Tables 8vo,  i  oo 

Maurer's  Technical  Mechanics 8vo,  4  oo 

Merriman  and  Woodward's  Higher  Mathematics 8vo,  5  oo 

Merriman's  Method  of  Least  Squares 8vo,  2  oo 

Rice  and  Johnson's  Elementary  Treatise  on  the  Differential  Calculus. .  Sm.  8vo,  3  oo 

Differential  and  Integral  Calculus.     2  vols.  in  one Small  8vo,    2  50 

Wood's  Elements  of  Co-ordinate  Geometry 8vo,    2  oo 

Trigonometry:  Analytical,  Plane,  and  Spherical i2mo,     i  oo 


MECHANICAL  ENGINEERING. 

MATERIALS  OF  ENGINEERING,  STEAM-ENGINES  AND  BOILERS. 

Bacon's  Forge  Practice i2mo,  i  50 

Baldwin's  Steam  Heating  for  Buildings i2mo,  2  50 

Barr's  Kinematics  of  Machinery 8vo,  2  50 

*  Bartlett's  Mechanical  Drawing 8vo,  3  oo 

*  "  "  "        Abridged  Ed 8vo,     150 

Benjamin's  Wrinkles  and  Recipes i2mo,    2  oo 

Carpenter's  Experimental  Engineering 8vo,    6  oo 

Heating  and  Ventilating  Buildings 8vo,    4  oo 

Cary's  Smoke  Suppression  in  Plants  using  Bituminous  Coal.     (In  Prepara- 
tion.) 

Clerk's  Gas  and  Oil  Engine Small  8vo,    4  oo 

Coolidge's  Manual  of  Drawing 8vo,  paper,     i  oo 

Coolidge  and  Freeman's  Elements  of  General  Drafting  for  Mechanical  En- 
gineers  Oblong  4to,    2  50 

11 


Cromwell's  Treatise  on  Toothed  Gearing tamo,  i  50 

Treatise  on  Belts  and  Pulleys i2mo,  i  50 

Durley's  Kinematics  of  Machines 8vo,  4  oo 

Flather's  Dynamometers  and  the  Measurement  of  Power i2mo,  3  oo 

Rope  Driving i2mo,  2  oo 

Gill's  Gas  and  Fuel  Analysis  for  Engineers i2mo,  i  25 

Hall's  Car  Lubrication 12:010,  i  oo 

Bering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,  2  50 

Button's  The  Gas  Engine 8vo,  5  oo 

Jamison's  Mechanical  Drawing 8vo,  2  50 

Jones's  Machine  Design : 

Part  I.     Kinematics  of  Machinery 8vo,  i  50 

Part  n.     Form,  Strength,  and  Proportions  of  Parts 8vo,  3  oo 

Kent's  Mechanical  Engineers'  Pocket-book i6mo,  morocco,  5  oo 

Kerr's  Power  and  Power  Transmission 8vo,  2  oo 

Leonard's  Machine  Shop,  Tools,  and  Methods " 8vo,  4  oo 

*Lorenz's  Modern  Refrigerating  Machinery.     (Pope,  Haven,  and  Dean.)  .  .8vo,  4  oo 

MacCord's  Kinematics;   or,  Practical  Mechanism. 8vo,  5  oo 

Mechanical  Drawing 4to,  4  oo 

Velocity  Diagrams 8vo,  i  50 

Mahan's  Industrial  Drawing.     (Thompson.) 8vo,  3  50 

Poole  s  Calorific  Power  of  Fuels 8vo,  3  oo 

Reid's  Course  in  Mechanical  Drawing 8vo,  2  oo 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design.  8vo,  3  oo 

Richard's  Compressed  Air i2mo,  i  50 

Robinson's  Principles  of  Mechanism 8vo,  3  oo 

Schwamb  and  Merrill's  Elements  of  Mechanism 8vo,  3  oo 

Smith's  Press-working  of  Metals 8vo,  3  oo 

Thurston's   Treatise   on   Friction  and   Lost   Work   in  Machinery   and   Mill 

Work 8vo,  3  oo 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics .  12 mo,  i  oo 

Warren's  Elements  of  Machine  Construction  and  Drawing ,  . .  .  8vo,  7  50 

Weisbach's    Kinematics    and    the    Power    of    Transmission.     (Herrmann — 

Klein.) 8vo,  5  oo 

Machinery  of  Transmission  and  Governors.     (Herrmann — Klein.).  .8 vo,  500 

Wolff's  Windmill  as  a  Prime  Mover 8vo,  3  oo 

Wood's  Turbines 8vo,  2  50 


MATERIALS   OF    ENGINEERING. 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo,  7  30 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering.    6th  Edition. 

Reset 8vo,  7  50 

Church's  Mechanics  of  Engineering 8vo,  6  oo 

Johnson's  Materials  of  Construction 8vo,  6  oo 

Keep's  Cast  Iron 8vo,  2  50 

Lanza's  Applied  Mechanics 8vo,  7  50 

Martens 's  Handbook  on  Testing  Materials.     (Henning.) 8vo,  7  50 

Merriman's  Mechanics  of  Materials.  8vo,  5  oo 

Strength  of  Materials I2mo,  i  oo 

Metcalf's  Steel.     A  manual  for  Steel-users I2mo.  2  oo 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish 8vo,  3  oo 

Smith's  Materials  of  Machines I2mo,  i  oo 

Thurston's  Materials  of  Engineering 3  vols.,  8vo,  8  oo 

Part  II.     Iron  and  Steel 8vo,  3  50 

Part  III.     A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo,  2  50 

Text-book  of  the  Materials  of  Construction 8wo,  5  oo 

12 


Wood's  (De  V.)  Treatise  on  the  Resistance  of  Materials  and  an  Appendix  on 

the  Preservation  of  Timber 8vo,    2  oo 

Wood's  (De  V.)  Elements  of  Analytical  Mechanics 8vo,    3  oo 

Wood's  (M.  P.)  Rustless  Coatings:    Corrosion  and  Electrolysis  of  Iron  and 

Ste«L 8vo,    4  oo 


STEAM-ENGINES  AND  BOILERS. 


Berry's  Temperature-entropy  Diagram I2mo,  25 

Carnot's  Reflections  on  the  Motive  Power  of  Heat.     (Thurston.) i2mo,  50 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.  .  .  .i6mo,  mor.,  oo 

Ford's  Boiler  Making  for  Boiler  Makers i8mo,  oo 

Goss's  Locomotive  Sparks 8vo,  oo 

Hemenway's  Indicator  Practice  and  Steam-engine  Economy i2mo,  oo 

Button's  Mechanical  Engineering  of  Power  Plants 8vo,  5  oo 

Heat  and  Heat-engines , 8vo,  5  bo 

Kent's  Steam  boiler  Economy 8vo,  4  oo 

Kneass's  Practice  and  Theory  of  the  Injector 8vo,  i  50 

MacCord's  Slide-valves „ 8vo,  2  oo 

Meyer's  Modern  Locomotive  Construction 4to,  10  oo 

Peabody's  Manual  of  the  Steam-engine  Indicator lamo.  i  50 

Tables  of  the  Properties  of  Saturated  Steam  and  Other  Vapors 8yo,  i  oo 

Thermodynamics  of  the  Steam-engine  and  Other  Heat-engines 8vo,  5  oo 

Valve-gears  for  Steam-engines 8vo,  2  50 

Peabody  and  Miller's  Steam-boilers .8vo,  4  oo 

Pray's  Twenty  Years  with  the  Indicator Large  8vo,  2  50 

Pupin's  Thermodynamics  of  Reversible  Cycles  in  Gases  and  Saturated  Vapors. 

(Osterberg.). i2mo,  i  25 

Reagan's  Locomotives:  Simple   Compound,  and  Electric ! . i2mo,  2  50 

Rontgen's  Principles  of  Thermodynamics.     (Du  Bois.) 8vo,  5  oo 

Sinclair's  Locomotive  Engine  Running  and  Management I2mo,  2  oo 

Smart's  Handbook  of  Engineering  Laboratory  Practice i2mo,  2  50 

Snow's  Steam-boiler  Practice 8vo,  3  oo 

Spangler's  Valve-gears 8vo,  2  50 

Notes  on  Thermodynamics i2mo,  i  oo 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering 8vo,  3  oo 

Thurston's  Handy  Tables 8vo,  i  50 

Manual  of  the  Steam-engine 2  vols.,  8vo,  10  oo 

Part  I.     History,  Structure,  and  Theory 8vo,  6  oo 

Part  II.     Design,  Construction,  and  Operation 8vo,  6  oo 

Handbook  of  Engine  and  Boiler  Trials,  and  the  Use  of  the  Indicator  and 

the  Prony  Brake 8vo,  5  oo 

Stationary  Steam-engines 8vo,  2  50 

Steam-boiler  Explosions  in  Theory  and  in  Practice i2tno,  i  50 

Manual  of  Steam-boilers,  their  Designs,  Construction,  and  Operation 8vo,  5  oo 

Weisbach's  Heat,  Steam,  and  Steam-engines.     (Du  Bois.) 8vo,  5  oo 

Whitham's  Steam-engine  Design 8vo,  5  oo 

Wilson's  Treatise  on  Steam-boilers.     (Flather.) i6mo,  2  50 

Wood's  Thermodynamics,  Heat  Motors,  and  Refrigerating  Machines.  .  .8vo,  4  oo 


MECHANICS  AND  MACHINERY. 

Barr's  Kinematics  of  Machinery 8vo,  2  50 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo,  7  50 

Chase's  The  Art  of  Pattern-making *•  •  •  •  I2mo,  2  50 

Churchls  Mechanics  of  Engineering 8ro,  6  oo 

13 


Church's  Notes  and  Examples  in  Mechanics 8vo»  oo 

Compton's  First  Lessons  in  Metal-working i2mo,  50 

Comptcm  and  De  Groodt's  The  Speed  Lathe i2mo,  50 

Cromwell's  Treatise  on  Toothed  Gearing i2mo,  50 

Treatise  on  Belts  and  Pulleys i2mo,  50 

Dana's  Text-book  of  Elementary  Mechanics  for  Colleges  and  Schools. . i2mo,  50 

Dingey's  Machinery  Pattern  Making i2mo,  oo 

Dredge's  Record  of  the  Transportation  Exhibits  Building  of  the  World's 

Columbian  Exposition  of  1893 4to  half  morocco,  5  oo 

Du  Bois's  Elementary  Principles  of  Mechanics: 

Vol.     I.     Kinematics 8vo,  3  50 

Vol.    II.     Statics 8vo,  4  oo 

Vol.  in.     Kinetics 8vo,  3  50 

Mechanics  of  Engineering.     Vol.    I Small  4to,  7  So 

Vol.  II Small  4to,  10  oo 

Durley's  Kinematics  of  Machines 8vo,  4  oo 

Fitzgerald's  Boston  Machinist i6mo,  i  oo 

Flather's  Dynamometers,  and  the  Measurement  of  Power i2mo,  3  oo 

Rope  Driving i2mo,  2  oo 

Goss's  Locomotive  Sparks. 8vo,  2  oo 

Hall's  Car  Lubrication i2mo,  i  oo 

Holly's  Art  of  Saw  Filing i8mo,  75 

James's  Kinematics  of  a  Point  and  the  Rational  Mechanics  of  a  Particle.  Sm.8vo,2  oo 

*  Johnson's  (W.  W.)  Theoretical  Mechanics i2mo,  3  oo 

Johnson's  (L.  J.)  Statics  by  Graphic  and  Algebraic  Methods 8vo,  2  oo 

Jones's  Machine  Design : 

Part   I.     Kinematics  of  Machinery 8vo,  i  50 

Part  II.     Form,  Strength,  and  Proportions  of  Parts 8vc,  3  oo 

Kerr's  Power  and  Power  Transmission 8vo,  2  oo 

Lanza's  Applied  Mechanics 8vo,  7  50 

Leonard's  Machine  Shop,  Tools,  and  Methods 8vo,  4  oo 

*Lorenz's  Modern  Refrigerating  Machinery.      (Pope,  Haven,  and  Dean.). 8vo,  4  oo 

MacCord's  Kinematics;  or,  Practical  Mechanism 8vo,  5  oo 

Velocity  Diagrams 8vo,  i  50 

Maurer's  Technical  Mechanics 8vo,  4  oo 

Merriman's  Mechanics  of  Materials 8vo,  5  oo 

*  Elements  of  Mechanics i2mo,  i  oo 

*  Michie's  Elements  of  Analytical  Mechanics 8vo,  4  oo 

Reagan's  Locomotives:  Simple,  Compound,  and  Electric i2mo,  2  50 

Reid's  Course  in  Mechanical  Drawing 8vo,  2  oo 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design. 8vo,  3  oo 

Richards's  Compressed  Air i2mo,  i  50 

Robinson's  Principles  of  Mechanism 8vo,  3  oo 

Ryan,  Norris,  and  Hoxie's  Electrical  Machinery.     VoL  1 8vo,  2  50 

Schwamb  and  Merrill's  Elements  of  Mechanism 8vo,  3  oo 

Sinclair's  Locomotive-engine  Running  and  Management i2mo,  2  oo 

Smith's  (0.)  Press-working  of  Metals 8vo,  3  oo 

Smith's  (A.  W.)  Materials  of  Machines i2mo,  i  oo 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering 8vo,  3  oo 

Thurston's  Treatise  on  Friction  and  Lost  Y/ork  in    Machinery  and    Mill 

Work 8vo,  3  oo 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics. 

i2mo,  i  oo 

Warren's  Elements  of  Machine  Construction  and  Drawing 8vo,  7  so 

Weisbach's  Kinematics  and  Power  of  Transmission.   (Herrmann — Klein. ) .  8vo.  5  oo 

Machinery  of  Transmission  and  Governors.      (Herrmann — Klein. ).8vo,  5  oo 

Wood's  Elements  of  Analytical  Mechanics 8vo,  3  oo 

Principles  of  Elementary  Mechanics I2mo,  i  25 

Turbines 8vo,  2  50 

The  World's  Columbian  Exposition  of  1893 4to,  I  oo 

14 


METALLURGY. 

Egleston's  Metallurgy  of  Silver,  Gold,  and  Mercury: 

Vol.    L     Silver 8vo,  7  So 

Vol.  IL     Gold  and  Mercury 8vo,  7  SO 

**  Iles's  Lead-smelting.     (Postage  9  cents  additional.) 1210.0,  2  50 

Keep's  Cast  Iron 8vo,  2  50 

Kunhardt's  Practice  of  Ore  Dressing  in  Europe 8vo,  i  go 

Le  Chatelier's  High- temperature  Measurements.  (Boudouard — Burgess.  )i2mo,  3  oo 

Metcalf's  Steel.     A  Manual  for  Steel-users     iimo,  2  oo 

Smith's  Materials  of  Machines I2mo>  i  oo 

Thurston's  Materials  of  Engineering.     In  Three  Parts 8vo  8  oo 

Part    II.     Iron  and  Steel 8vo,  3  50 

Part  III.     A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo,  2  50 

Ulke's  Modern  Electrolytic  Copper  Refining 8vo,  3  oo 

MINERALOGY. 

Barringer's  Description  of  Minerals  of  Commercial  Value.    Oblong,  morocco,  2  50 

Boyd's  Resources  of  Southwest  Virginia 8vo,  3  oo 

Map  of  Southwest  Virignia Pocket-book  form.  2  oo 

Brush's  Manual  of  Determinative  Mineralogy.     (Penfield.) 8vo,  4  oo 

Chester's  Catalogue  of  Minerals 8vo,  paper,  i  oo 

Cloth,  i  25 

Dictionary  of  the  Names  of  Minerals 8vo,  3  50 

Dana's  System  of  Mineralogy Large  8vo,  half  leather,  12  50 

First  Appendix  to  Dana's  New  "  System  of  Mineralogy." Large  8vo,  i  oo 

Text-book  of  Mineralogy 8vo,  4  oo 

Minerals  and  How  to  Study  Them I2mo,  i  50 

Catalogue  of  American  Localities  of  Minerals Large  8vo,  i  oo 

Manual  of  Mineralogy  and  Petrography i2mo ,  2  oo 

Douglas's  Untechnical  Addresses  on  Technical  Subjects i2mo,  i  oo 

Eakle's  Mineral  Tables 8vo,  i  25 

Egleston's  Catalogue  of  Minerals  and  Synonyms 8vo,  2  50 

Hussak's  The  Determination  of  Rock-forming  Minerals.    (Smith*. ). Small  8vo,  2  oo 

Merrill's  Non-metallic  Minerals;   Their  Occurrence  and  Uses 8vo,  4  oo 

*  Penfield's  Notes  on  Determinative  Mineralogy  and  Record  of  Mineral  Tests. 

8vo   paper,  o  50 
Rosenbusch's   Microscopical   Physiography   of   the   Rock-making  Minerals. 

(Iddings.) 8vo,  5  oo 

*  Tillman's  Text-book  of  Important  Minerals  and  Rocks ,  .8vo.  2  oo 

Williams's  Manual  of  Lithology 8vo,  3  oo 

MINING. 

Beard's  Ventilation  of  Mines I2mo,  2  50 

Boyd's  Resources  of  Southwest  Virginia 8vo.  3  oo 

Map  of  Southwest  Virginia Pocket  book  form,  2  oo 

Douglas's  Untechnical  Addresses  on  Technical  Subjects I2mo.  i  oo 

*  Drinker's  Tunneling,  Explosive  Compounds,  and  Rock  Drills.  .4to.hf.mor  25  oo 

Eissler's  Modern  High  Explosives 8vo  4  co 

Fowler's  Sewage  Works  Analyses. 12010,  2  oo 

Goodyear's  Coal-mines  of  the  Western  Coast  of  the  United  States i2mo.  2  50 

Ihlseng's  Manual  of  Mining , Svcv.  5  oo 

**  Iles's  Lead-smelting.     (Postage  pc.  additional.) iamo.  2  50 

Kunhardt's  Practice  of  Ore  Dressing  in  Europe SYO,  i  50 

O'Driscoll's  Notes  on  the  Treatment  of  Gold  Ores .8vo,  2  oo 

*  Walke's  Lectures  on  Explosives Svo,  4  oo 

Wilson's  Cyanide  Processes , .  12010.,  i  50 

Chlorination  Process I2mo,  i  50 

15 


Wilson's  Hydraulic  and  Placer  Mining 12010,  2  oo 

Treatise  on  Practical  and  Theoretical  Mine  Ventilation I2mo,  i   25 

SANITARY  SCIENCE. 

Bashore's  Sanitation  of  a  Country  House lamo,  i  oo 

FolwelTs  Sewerage.     (Designing,  Construction,  and  Maintenance.) 8vo,  3  o» 

Water-supply  Engineering 8vo,  4  oo 

Fuertes's  Water  and  Public  Health i2mo,  i  50 

Water-filtration  Works I2mo,  2  50 

Gerhard's  Guide  to  Sanitary  House-inspection i6mo,  i  oo 

Goodrich's  Economic  Disposal  of  Town's  Refuse Demy  8vo,  3  50 

Hazen's  Filtration  of  Public  Water-supplies 8vo,  3  oo 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control 8vo,  7  50 

Mason's  Water-supply.  (Considered  principally  from  a  Sanitary  Standpoint)  8vo,  4  oo 

Examination  of  Water.     (Chemical  and  Bacteriological.) i2mo,  i  25 

Merriman's  Elements  of  Sanitary  Engineering 8vo,  2  oo 

Ogden's  Sewer  Design i2mo,  2  oo 

Prescott  and  Winslow's  Elements  of  Water  Bacteriology,  with  Special  Refer- 
ence to  Sanitary  Water  Analysis I2mo,  i  25 

*  Price's  Handbook  on  Sanitation i2mo,  i  50 

Richards's  Cost  of  Food.     A  Study  in  Dietaries I2mo,  i  oo 

Cost  of  Living  as  Modified  by  Sanitary  Science i2mo,  i  oo 

Richards  and  Woodman's  Air,  Water,  and  Food  from  a  Sanitary  Stand- 
point. .  .    8vo,  2  oo 

*  Richards  and  W   .iams's  The  Dietary  Computer 8vo,  i  50 

Rideal's  Sewage  and  Bacterial  Purification  of  Sewage 8vo,  3  50 

Turneaure  and  Russell's  Public  Water-supplies 8vo,  5  oo 

Von  Behring's  Suppression  of  Tuberculosis.     (Bolduan.) i2mo,  i  oo 

Whipple's  Microscopy  of  Drinking-water , 8vo,  3  50 

Woodhull's  Notes  on  Military  Hygiene i6mo,  i  50 

MISCELLANEOUS. 

De  Fursac's  Manual  of  Psychiatry.     (Rosanoff  and  Collins.).  ••  .Large  i2mo,  2  50 
Emmons's  Geological  Guide-book  of  the  Rocky  Mountain  Excursion  of  the 

International  Congress  of  Geologists Large  8vo,  i  50 

Ferrel's  Popular  Treatise  on  the  Winds 8vo.  4  oo 

Haines's  American  Railway  Management I2mo,  2  50 

Mott's  Composition,  Digestibility,  and  Nutritive  Value  of  Food.  Mounted  chart,  i  25 

Fallacy  of  the  Present  Theory  of  Sound i6mo,  i  oo 

Ricketts's  History  of  Rensselaer  Polytechnic  Institute,  1824-1894.  .Small  8vo,  3  oo 

Rostoski's  Serum  Diagnosis.     (Bolduan.) 12010,  i  oo 

Rotherham's  Emphasized  New  Testament Large  8vo,  2  oo 

Steel's  Treatise  on  the  Diseases  of  the  Dog 8vo,  3  50 

Totten's  Important  Question  in  Metrology 8vo,  2  50 

The  World's  Columbian  Exposition  of  1893 4*0,  i  oo 

Von  Behring's  Suppression  of  Tuberculosis.     (Bolduan.) i2mo,  i  oo 

Winslow's  Elements  of  Applied  Microscopy i2mo,  i  50 

Worcester  and  Atkinson.     Small  Hospitals,  Establishment  and  Maintenance; 

Suggestions  for  Hospital  Architecture :  Plans  for  Small  Hospital .  i2mo,  i  25 

HEBREW  AND  CHALDEE  TEXT-BOOKS. 

Green's  Elementary  Hebrew  Grammar i2mo,  i  25 

Hebrew  Chrestomathy 8vo,  a  oo 

Gesenius's  Hebrew  and  Chaldee  Lexicon  tc   the  Old  Testament  Scriptures. 

(Tregelles.) Small  4to,  half  morocco,  5  oo 

Lettepis's  Hebrew  Bible 8vo,  2  2§ 

16 


OCT  9     1947 

14 


WMf49 


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UNIVERSITY  OF  CALIFORNIA  LIBRARY 


